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Here we are providing Class 11 chemistry Important Extra Questions and Answers Chapter 6 Thermodynamics. Chemistry Class 11 Important Questions are the best resource for students which helps in Class 11 board exams.

## Class 11 Chemistry Chapter 6 Important Extra Questions Thermodynamics

### Thermodynamics Important Extra Questions Very Short Answer Type

Question 1.
Under what conditions the heat evolved or absorbed is equal to the internal energy change?
At constant volume.

Question 2.
What is the sign of AH for endothermic reactions and why?
AH is positive as ΔH = Hp – Hr and Hr < Hp.

Question 3.
What is the relationship between the standard enthalpy of formation and the enthalpy of a compound?
They are equal.

Question 4.
Why enthalpy of neutralization of HF is greater than 57.1 kJ mol-1?
This is due to the high hydration energy of fluoride ions.

Question 5.
What are the specific heat capacity and molar heat capacity for water?
Specific heat capacity for H2O = 4.18 JK-1 g-1
Molar heat capacity for H2O = 4.18 × 18 = 75.24 JK-1 mol-1.

Question 6.
Why enthalpy of neutralization is less if either the acid or the base or both are weak?
A part of the heat is used up for dissociation of the weak acid or weak base or both.

Question 7.
What do you mean by a system?
A specified part of the universe that is under thermodynamic observation is called a system.

Question 8.
Define a cyclic process.
A process in which a system undergoes a series of changes and ultimately returns to the original state is called a cyclic process. For a cyclic process; ΔU = 0.

Question 9.
Why the entropy of a diamond is less than that of graphite?
Diamond is more compact than graphite.

Question 10.
Under what conditions, ΔH of a process is equal to ΔU?
At constant temperature and constant volume.

Question 11.
Is the enthalpy of neutralization of HCl is same as that of H2S04? If so, why?
Yes. Because both are strong acids and ionized almost completely in aqueous solutions.

Question 12.
Which is a better fuel for the animal body: proteins or carbohydrates?
Carbohydrates. They have high calorific value.

Question 13.
Is the experimental determination of enthalpy of formation of CH4 possible?
No.

An online enthalpy calculator is specially designed to calculate exact amount of enthalpy generated in a thermodynamic system.

Question 14.
Can we calculate ΔH of every process from the bond energy data of reactants and products?
No.

Question 15.
Define enthalpy of fusion.
It is defined as the heat change when one mole of solid changes to its liquid form at its melting point. ‘

Question 16.
For the reaction NaCl (aq) + AgNO3 (aq) → AgCl(s) + NaNO3(aq), will ΔH be greater than, equal to or less than ΔE?
ΔH will be ΔE.

Question 17.
Write the mathematical relationship between heat, internal energy, and work done on the system.
ΔE = q + w.

Question 18.
What is the limitation of the I law of thermodynamics?
It cannot tell us about the direction of the process

Question 19.
What is the relationship between ΔH and ΔE?
ΔH = ΔE + PΔV = ΔE + ΔngRT.

Question 20.
State the I law of thermodynamics.
OR
State the Law of conservation of energy.
The energy of an isolated system remains conserved.
OR
The energy can neither be created nor destroyed, though it can be converted from one form into another.

Question 21.
Which of the following is a state function?
(i) height of a hill
(ii) distance traveled in climbing the hill
(iii) energy consumed in climbing the hill.
Energy consumed in climbing the hill.

Question 22.
What is the internal energy of one mole of a noble gas?
U = $$\frac{3}{2}$$RT.

Question 23.
Which of the following are state functions?
(i) q
(ii) heat capacity
(iii) specific heat capacity
(iv) ΔH and ΔU.
(iii) specific heat capacity and
(iv) ΔH and ΔU

Question 24.
Name the two most common modes by which a system and surroundings exchange their energy?
Heat and work.

Question 25.
What will happen to internal energy if work is done by the system?
The internal energy of the system will decrease.

Question 26.
How many times is the molar heat capacity greater than the specific heat capacity of water?
18 times.

Question 27.
Neither q nor w is a state function but q + w is a state function. Why?
q + w is equal to ΔU, which is a state function.

Question 28.
Name the state function Which remains constant during an isothermal change.
Temperature.

Question 29.
What is the value of enthalpy of neutralization of a strong acid and a strong base?
– 57.1 kJ/gm equivalent.

Question 30.
Out of 1 mole of H2O(g) and 1 mole of H2O(l) which one will have greater entropy?
H2O(g) will have greater entropy.

Question 31.
When a substance is said to be in its standard state?
When it is present at 298 K and under one atmospheric pressure.

Question 32.
What is the entropy of the formation of an element in its standard state?
By convention, the enthalpies (heats) of the formation of all elements in their most stable form (standard state) is taken as zero.

Question 33.
What is the physical significance of free energy change of a system?
The decrease in free energy (- ΔG) is a measure of useful work or network obtainable during the process at constant temp and pressure.

Question 34.
Why is it essential to mention the physical state of reactants and products in thermochemical reactions?
It is because the physical state of reactants and products also contributes significantly to the value ΔU or ΔH.

Question 35.
Name two intensive and extensive properties of a system.

• Intensive properties: Viscosity, refractive index.
• Extensive properties: Mass, volume, heat capacity, etc.

Question 36.
What is fuel value or calorific value?
It is defined as the amount of heat released when 1 gm of fuel or food is burnt completely in air or oxygen.

Question 37.
Is the process of diffusion of gases enthalpy driven or entropy-driven?
It is an entropy-driven process.

Question 38.
The value of ΔHsol of NaNO3 is positive, yet the dissolution is spontaneous, why?
There is a large increase in entropy i.e., it is an entropy-driven process.

Question 39.
Write a thermochemical equation of enthalpy of combustion of methanol.
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(1); ΔH = – Q kJ
where Q kJ mol-1 is the enthalpy of combustion.

Question 40.
What is the enthalpy of the formation of Cl2?
Enthalpy of formation of a homonuclear molecule like Cl2 is zero.

Question 41.
For a reaction also ΔH and ΔS are positive. What is the condition that this reaction occurs spontaneously?
To make ΔG negative TΔS > ΔH.

Question 42.
Arrange the following fuels in order of increasing fuel efficiency; kerosene, diesel oil, wax, natural gas.
Lower hydrocarbons have higher calorific value and thus are more efficient.
wax < diesel oil < kerosene < natural gas.

Question 43.
What is the sign of ΔS when N2 and H2 combine to form NH3?
ΔS is negative as the number of molecules decreases.

Question 44.
How will compare the efficiency of the three given fuels?
By comparing their calorific values. Larger the calorific value, the greater the fuel efficiency.

Question 45.
How is a non-spontaneous process made spontaneous?
By continuously supplying energy to it from outside.

Question 46.
What is the expression for entropy change for a phase transition? ‘
ΔS = $$\frac{q_{\mathrm{rev}}}{\mathrm{T}}$$

Question 47.
Give an example of an isolated system.
Thermos.

Question 48.
Is the enthalpy of formation of SnCl2(s) the same as that of ZnCl2(s)?
No. They are different.

Question 49.
Why we usually study enthalpy change and not internal energy change?
Most of the processes including reactions are carried out in open vessels at constant pressure.

Question 50.
What is the enthalpy of one mole of a noble gas?
H = U + PV = $$\frac{3}{2}$$ RT + RT = $$\frac{5}{2}$$ RT.

Question 51.
Which of the thermodynamic properties out of U, S, T, P, V, H, and G are intensive properties and why?
T and P, because they depend only upon the nature of the substance.

Question 52.
What is the relationship between qp and qv?
qp = qv + Δng RT, where Δn = np– nr (gaseous).

Question 53.
What is the Gibb’s Helmholtz equation?
ΔG = ΔH – TΔS, where ΔG, ΔH, and ΔS are free energy change, enthalpy change, and entropy change respectively.

Question 54.
Comment on the bond energies of four C-H bonds present in CH4?
The bond energies of 1st, 2nd, 3rd, and 4th C-H bonds are not equal and so average values are taken.

Question 55.
What is the main limitation of the first law of thermodynamics?
It cannot predict the spontaneity of a process.

Question 56.
What is entropy?
Entropy is a measure of the randomness/disorder of a system.

Question 57.
A reversible reaction has ΔG° negative for forwarding reaction. What will be the sign of ΔG° for the backward reaction?
Negative.

Question 58.
What is the effect of increasing temperature on the entropy of a substance?
It increases.

Question 59.
When is the entropy of a perfectly crystalline solid zero?
At absolute zero (O.K).

Question 60.
What is an adiabatic process?
In which no eat flow between the system and surroundings.

### Thermodynamics Important Extra Questions Short Answer Type

Question 1.
Ice is lighter than water, but the entropy of ice is less than that of water. Explain.
Water is the liquid form while ice is its solid form. Molecular motion in ice is restricted than in water, i.e., a disorder in ice is restricted than water, i.e., a disorder in ice is less than in water.

Question 2.
Define spontaneity or-feasibility of a process.
Spontaneity or feasibility of a process means its inherent tendency to occur on its own in a particular direction under a given set of conditions.

Question 3.
Enthalpy of neutralization of CH3COOH and NaOH is 55.9 kJ. What is the value of ΔH for ionization of CH3COOH?
The heat of neutralization of strong acid and strong base + ΔH of ionization of CH3COOH = Enthalpy of neutralization of CH3COOH and NaOH
∴ – 57.1 kJ + ΔH of ionisation of CH3COOH = – 55.9 kJ
∴ ΔH of ionisation of CH3COOH = (- 55.9 + 57.1) kJ
= 1.2 kJ.

Question 4.
When 1 gm of liquid naphthalene (C10H8) solidifies, 150 J of heat is evolved. What is the enthalpy of fusion of C10H8?
ΔHsolidifcation = – 150 × 128 = – 19200 J = – 19.2 kJ
[∵ M.wt.of C10H8 = 128]

Question 5.
Why most of the exothermic processes (reactions) are spontaneous?
ΔG = ΔH – TΔS; For exothermic reactions,
ΔH is -ve For a spontaneous process ΔG is to be -ve.

Thus decrease in enthalpy (- AH) contributes significantly to the driving force (To make ΔG negative).

Question 6.
What is meant by the term state function? Give examples.
A state function is a thermodynamic property that depends upon the state of the system and is independent of the path followed to bring about the change. Internal energy change (ΔU), enthalpy change (ΔH) entropy change (ΔS), and free energy change (ΔG) are examples.

Question 7.
The enthalpy of combustion of sulfur is 297 kJ.
Write the thermochemical equation for the combustion of sulfur. What is the value of ΔfH of SO2?
S(s) + O2(g) → SO2(g); ΔfH = – 297 kJ
∴ ΔH of SO2 = – 297 kJ mol-1

Question 8.
What would be the heat released when 0.35 mol of HC1 in solution is neutralized by 0.25 mol of NaOH solution?
HCl and NaOH being strong acid and strong- base is completely ionization in dilute aqueous solutions. The net reaction is H + (0.25 mol) + OH (0.25. mol) → H2O (0.25 mol)

Now the heat of neutralization of 1 mole of a strong acid is – 57.1 kJ
∴ The heat released will be 57.1 × 0.25 kJ = 14.27 kJ.

Question 9.
Predict which of the following entropy increases/ decreases?
(i) A liquid crystallizes into a solid.
After crystallization molecules attain an ordered state and therefore entropy decreases.

(ii) Temperature bf a crystalline solid is raised from 0 K to 115 K.
When the temperature is raised, a disorder in molecules increases, and therefore entropy increases.

(iii) 2NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)
The reactant is solid and hence has low entropy. Among the products there are two gases and be solid, therefore products represent a condition of higher entropy.

(iv) H2(g) → 2H(g).
Here 2 moles of H atoms have higher entropy than one mole of the hydrogen molecule.

Question 10.
Discuss the effect of temperature on the spontaneity of reactions.
Effect of temperature on the spontaneity of reactions:

The terms low temperature and high temperature are relative. For a particular reaction, the high temperature could even mean room temperature.

Question 11.
What is the most important condition for a process to be reversible in thermodynamics?
The process should be carried out infinitesimally slowly or the driving force should be infinitesimally greater than the opposing force.

Question 12.
Why heat is not a state function?
According to first law of Thermodynamics; ΔU = q + w or q – = ΔU – w. As ΔU is a state function, but w is not a state function, therefore q is also not a state function.

Question 13.
Why the absolute value of enthalpy cannot be determined?
As H = U + PV
The absolute value of U – the internal energy cannot be determined as it depends upon various factors whose value, cannot be determined.
∴ The absolute value of H cannot be determined.

Question 14.
Which of the following is/are exothermic and which are endothermic?
(i) Ca(g) → Ca2+(g) + 2e
Endothermic (Ionisation enthalpy is required)

(ii) O(g) + e → O(g) .
Exothermic (first electron affinity-energy is released)

(iii) N2-(g) + e- → N3-(g).
Endothermic (higher electron affinities are required).

Question 15.
Calculate Δr Gθ for the conversion of oxygen to ozone, $$\frac{3}{2}$$O2(g) → O3(g) at 298 K, if Kp for this conversion is 2.47 × 10-29
Δr Gθ = – 2.303RT log Kp
and R = 8.314 JK-1 mol-1
∴ Δr Gθ = – 2.303 × 8.314 × 298 log 2.47 × 10-29
= 163000 j mol-1 = 163 kJ mol-1.

Question 16.
Find the value of the equilibrium constant for the following conversion reaction at 298 K.

Δr Gθ = – 13.6 kJ mol-1.
r Gθ = 2.303 RT log K

Question 17.
At 60°C, N2O4 is 50% dissociated. Calculate Δr Gθ this temperature and at one atmospheric pressure.
N2O4(g) ⇌ 2NO2(g)
If N2O4 is 50% dissociated, the mole fraction of both the substances are given by

Question 18.
Calculate the heat released when 0.5 moles of the nitric acid solution is mixed with 0.2 moles of potassium hydroxide solution.
HNO3 is a strong acid and KOH is a strong base. Therefore, both are completely ionized in water. The net reaction is

∴ Heat evolved = – 57.1 × 0.2 = – 11.42 kJ.

Question 19.
Define Hess’s Law of Constant Heat Summation.
The enthalpy change for a reaction remains the same whether it proceeds in one step or in series of steps all’ measurements being done under similar conditions of temperature.

This law is a corollary from I Law of Thermodynamics.

Question 20.
What are the applications of Hess’s Law of constant heat summation?

1. It helps to calculate the enthalpies of formation of those compounds which Cannot be determined experimentally.
2. It helps to determine the enthalpy of allotropic transformations like C(graphite) → C (diamond).
3. It helps to calculate the enthalpy of hydration.

Question 21.
Calculate the maximum work obtained when 0.75 mol of an ideal gas expands isothermally and reversible at 27°C from a volume of 15 L to 25 L.
For an isothermal reversible expansion of an ideal gas
w = – nRT log $$\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$$ = – 2.303 nRT log $$\frac{\mathrm{V}_{2}}{\mathrm{~V}_{1}}$$

Putting n = 0.75 mol; V1 = 15 L; V2 = 25 L, T = 27 + 273 = 300 K R = 8.314 JK-1 mol-1.
w = – 2.303 × 0.75 × 8.314 × 300 log $$\frac{25}{15}$$
= – 955.5 J.

Question 22.
What are heat capacities at constant volume and constant pressure? What is the relationship between them?
Heat capacity at constant volume (Cv): Heat supplied to a system to raise its temperature through 1°C keeping the volume of the system constant is called its heat capacity at constant volume (Cv).

Heat capacity at constant pressure (Cp): Heat supplied to a system to raise its temperature through 1°C keeping the external pressure constant is called its heat capacity at constant pressure (Cp).

Relationship between Cp and Cv: Cp – Cv = R.

Question 23.
Explain what do you mean by a reversible process.
A process or a change is said to be reversible if it can be reversed at any moment by an infinitesimal change. It proceeds infinitesimally slowly by a series of equilibrium states such that the system and surroundings are always in near equilibrium with each other.

Question 24.
Two liters of an ideal gas at a pressure of 10 atm expands isothermally into a vacuum until its. total volume is 10 liters. How much heat is absorbed and how much work is done in the expansion?
We have q = – w: pext(10 – 2) = 0 × 8 = 0
No. work is done; No heat is absorbed.

Question 25.
Define (i) Molar enthalpy of fusion
The enthalpy change that accompanies the melting of one mole of a solid substance at its melting point is called the articular enthalpy of fusion.

(ii) Molar enthalpy of vaporisation.
The amount of heat required to vaporize one mole of a liquid at constant temperature and under standard pressure (1 bar) is called molar enthalpy of vaporization.

Question 26.
How will you arrive at the relationship qp = qv + ΔngRT?
Enthalpy change ΔH = qp; where qp = heat change at constant pressure,
Internal energy change ΔU = qv; where qv = heat change at. constant volume.

Now ΔH = ΔU + PΔV
For ideal gases PV = nRT
∴ ΔH = ΔU + (PV2 – PV1)
= ΔU + P(V2 – V1) = ΔU + (n2RT – n1RT)
= ΔU + RT(n2 – n1) = ΔU + ΔngRT
or
qp = qv + ΔngRT

Question 27.
Define
(i) Specific heat capacity
Specific heat capacity: It is the amount of heat required to raise the temperature of 1 gram of the substance through 1°C.

(ii) molar heat capacity.
Molar heat capacity: It is the amount of heat required to raise the temperature of one mole of the substance through 1 °C.

Question 28.
Derive the relationship Cp – Cv = R.

Question 28.
Derive the expression -ΔG = w non-expansion
From I law of thermodynamics

Question 29.
How does the sign of G help in predicting the spontaneity/non-spontaneity of a process?

1. IF ΔG is negative, the process is spontaneous.
2. If ΔG = O, the process does not occur and the system is in equilibrium.
3. If ΔG is positive, the process does not proceed in the forward direction.

Question 30.
An exothermic reaction A B is spontaneous in the backward direction. What will be the sign of ΔS for the forward direction?
The backward reaction will be endothermic. Thus, the energy factor opposes the backward reaction., As the backward reaction is spontaneous, the randomness factor must favor i.e., ΔS will be positive for the backward reaction or it will be negative for the forward direction.

### Thermodynamics Important Extra Questions Long Answer Type

Question 1.
Define
(i) Standard enthalpy of formation.
Standard enthalpy of formation: The heat change accompanying the formation of 1 mole df a substance from its elements in their most stable state of aggregation is called its standard enthalpy of formation.
H2(g) + $$\frac{1}{2}$$O2(g) H2O(l); Δf He = 285.8 kJ mol-1

(ii) Standard enthalpy of combustion
Standard enthalpy of combustion: It is the heat change accompanying the complete combustion or burning of one mole of a substance in its standard state in excess of air or oxygen.
C4H10(g) + $$\frac{13}{2}$$O2(g) → 4CO2(g) + 5H2O(1); ΔHθ = – 2658.0 kJ mol-1.

(iii) Enthalpy of atomization
Enthalpy of atomization: It is defined as the enthalpy change accompanying the breaking of one mole of a substance completely into its atoms in the gas phase.
H2(g) → 2H(g) ΔcHe = 435.0 kJ mol-1

(iv) Enthalpy of solution
Enthalpy of solution: It is defined as the heat change when one mole of a substance dissolves in a specified amount of the solvent. The enthalpy of solution at infinite dilution is the enthalpy change observed on dissolving 2 moles of the substance in an infinite amount of the solvent.

(v) Lattice enthalpy
Lattice Enthalpy: The lattice enthalpy of an ionic compound is the enthalpy change that occurs when one mole of an ionic compound dissociates into its ions in a gaseous state.

(vi) Thermochemical equation.
Thermochemical Equation: A balanced chemical equation together with the value of its A^H is called a thermochemical equation.

The above equation describes the combustion of liquid ethanol. The negative sign indicates that tills are an exothermic reaction. We specify the physical state along with the allotropic state of the substance in a thermochemical equation.

### Thermodynamics Important Extra Questions Numerical Problems

Question 1.
For the equilibrium PCl5(g) ⇌ PCl3(g) + Cl2(g) at 298 K, Kc = 1.8 × 10-7. What is ΔG° for the reaction? (R = 8.314 JK-1 mol-1).

Question 2.
Calculate the equilibrium constant, K, for the following reaction at 400 K?
2NOCl(g) ⇌ 2NO(g) + Cl2(g)
Given that ΔrH° = 80.0 kJ mol-1 and ΔrS° = 120 JK-1 mol-1.

Question 3.
Calculate the standard entropy change for the reaction X ⇌ Y if the value of ΔH° = 28.40 kJ and equilibrium constant is 1.8 × 10-7 at 298 K and ΔrG° = 38.484 kJ.

= -33.8 JK-1 mol-1

Question 4.
Calculate the enthalpy of formation of methane, given that the enthalpies of combustion of methane, graphite, and hydrogen are 890.2 kJ, 393.4 kJ, and 285.7 kJ mol-1 respectively.

Multiply equation (iii) by 2, add it to equation (ii) and subtract equation (i) from their sum
C + 2H2 → CH4 .
ΔH = – 393.4 + 2(-285.7) – (-890.2)
= – 74.6 kJ mol-1
Hence the heat of formation of methane (CH4) is
ΔfH = – 74.6 kJ mol-1.

Question 5.
CO is allowed to expand isothermally and reversibly from 10 m3 to 20 m3 at 300 K and work obtained is 4.75 KJ. Calculate the number of moles of carbon monoxide (CO). R = 8.314 JK-1 mol-1.

Question 6.
Two moles of an ideal gas initially at 27°C and one atm pressure are compressed isothermally and reversible till the final pressure of the gas is 10 atm. Calculate q, w, and AU for the process.
Here n = 2; p1 = 1 atm; P2 = 10 atm; T = 300 K
w = 2.303 nRT log $$\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}$$
= 2.303 × 2 × 8.314 JK mol-1 × 300 K × log $$\frac{10}{1}$$
= 11488 J .

For isothermal compression of ideal gas
ΔU = 0
Further, ΔU = q + w
∴ q = -w = 11488 J.

Question 7.
The heat of combustion of benzene in a bomb calorimeter (i.e. at constant volume) was found to be 3263.9 kJ mol-1 at 25°C. Calculate the heat of combustion of benzene at constant pressure.
The given reaction is
C6H6(l) + $$\frac{15}{2}$$ O2(g) → 6CO2(g) + 3H2O(l)
C6H6 is a liquid and H20 is a liquid at 25°C.

Question 8.
When 0.532 g of benzene (C6H6) with boiling point 353 K is burnt with an excess of O2 in a calorimeter, 22.3 kJ of heat is given out. Calculate ΔH for the combustion process (R = 8.31 JK-1 mol-1)

Question 9.
Specific heat of an elementary gas is found to be 0.313 Jat constant volume. If the molar mass of the gas is 40 g mol-1, what is the atomicity of the gas? R = 8.31 JK-1 mol-1.

Question 10.
Calculate the energy needed to raise the temperature of 10.0 g of iron from 25° C to 500°C if the specific heat of iron is 0.45 J (0°C)-1g-1.
Energy needed (q) = m × C × ΔT
= 10.0 × 0.45 × (500 – 25) J
= 2137.5 J

Question 11.
Calculate the enthalpy of formation of carbon disulfide given that the enthalpy of combustion of it is 110.2 kJ mol-1 and those of sulfur and carbon are 297.4 kJ and 394.5 kJ/g atoms respectively.
Our aim is C(s) + 2S(s) → CS2(l); ΔH =?
Given (i) CS2(l) + 3O2(g) → CO2(g) + 2SO2(g); ΔH = – 110.2 kJ mol-1
(ii) S(s) + O2(g) → SO2(g); ΔH = – 297.4 kj mol-1
(iii) C(s) + O2(g) → CO2(g); ΔH= – 394.5 kj mol-1

Add (iii) + 2(ii) and subtract (i), it gives, on rearranging
C(s) + 2S(s) → CS2(l);
ΔH = (- 394.5) +- 2(- 297.4) – (- 110.2)
= -879.1 kj mol-1

Thus the enthalpy of formation of CS2 = – 879.1 kJ mol-1

Question 12.
There are two crystalline forms of PbO; one is yellow and the other is red. The standard enthalpies of formation of these two forms are – 217.3 and – 219.0 kJ mol-1 respectively. Calculate the enthalpy change for the solid-solid phase transition:
PbO (yellow) → PbO(red)
Our aim is PbO (yellow) → PbO (rpd); ΔH =?
Given (i) PbO(s) + $$\frac{1}{2}$$02(g) ) → PbO (yellow); ΔH = – 217.3 kJ mol-1
(ii) Pb(s) + $$\frac{1}{2}$$O2(g) ) → PbO(red); ΔH = —219.0 kJ mol-1

Subtracting (1) from (ii), we get
PbO (yellow) → PbO(red); ΔH =- 219.0 – (- 2173) = – 1.7 kJ mol-1.

Question 13.
The thermite reaction used for welding of metals involves the reaction 2Al (s) + Fe2O3(s) → Al2O3(s) + 2Fe(s)
What is Δr, H° at 25°C for this reaction? Given that the standard heats of formation of Al2O3 and Fe2O3 are – 1675.7 k J and – 828.4 kJ mol-1 respectively.
Our aim is 2Al(s) + Fe2O3(s) → Al2O3(s) + 2Fe(s); ΔrH° =?

Question 14.
Calculate the bond enthalpy of HCI. Given that the bond enthalpies of H2 and Cl are 430 and 242 kJ mol-1 respectively and for HCI is -91 kJ mol-1.

Question 15.
CaIculat the enthalpy of hydrogenation of C2H2(g) to C2H4(g). Given bond energies: C—H = 414.50 kJ mol-1; C≡C is 827.6 kJ mol-1, C=C is 606.0 kJ mol-1; H—H = 430.5 kJ mol-1.

= [827.6 + 2 × 414.0 + 430.5] – [606.0 + 4 × 414.0]
= 175.9 kJ mol-1.

Question 16.
The entropy change for the vaporization of water is 109 JK-1 mol-1. Calculate the enthalpy change for the vaporization of water at 373 K.
Δvap H = TΔvap S
= 373 × 109 = 40657 Jmol-1
= 40.657 kJ mol-1

Question 17.
Enthalpy and entropy changes of a reaction are 40.63 kJ mol-1 and 108.8 JK-1 mol-1 respectively. Predict the feasibility of the reaction at 27°C.
ΔH = 40.63 kJ mol-1 = 40630 Jmol-1
ΔS = 108.8 JK-1 mol-1
T = 27°C =27 + 273 = 300 K
Now ΔG = ΔH – TΔS
= 40630 – 3 × 108.8 = 7990 J mol-1

Since AG comes out to be positive (i.e., ΔG > 0), the reaction is not feasible at 27°C in the forward direction.

Question 18.
At 0°C ice and water are in equilibrium and ×H = 6.0 kJ mol-1 for the process H2O(s) → H2O(l). What will be ΔS and
ΔG for the conversion of ice to liquid water.?
Since the given process is in equilibrium
ΔG = 0
∴ from the equation ΔG = ΔH – TΔS
ΔH becomes = TΔS

∴ ΔS = $$\frac{6000 \mathrm{~J} \mathrm{~mol}^{-1}}{273 \mathrm{~K}}$$ = 21.98 JK-1 mol-1.

Question 19.
Calculate the standard- free energy for the reaction
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)
Given that the standard free energies of formation (ΔfG°) for NH3(g), NO(g), H2O(l) are – 16.8, + 86.7, and – 237.2 kJ mol-1 respectively. Predict the feasibility of the above reaction at the standard state.
Given ΔfG°(NH3) = – 16.8 kJ mol-1, ΔfG°(NO) = + 86.7 kj mol-1; ΔfG°(H2O) = – 237.2 kJ mol-1

= – [4 × ΔfG°(NH3) + 5 × ΔfG°(O2)]
= [4 × 86.7 + 6(- 237.2)] – [4 × (- 16.8) + 5 × 0]
= – 1009.2 kJ
Since ΔrG° is negative, the reaction is feasible.

Question 20.
The value of K for the water gas reaction
CO + H2O ⇌ CO2 + H2 is 1.06 × 105 at 25°C. Calculate the standard free energy change for the reaction at 25°C.
ΔrG° = – 2.303 RT log K
= – 2.303 × 8.314 × 298 × log (1.06 × 105)
On solving, ΔrG° = – 28.38 KJ mol-1.

Question 21.
Calculate ΔrG° for the conversion of oxygen to ozone. $$\frac{3}{2}$$O2(g) → O3(g) at 298 K. If Kp for the reaction is 2.47 × 10-29
ΔrG° = – 2.303 RT log K
R = 8.314 JK-1 mol-1
∴ ΔfG° = – 2.303 (8-314 JK-1 mol-1) × 298 × (log 2.47 × 10-29)

On solving
ΔrG° = 163000 J mol-1 = 163 kJ mol-1

Question 22.
Find out the value of the equilibrium constant for the following reaction at 298 K. Standard Gibbs energy change, ΔrG° at the given temperature is – 13.6 kJ mol-1.
rG° = 2.303 RT log K

Question 23.
For oxidation of iron, 4Fe(s) + 3O2(g) → 2Fe2O3 (s) entropy change is – 549.4 JK-1 mol-1 at 298 K. Inspite .of negative entropy change of this reaction, why is the reaction spontaneous? (ΔrH° = -1648 × 103 J mol-1).
The spontaneity of a reaction is determined from ΔStotal
(which is = Δsys + Δsutu)

Now ΔSsurr = – $$\frac{\Delta_{r} \mathrm{H}^{0}}{\mathrm{~T}}$$ at constant pressure
= $$\frac{-1648 \times 10^{3}}{298 \mathrm{~K}}$$ J mol-1 = 5530 JK-1 mol-1

Thus total entropy change for the reaction is
Δr Stotal = 5530 + (- 549.4) = 4980.6 JK-1 mol-1.
Since the total entropy change is positive, therefore, the above reaction is spontaneous.

Question 24.
A swimmer coming from a pool is covered with a film of water weighing about 18 g. How much heat must be supplied to evaporate this water at 298 K? Calculate the internal energy of vaporization at 100°C. AvapH° for water.

No. of moles in 18 g H2O(l) = $$\frac{18 \mathrm{~g}}{18 \mathrm{~g} \mathrm{~mol}^{-1}}$$ = 1 mol.
Δvap U = Δvap H° – pΔV = ΔvapH – Δng RT (assuming steam behaving as an ideal gas)
Δvap H° – Δng RT = 40.66 kJ mol-1 – (1)(8.314 JK-1 mol-1)(373 K) (10-3 kJ) ,
∴ Δvap U° = 40.66 kJ mol-1 – 3.10 kJ mol-1 = 37.56 kJ mol-1

Question 25.
1 g or graphite is burnt in a bomb calorimeter in excess of oxygen at 298 K and 1 atm pressure according to the reaction
C(graphite) + O2(g) → CO2(g). During the reaction, the temperature rises from 298 to 299 K. If the heat, the capacity of the bomb calorimeter is 20.7 kJ K-1 what is the enthalpy change for the above reaction at 298 K and 1. atm?
= – $$\frac{12.0 \times(-20.7)}{1}$$ = – 2.48 × 102 kJ mol-1


text:
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A modified May–Holling–Tanner predator-prey model with multiple Allee effects on the prey and an alternative food source for the predator
February  2021, 26(2): 963-985. doi: 10.3934/dcdsb.2020149

## Asymmetric diffusion in a two-patch mutualism system characterizing exchange of resource for resource

 School of Mathematics, Sun Yat-sen University, Guangzhou 510275, P.R. China

Received  May 2019 Revised  February 2020 Published  May 2020

Fund Project: The author is supported by NSF of P.R. China (11571382)

This paper considers a two-patch mutualism system derived from exchange of resource for resource, where the obligate mutualist can diffuse asymmetrically between patches. First, we give a complete analysis on dynamics of the system without diffusion, which exhibit how resource production of the obligate mutualist leads to its survival/extinction. Using monotone dynamics theory, we show global stability of a positive equilibrium in the three-dimensional system with diffusion. A novel finding of this work is that the obligate species' final abundance is explicitly expressed as a function of the diffusion rate and asymmetry, which demonstrates precise mechanisms by which the diffusion and asymmetry lead to the abundance higher than if non-diffusing, even though the facultative species declines. It is shown that for a fixed diffusion rate, intermediate asymmetry is favorable while extremely large asymmetry is unfavorable; For a fixed asymmetry, small diffusion is favorable while extremely large asymmetry is unfavorable. Initial densities of the species are also shown to be important in species' persistence and abundance. Numerical simulations confirm and extend our results.

Citation: Yuanshi Wang. Asymmetric diffusion in a two-patch mutualism system characterizing exchange of resource for resource. Discrete & Continuous Dynamical Systems - B, 2021, 26 (2) : 963-985. doi: 10.3934/dcdsb.2020149
##### References:
 [1] R. Arditi, C. Lobry and T. Sari, Is dispersal always beneficial to carrying capacity? New insights from the multi-patch logistic equation, Theor. Popul. Biol., 106 (2015), 45-59.  doi: 10.1016/j.tpb.2015.10.001.  Google Scholar [2] R. Arditi, C. Lobry and T. Sari, Asymmetric dispersal in the multi-patch logistic equation, Theor. Popul. Biol., 120 (2018), 11-15.  doi: 10.1016/j.tpb.2017.12.006.  Google Scholar [3] J. $\hat{A}$str$\ddot{o}$m and T. P$\ddot{a}$rt, Negative and matrix-dependent effects of dispersal corri- dors in an experimental metacommunity, Ecology, 94 (2013), 1939-1970.   Google Scholar [4] C. J. Briggs and M. F. Hoopes, Stabilizing effects in spatial parasitoid-host and predator-prey models: A review, Theor. Popul. Biol., 65 (2004), 299-315.  doi: 10.1016/j.tpb.2003.11.001.  Google Scholar [5] G. J. Butler, H. I. Freedman and P. Waltman, Uniformly persistent systems, Proc. Amer. Math. Sco., 96 (1986), 425-430.  doi: 10.1090/S0002-9939-1986-0822433-4.  Google Scholar [6] L. Fahrig, Effect of habitat fragmentation on the extinction threshold: A synthesis, Ecol. Appl., 12 (2002), 346-353.   Google Scholar [7] H. I. Freedman and D. Waltman, Mathematical models of population interactions with dispersal. I. Stability of two habitats with and without a predator, SIAM J Appl Math., 32 (1977), 631-648.  doi: 10.1137/0132052.  Google Scholar [8] J. Hofbauer and K. Sigmund, Evolutionary Games and Population Dynamics, Cambridge University Press, Cambridge, UK, 1998.  doi: 10.1017/CBO9781139173179.  Google Scholar [9] R. D. Holt, Population dynamics in two-patch environments: Some anomalous consequences of an optimal habitat distribution, Theor. Popul. Biol., 28 (1985), 181-208.  doi: 10.1016/0040-5809(85)90027-9.  Google Scholar [10] J. N. Holland and D. L. DeAngelis, A consumer-resource approach to the density-dependent population dynamics of mutualism, Ecology, 91 (2010), 1286-1295.   Google Scholar [11] V. Hutson, Y. Lou and K. Mischaikow, Convergence in competition models with small diffusion coefficients, J. Diff. Equa., 211 (2005), 135-161.  doi: 10.1016/j.jde.2004.06.003.  Google Scholar [12] V. A. A. Jansen, The dynamics of two diffusively coupled predator-prey populations, Theor. Popul. Biol., 59 (2001), 119-131.  doi: 10.1006/tpbi.2000.1506.  Google Scholar [13] J. Jiang, Three- and four-dimensional cooperative systems with every equilibrium stable, J. Math. Anal. Appl., 188 (1994), 92-100.  doi: 10.1006/jmaa.1994.1413.  Google Scholar [14] Y. Lou, On the effects of migration and spatial heterogeneity on single and multiple species, J. Diff. Equa., 223 (2006), 400-426.  doi: 10.1016/j.jde.2005.05.010.  Google Scholar [15] T. A. Revilla, Numerical responses in resource-based mutualisms: A time scale approach, J. Theor. Biol., 378 (2015), 39-46.  doi: 10.1016/j.jtbi.2015.04.012.  Google Scholar [16] S. Rinaldi and M. Scheffer, Geometric analysis of ecological models with slow and fast processes, Ecosystems, 3 (2000), 507-521.  doi: 10.1007/s100210000045.  Google Scholar [17] A. Ruiz-Herrera and P. J. Torres, Effects of diffusion on total biomass in simple metacommunities, J. Theor. Biol., 447 (2018), 12-24.  doi: 10.1016/j.jtbi.2018.03.018.  Google Scholar [18] H. L. Smith, Monotone Dynamical Systems: An Introduction to the Theory of Competitive and Cooperative Systems, Amer. Math. Soci. Press, New York, USA, 1995.   Google Scholar [19] H. L. Smith and P. Waltman, The Theory of the Chemostat, New York: Cambridge University Press, 1995.  doi: 10.1017/CBO9780511530043.  Google Scholar [20] G. Takimoto and K. Suzuki, Global stability of obligate mutualism in community modules with facultative mutualists, OIKOS, 125 (2015), 535-540.  doi: 10.1111/oik.02741.  Google Scholar [21] J. J. Tewksbury et al., Corridors affect plants, animals, and their interactions in fragmented landscapes, Proc. Natl. Acad. Sci. U.S.A., 99 (2002), 12923-12926. Google Scholar [22] Y. Wang and D. L. DeAngelis, Comparison of effects of diffusion in heterogeneous and homogeneous with the same total carrying capacity on total realized population size, Theor. Popul. Biol., 125 (2019), 30-37.   Google Scholar [23] Y. Wang, H. Wu and D. L. DeAngelis, Global dynamics of a mutualism-competition model with one resource and multiple consumers, J. Math. Biol., 78 (2019), 683-710.  doi: 10.1007/s00285-018-1288-9.  Google Scholar [24] B. Zhang, K. Alex, M. L. Keenan, Z. Lu, L. R. Arrix, W.-M. Ni, D. L. DeAngelis and J. D. Dyken, Carrying capacity in a heterogeneous environment with habitat connectivity, Ecology Letters, 20 (2017), 1118-1128.  doi: 10.1111/ele.12807.  Google Scholar

show all references

##### References:
 [1] R. Arditi, C. Lobry and T. Sari, Is dispersal always beneficial to carrying capacity? New insights from the multi-patch logistic equation, Theor. Popul. Biol., 106 (2015), 45-59.  doi: 10.1016/j.tpb.2015.10.001.  Google Scholar [2] R. Arditi, C. Lobry and T. Sari, Asymmetric dispersal in the multi-patch logistic equation, Theor. Popul. Biol., 120 (2018), 11-15.  doi: 10.1016/j.tpb.2017.12.006.  Google Scholar [3] J. $\hat{A}$str$\ddot{o}$m and T. P$\ddot{a}$rt, Negative and matrix-dependent effects of dispersal corri- dors in an experimental metacommunity, Ecology, 94 (2013), 1939-1970.   Google Scholar [4] C. J. Briggs and M. F. Hoopes, Stabilizing effects in spatial parasitoid-host and predator-prey models: A review, Theor. Popul. Biol., 65 (2004), 299-315.  doi: 10.1016/j.tpb.2003.11.001.  Google Scholar [5] G. J. Butler, H. I. Freedman and P. Waltman, Uniformly persistent systems, Proc. Amer. Math. Sco., 96 (1986), 425-430.  doi: 10.1090/S0002-9939-1986-0822433-4.  Google Scholar [6] L. Fahrig, Effect of habitat fragmentation on the extinction threshold: A synthesis, Ecol. Appl., 12 (2002), 346-353.   Google Scholar [7] H. I. Freedman and D. Waltman, Mathematical models of population interactions with dispersal. I. Stability of two habitats with and without a predator, SIAM J Appl Math., 32 (1977), 631-648.  doi: 10.1137/0132052.  Google Scholar [8] J. Hofbauer and K. Sigmund, Evolutionary Games and Population Dynamics, Cambridge University Press, Cambridge, UK, 1998.  doi: 10.1017/CBO9781139173179.  Google Scholar [9] R. D. Holt, Population dynamics in two-patch environments: Some anomalous consequences of an optimal habitat distribution, Theor. Popul. Biol., 28 (1985), 181-208.  doi: 10.1016/0040-5809(85)90027-9.  Google Scholar [10] J. N. Holland and D. L. DeAngelis, A consumer-resource approach to the density-dependent population dynamics of mutualism, Ecology, 91 (2010), 1286-1295.   Google Scholar [11] V. Hutson, Y. Lou and K. Mischaikow, Convergence in competition models with small diffusion coefficients, J. Diff. Equa., 211 (2005), 135-161.  doi: 10.1016/j.jde.2004.06.003.  Google Scholar [12] V. A. A. Jansen, The dynamics of two diffusively coupled predator-prey populations, Theor. Popul. Biol., 59 (2001), 119-131.  doi: 10.1006/tpbi.2000.1506.  Google Scholar [13] J. Jiang, Three- and four-dimensional cooperative systems with every equilibrium stable, J. Math. Anal. Appl., 188 (1994), 92-100.  doi: 10.1006/jmaa.1994.1413.  Google Scholar [14] Y. Lou, On the effects of migration and spatial heterogeneity on single and multiple species, J. Diff. Equa., 223 (2006), 400-426.  doi: 10.1016/j.jde.2005.05.010.  Google Scholar [15] T. A. Revilla, Numerical responses in resource-based mutualisms: A time scale approach, J. Theor. Biol., 378 (2015), 39-46.  doi: 10.1016/j.jtbi.2015.04.012.  Google Scholar [16] S. Rinaldi and M. Scheffer, Geometric analysis of ecological models with slow and fast processes, Ecosystems, 3 (2000), 507-521.  doi: 10.1007/s100210000045.  Google Scholar [17] A. Ruiz-Herrera and P. J. Torres, Effects of diffusion on total biomass in simple metacommunities, J. Theor. Biol., 447 (2018), 12-24.  doi: 10.1016/j.jtbi.2018.03.018.  Google Scholar [18] H. L. Smith, Monotone Dynamical Systems: An Introduction to the Theory of Competitive and Cooperative Systems, Amer. Math. Soci. Press, New York, USA, 1995.   Google Scholar [19] H. L. Smith and P. Waltman, The Theory of the Chemostat, New York: Cambridge University Press, 1995.  doi: 10.1017/CBO9780511530043.  Google Scholar [20] G. Takimoto and K. Suzuki, Global stability of obligate mutualism in community modules with facultative mutualists, OIKOS, 125 (2015), 535-540.  doi: 10.1111/oik.02741.  Google Scholar [21] J. J. Tewksbury et al., Corridors affect plants, animals, and their interactions in fragmented landscapes, Proc. Natl. Acad. Sci. U.S.A., 99 (2002), 12923-12926. Google Scholar [22] Y. Wang and D. L. DeAngelis, Comparison of effects of diffusion in heterogeneous and homogeneous with the same total carrying capacity on total realized population size, Theor. Popul. Biol., 125 (2019), 30-37.   Google Scholar [23] Y. Wang, H. Wu and D. L. DeAngelis, Global dynamics of a mutualism-competition model with one resource and multiple consumers, J. Math. Biol., 78 (2019), 683-710.  doi: 10.1007/s00285-018-1288-9.  Google Scholar [24] B. Zhang, K. Alex, M. L. Keenan, Z. Lu, L. R. Arrix, W.-M. Ni, D. L. DeAngelis and J. D. Dyken, Carrying capacity in a heterogeneous environment with habitat connectivity, Ecology Letters, 20 (2017), 1118-1128.  doi: 10.1111/ele.12807.  Google Scholar
Phase-plane diagram of system (5). Stable and unstable equilibria are identified by solid and open circles, respectively. Vector fields are shown by green arrows. Isoclines of species $u,v_1$ are represented by red and blue lines, respectively. Fix $r = r_1 = c = 1$. (a) Let $a_{12} = 2.5, a_{21} = 1.3$. Equilibrium $E^+(2.7,2.5)$ is globally asymptotically stable. (b) Let $a_{12} = 4.5, a_{21} = 0.9$. There are two positive equilibria $E^-(1.25,0.14)$ and $E^+(2.79,1.5)$. (c) Let $a_{12} = 3.5, a_{21} = 0.9$. The equilibria $E^-$ and $E^+$ coincide and form a saddle-node point $E^\pm(1.6,0.45)$. In the cases of (b-c), the separatrices (the black line) of $E^-$ subdivide the first quadrant into two regions: one is the basin of attraction of $E_1$ while the other is that of $E^+$. (d) Let $a_{12} = 2.5, a_{21} = 0.5$. All positive solutions converge to equilibrium $E_1(1,0)$
Comparison of $T_1(s, 0)$ and $T_2(s, D)$ when there is diffusion $D$ as shown in Theorem 4.1(ⅰ). The solid blue line represents $T_2(s, D)$, while the dash-dot red line represents $T_1(s, 0)$. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = b_1 = 1, s = 1$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1$. (a) When $s = 1$, we have $T_2>T_1$ for $D>0$. (b) When $s = 10$, we have $T_2>T_1$ as $0<D< 0.0135$ while $T_2<T_1$ as $D> 0.0135$
Comparison of $T_1(s, 0)$ and $T_2(s,100)$ when there is asymmetry $s$ in large diffusion as shown in Theorem 4.2(ⅰ). The solid blue line represents $T_2(s,100)$, while the dash-dot red line represents $T_1(s, 0)$. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = b_1 = 1, D = 100$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1$. Then we have $T_1(s, 0) = 1.9233$ and $T_2(s,100)>T_1(s, 0)$ as $0.1935< s< 9.3201$. Numerical computation shows that the function $T_2 = T_2(s,100)$ is convex upward with $T_2(s,100) = 0$ as $s\ge 12.1537$
Comparison of $T_1(s, D)$ and $T_1(s, 0), T_2(s, D)$ and $T_2(s, 0)$ when there is a small diffusion rate $D$, as shown in Theorem 4.1(ⅰ). The solid red and blue lines represent $T_1(s, D)$ and $T_2(s, D)$, while the dash-dot red and blue lines represent $T_1(s, 0)$ and $T_2(s, 0)$, respectively. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = b_1 = 1, D = 0.1, s = 1$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1$. Then we have $T_1(1, 0.1)<T_1(1,0)$ but $T_2(1,0.1)>T_2(1,0)$
Comparison of $T_1(s, D)$ and $T_1(s, 0), T_2(s, D)$ and $T_2(s, 0)$ when there is a large diffusion rate $D$, as shown in Theorem 4.2(ⅰ). The solid red and blue lines represent $T_1(s, D)$ and $T_2(s, D)$, while the dash-dot red and blue lines represent $T_1(s, 0)$ and $T_2(s, 0)$, respectively. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = b_1 = 1, D = 100, s = 0.1$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1$. Then we have $T_1(0.1,100)<T_1(0.1,0)$ but $T_2(0.1,100)>T_2(0.1,0)$
The surface of $T_2 = T_2(s,D)$ when both of $s$ and $D$ varies. Let $r = 0.2, c = 0.05, a_{12} = 0.1, b = 4, b_1 = 1$, $a_{21} = 0.5, r_1 = 0.8, r_2 = 0.1, 0<s<6, 0<D<6$. Numerical computation shows that for fixed $s$, the surface decreases monotonically, which is consistent with Fig. 4. For fixed $D$, the surface is convex upward, which is consistent with Fig. 5
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Lakhmir Singh Manjit Kaur Physics 2020 2021 Solutions for Class 9 Science Chapter 6 Multiple Choice Questions Mcqs are provided here with simple step-by-step explanations. These solutions for Multiple Choice Questions Mcqs are extremely popular among Class 9 students for Science Multiple Choice Questions Mcqs Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Lakhmir Singh Manjit Kaur Physics 2020 2021 Book of Class 9 Science Chapter 6 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Lakhmir Singh Manjit Kaur Physics 2020 2021 Solutions. All Lakhmir Singh Manjit Kaur Physics 2020 2021 Solutions for class Class 9 Science are prepared by experts and are 100% accurate.

#### Question 1:

While performing an experiment on verifying the laws of reflection of sound, a student is to choose between
(i) a narrow or a wide tube, and
(ii) a strong or a faint source of the sound.
The observed experimental difference, between the values of angle of incidence and angle of reflection, is likely to be minimum when he chooses a :
(1) narrow tube and a faint source
(2) wide tube and a faint source
(3) narrow tube and a strong source
(4) wide tube and a strong source

The observed experimental difference, between the values of angle of incidence and angle of reflection, is likely to be minimum when he chooses a narrow tube and a strong source. Because narrow tube allows the selected sound waves to travel clearly so that more distinct sound can be heard. Also, a strong source is always preferred over a faint source because the sound should be clear and audible to verify the laws of reflection of sound.

Hence, the correct answer is option 3.

#### Question 2:

For verifying the laws of reflection of sound, a student has to choose from :
(i) a black polished metal sheet or a white thermocol sheet.
(ii) a 0.5 m long tube of diameter 3 cm or a 1.5 m long tube of diameter 20 cm
He should prefer to choose the :
(1) metal sheet and the 0.5 m long tube
(2) metal sheet and the 1.5 m long tube
(3) thermocol sheet and the 0.5 m long tube
(4) thermocol sheet and the 1.5 m long tube

Metal surfaces reflect the highest percentage of the incident sound waves, whereas the thermocol sheets absorb the sound waves. Also, a narrow tube allows the selected sound wave to travel more clearly. That's why to verify the laws of reflection, the student should choose a metal sheet with a 0.5 m long tube of 3 cm diameter.

Hence, the correct answer is option 1.

#### Question 3:

A student sets up a slinky on a smooth tabletop in the manner shown here.

How can he produce transverse waves in the slinky by moving its free end Q?
(1) at an angle of 45º with the table top
(2) backward and forward along the length of the slinky
(3) up and down
(4) left and right

In a transverse wave, the particles are displaced perpendicular to the direction the wave travels. The student can generate a transverse wave by moving the free end Q of the slinky, vertically up and down.

Hence, the correct answer is option 3.

#### Question 4:

A student lists the following precautions for the experiment on determining the velocity of a pulse propagated through a stretched string :
(A) The string should not be stretched too tight.
(B) The counting of the pulse journeys must start from zero and not from one.
(C) The string should be stretched straight in contact with the table.
(D) The amplitude of the pulse should be kept appreciably high.
The incorrect entry, in this list of precautions, is the precaution listed as :
(1) A
(2) B
(3) C
(4) D

The incorrect entry in the above list of precautions is C, because the table should not in contact with the string, as it would cause a disturbance in the pulse propagation. when a pulse travels on a string, particles of the string move perpendicularly up and down, so there should be enough space both side the rope to particle moves up and down.

Hence, the correct answer is option 3.

#### Question 5:

For plotting temperature-time graph for a hot body, as it cools to room temperature, a student is to choose one each from each of the following pairs.
A : Calorimeter
(i) blackened from outside
(ii) polished from outside

B : Base for keeping the calorimeter
(i) insulated
(ii) metallic
In order to get the correct graph he should prefer to choose :
(1) A (i), B (ii)
(2) A (ii), B (ii)
(3) A (i) , B (i)
(4) A (ii), B (i)

For plotting temperature-time graph for a hot body, the student needs to choose a calorimeter which is blackened from outside as it would absorb more heat and the temperature change could be observed more accurately, also the base for keeping the calorie meter should be insulated so that the temperature of the surface where it is put does not interfere with the experiment readings.

Hence, the correct answer is option 3.

#### Question 6:

The magnitude of zero error of the spring balance and least count of the measuring cylinder, shown here, are, respectively :

(1) 2.5g and 0.1 mL
(2) 5.0g and 0.1 mL
(3) 2.5g and 0.2 mL
(4) 5.0g and 0.2 mL

Total number of divisions on the scale of spring balance = 100

Least count of the spring balance

Zero mark lags behind by 2 divisions on the main scale, so the zero error =

Total number of divisions on the scale of measuring cylinder = 50

Least count

Hence, the correct answer is option 4.

#### Question 7:

A student takes some water in a beaker and heats it over a flame for determining its boiling point. He keeps on taking its temperature readings. He would observe that the temperature of water :
(1) keeps on increasing regularly
(2) keeps on increasing irregularly
(3) first increases slowly, then decreases rapidly and eventually becomes constant
(4) first increases gradually and then becomes constant.

When the water is heated over the flame, it's temperature increases continuously, until it starts boiling, after reaching boiling point, the temperature of the water does not change, it becomes constant, as water starts vapourising.

Hence, the correct answer is option 4.

#### Question 8:

The zero-error in the spring balance shown and the correct weight of the solid, suspended from it, are equal, respectively, to :

(1) + 2g wt; 19g wt
(2) + 2g wt; 15g wt
(3) − 2g wt; 19g wt
(4) − 2g wt; 15g wt

Total number of the divisions on the scale = 30

Least count

As the zero mark of the scale is leading from its main position by two divisions, so the zero error of the scale =

Weight according to the reading = 17 g wt

Actual weight = Main reading $-$ zero error = 17 g wt $-$ 2 g wt = 15 g wt

Hence, the correct answer is option 2.

#### Question 9:

Four students A, B, C, and D while performing an experiment on establishing the relation between the loss of weight of a small solid when fully immersed in tap water, and the weight of water displaced by it, used four different shapes of overflow cans containing water as shown.

The arrangement, that would give correct results, is that of student :
(1) (A)
(2) (B)
(3) (C)
(4) (D)

In the given experiment, when the body is immersed in water, the water displaced should be moved out of the small made in the can, so its weight can be measured for experiment purposes. But in cans A, B, and D, the water level is not matching with the height of the hole. So when the body will be immersed in those cans it will not release the exact amount of water which the body displaced, So all the cans except C would give false results, but C would give correct results, as the water level is exactly matching with the height of the hole.

Hence, the correct answer is option 3.

#### Question 10:

Two students X and Y have to do their experiment on plotting the temperature-time graph for a hot body as it cools to room temperature. They do their experiments in the same lab, using completely identical apparatus, take equal amounts of tap water heated up to the same temperature, start their observations simultaneously and note the temperature values at identically spaced intervals of time. Their temperature-time graphs, plotted on a given graph paper, with the same give choices of scales along the axis, are however, as shown :

The following could be the reason for this difference :
(A) Use of an overhead fan by student X.
(B) Use of an overhead fan by student Y.
(C) Less frequent stirring of water by student X.
(D) Less frequent stirring of water by student Y.
The most likely reason is :
(1) A
(2) B
(3) C
(4) D

As we can see from the graph made by student X, the body cools down fast in comparison with the graph made by student Y. The body cools down slowly in the case of student B. As both students performed the same experiment under the same laboratory, the only factor of faster cooling in the case of student X can be the use of an overhead fan. So due to the use of an overhead fan by student X, the temperature of the body falls faster, it cools down quicker.

Hence, the correct answer is option 1.

#### Question 11:

In the set up shown the weight of the body was measured in air and in water. The reading of mark X in the spring balance would be :

(1) 36g wt
(2) 30g wt
(3) 24g wt
(4) 6g wt

When a body is immersed in a liquid, it feels an upward force, which is known as the buoyant force. The measure of the buoyant force is equal to the weight of the liquid displaced by the body. As we can see from the given figure, the volume of the liquid displaced is 6 mL, so the weight of the water displaced it 6 g wt. (Density of the water = 1 g/mL)

When the body is immersed in water, the spring balance will show the net weight of the body.

Net weight of the body in liquid = Actual weight $-$ Buoyant force = 30 g wt $-$ 6 g wt. = 24 g wt.

Hence, the correct answer is option 3.

#### Question 12:

A student uses a spring balance of least count 10g wt and range 500g wt. He records the weight of small iron cube in air, in tap water and in a concentrated solution of common salt in water. If his three readings taken in this order are W1, W2 and W3, he is likely to observe that :
(1) W1 > W2 > W3
(2) W3 > W2  > W1
(3) W1 > W3 > W2
(4) WW> W3

When a body is immersed in liquid, it feels an upward force, which is known as buoyant force. The measure of the buoyant force is equal to the weight of the liquid displaced by the body.

When immersed in the liquid, the spring balance will show the net weight reading of the body. So due to buoyant force, the weight of the body in the liquid will be less than the weight of the body in the air. As the density of the salt water is more than the normal water, hence the weight of the displaced salt water will be more than the same volume of the normal tap water, hence the buoyant force will be more when the body will be immersed in the salt water.

The body will have the least weight when immersed in salt water, and most when it is in the air. So we can say, W1 > W2 >W3

Hence, the correct answer is option 1.

#### Question 13:

While performing an experiment on verifying the laws of reflection of sound, the 'reflected sound' can be detected better by keeping one ear :
(1) near the end of the tube and keeping the other ear closed
(2) near the end of the tube and keeping the other ear open
(3) at about 5 cm from the end of the tube and keeping the other ear closed
(4) at about 5 cm from the end of the tube and keeping the other ear open

The reflected sound can be detected better by keeping one ear near the end of the tube and the other ear closed. Because by doing so, other environmental sounds can be excluded and one can focus only on the sound coming from the tube.

Hence, the correct answer is option 1.

#### Question 14:

In an experiment to determine the densities, four solids A, B, C and D are immersed in a liquid contained in a measuring cylinder, one by one. The volumes of water displaced by the solids A, B, C and D are 100 cm3, 100 cm3, 80 cmand 80 cm3, respectively. When weighed in air, the masses of solids A, B, C, and D were found to be 80g, 100g, 100g and 80g, respectively. The two solids having identical densities are :
(1) A and C
(2) B and D
(3) C and D
(4) A and D

Weights of the liquid displaced by A, B, C, and D are 100 g, 100 g, 80 g, and 80 g respectively, as the density of the water is 1 g/cm3. Now we can see the solid B and D displaces the liquid equal to their own weights, that means buoyant force is equal to the weight of the body in case of solids b and D. By this we can conclude that the density of both B and D is equal to the density of water, which makes their density identical, that's why they are experiencing the same buoyant force equal to their own weight.

Hence, the correct answer is option 2.

#### Question 15:

A student is given an iron cube of side 1 cm, a measuring cylinder of range 100 mL and least count 1 mL, and a spring balance of range 100g wt and least count 1g wt. He can use these to measure :
(1) both the mass and the volume of the given iron cube
(2) neither the mass nor the volume of the given iron cube
(3) only the mass of the given iron cube but not its volume
(4) only the volume of the given iron cube but not its mass

The student can measure the mass of the cube by using spring balance. Also, he can measure the volume of the cube by immersing the whole spring balance system with the attached cube, inside the measuring cylinder. The volume of the water displaced will be equal to the volume of the body.

Hence, the correct answer is option 1.

#### Question 16:

Four beakers are labelled as P, Q, R and S. A student puts salt solutions of different concentrations in the four beakers without noting which beaker contains solution of which concentration. A solid suspended from the hook of a spring balance gives a reading of 150g in air. When this solid, while still suspended from the hook of spring balance, is fully immersed in beakers P, Q, R, and S, the readings shown by the spring balance in the four beakers P, Q, R and S are 110g, 130g, 140g and 120g respectively. The most concentrated solution is contained in the beaker labelled as :
(1) P
(2) Q
(3) R
(4) S

The buoyant force depends on the weight of the liquid displaced, as the density of the different salt solution is different, so the buoyant force acting on them will be different too. More the density of the liquid more will be the weight of the liquid displaced, hence more will be the buoyant force acting in an upward direction on the body. Therefore, the most concentrated solution will show the least reading on the spring balance when immersed in liquid.

Hence, the correct answer is option 1.

#### Question 17:

While doing an experiment on plotting the temperature-time graph of hot water as it cools, we can get a good graph :
(1) only by noting the temperature of hot water every 30 seconds throughout
(2) only by noting the temperature of hot water every 1 minute throughout
(3) by noting the temperature of hot water every 1 minute to start with and every 2 minutes later on
(4) by noting the temperature of hot water every 2 minutes to start with and every 1 minutes later on

When hot water cools down, Initially its temperature falls at a higher rate but later the rate of fall of temperature decreases. So, in starting the temperature will fall faster but after sometime, it will fall a little slower. Hence, to plot a temperature-time graph for cooling of hot water, we have to note the temperature of hot water every 1 minute to start with and every 2 minutes later on.

Hence, the correct answer is option 4.

#### Question 18:

A student heats some amount of tap water in a calorimeter to a temperature of nearly 50ºC above the room temperature. He then records the temperature of this water as it is cooling down at regular intervals of 2 minutes each. He tabulates his observations as follows :

 Time (in minutes) → 0 2 4 6 8 10 Temperature (in ºC) → t1 t2 t3 t4 t5 t6
He is likely to observe that :
(1) (t− t2) > (t3  t4) > (t5  t6)
(2)
(t− t2) = (t t4) = (t5  t6)
(3)
(t− t2) < (t t4) < (t5  t6)
(4)
(t− t2) < (t t4) > (t5  t6)

When hot water cools down, initially its temperature falls at a higher rate but later the rate of fall of temperature decreases. So, for every two minutes, the temperature difference would be greater in starting and will decrease with time. Therefore, we can say that (t− t2) > (t t4) > (t5  t6).

Hence, the correct answer is option 1.

#### Question 19:

When an object is fully immersed in a liquid, the apparent loss in weight :
(1) is less than the weight of liquid displaced by it
(2) is more than the weight of liquid displaced by it
(3) is equal to the weight of liquid displaced by it
(4) does not depend on the density of the liquid displaced by it

When a body is immersed in liquid, an upside force acts on it, which is known as the buoyant force. The measure of the buoyant force is equal to the weight of the liquid displaced by the body. As the apparent weight loss in the body happens due to the buoyant force acting in the opposite direction of its weight, so the apparent weight loss is equal to the weight of the liquid displaced by the body.

Hence, the correct answer is option 3.

#### Question 20:

Four metal balls A, B, C and D having radius of 2.5 cm each are made of copper, aluminium, gold and iron respectively. The densities of copper, aluminium, gold and iron are 8.9 g/cm3, 2.7 g/cm3, 19.3 g/cm3 and 7.8 g/cm3 respectively. When the balls A, B, C and D are tied to threads, suspended from the hook of a spring balance and immersed completely in strong salty water, one by one, the apparent loss in weight will be :
(1) maximum in ball B
(2) maximum in ball C
(3) minimum in ball B
(4) same in all the balls

When a body is immersed in a liquid, the apparent loss in the weight of the body occurs due to the Buoyant force acting on it. The buoyant force acts in an upward direction and is equal to the weight of water displaced by the body. All the four balls are of the same size, so they will replace the same amount of water. That's why the apparent loss in weight will be the same for all the balls.

Hence, the correct answer is option 4.

#### Question 21:

A glass ball hanging from the hook of a sensitive spring balance is completely submerged in highly salty water and tap water, one by one. If the readings of spring balance when the ball is in highly salty water and tap water are x and y respectively, then :
(1) x < y
(2) x > y
(3) x = y
(4) x = 2y

When a body is immersed in a liquid it feels an upward force, which is known as buoyant force. The buoyant force depends upon the weight of the liquid displaced by the body.

In the given case, the same glass ball is being submerged in both the liquids, so the volume of water displaced will be the same in both cases. But the density of the highly salty water is more than the normal tap water, so the weight of the liquid displaced in case of highly salty water will be more. As a result, the glass ball will feel a more upward force when immersed in salt water. So the spring balance reading will be less for salty water and more for tap water. i.e. x < y.

Hence, the correct answer is option 1.

#### Question 22:

You are given four salt solutions W, X, Y and Z of different concentrations. W is a 20% salt solution, X is a 35% salt solution, Y is a 10% salt solution whereas Z is a 50% salt solution. When a solid suspended from the hook of a spring balance is fully submerged in all these salt solutions, one by one, then the spring balance will show the minimum reading when the solid is immersed in :
(1) solution W
(2) solution X
(3) solution Y
(4) solution Z

When a body is immersed in a liquid it feels an upward force, which is known as buoyant force. The buoyant force depends upon the weight of the liquid displaced by the body.

In the given case, the same body is being submerged in all the solutions, so the volume of water displaced will be the same for all. But all the solutions have different densities. More the density of the solution, more the weight of the liquid displaced. As we can see from the given data that solution Z is more concentrated, so has more density as compared to other solutions. Weight of the liquid displaced will be maximum when the body will be submerged in the solution Z. So, the body will feel the maximum upward force in solution Z and will show the minimum reading on the spring balance.

Hence, the correct answer is option 4.

#### Question 23:

An aluminium ball is fully immersed in distilled water, well water, sea water and inland lake water, one by one. The aluminium ball will appear to suffer the maximum loss in weight when immersed in :
(1) distilled water
(2) sea water
(3) well water
(4) lake water

When a body is immersed in a liquid it feels an upward force, which is known as buoyant force. The buoyant force depends upon the weight of the liquid displaced by the body.

In the given case, an aluminum ball is immersed in different liquids of the different densities. More the density of the liquid, more the weight of the liquid displaced by the ball. When the weight of the liquid displaced will be maximum, the ball will feel the maximum upward force, hence maximum loss in weight. Seawater is known to have the maximum density among all types of waters, so when the ball will be immersed in sea water, it will experience the maximum weight loss.

Hence, the correct answer is option 2.

#### Question 24:

When an object is fully submerged in strong salty water, it undergoes an apparent :
(1) loss in mass
(2) loss in volume
(3) loss in density
(4) loss in weight

When the object will be fully submerged in strongly salty water it will feel an upward force, which is known as the buoyant force. Due to the presence of this upward buoyant force, the object will undergo an apparent loss in weight.

Hence, the correct answer is option 4.

#### Question 25:

If a body tied to a spring balance is fully immersed in a liquid, the apparent loss in its weight :
(1) is more in a denser liquid
(2) is less in a denser liquid
(3) does not depend on density of liquid
(4) is equal to apparent gain in weight of liquid

When a body is immersed in a liquid it feels an upward force, which is known as buoyant force. The buoyant force depends upon the weight of the liquid displaced by the body. Due to the presence of this upward buoyant force, the body undergoes in an apparent weight loss. More the density of the liquid more will be the weight of the liquid displaced by the body, so more apparent weight loss.

Hence, the correct answer is option 1.

#### Question 26:

In an experiment on  determining the velocity of a pulse propagating through a stretched string, the stopwatch should be started and stopped at instants corresponding to the ones shown in :

(1) Fig. 1 and Fig. 2
(2) Fig. 1 and Fig. 3
(3) Fig. 2 and Fig. 1
(4) Fig. 2 and Fig. 3

As given in the figure, a wave pulse starts traveling from end A towards the end B. To calculate the velocity of the given wave, the stopwatch should be started at the instant corresponding to the figure A and should be stopped at the instant when the pulse reaches the other end, which corresponds to the figure 3. By dividing the distance travelled by the pulse from end A to end B with the time reading obtained by the stopwatch, we can easily determine the velocity of the given wave.

Hence, the correct answer is option 2.

#### Question 27:

When a fresh egg is put into a beaker filled with water, it sinks. When the same egg is put in a strong salty water, then it floats. Which of the following is the incorrect statement in this context ?
(1) salt water enters into egg by osmosis and makes it lighter
(2) salt water exerts more buoyant force
(3) salt water is denser than tap water
(4) upthrust exerted by a liquid depends on its density

The density of an egg is higher than the density of the water, so a normal egg sinks when it's put in the water. But if the same egg is put in strongly salty water, it floats. Because the density of the salty water is more than the egg, so it exerts more buoyant force on the egg which balances its weight.

Hence, the correct answer is option 2.

#### Question 28:

When two balls, one of iron and the other of aluminium, are completely immersed in strong salty water, they undergo an equal loss in weight. This shows that iron and aluminium balls have :
(1) the same densities
(2) the same masses
(3) the same volumes
(4) the same weights

When a body is immersed in a liquid, it undergoes an apparent weight loss due to an upward buoyant force acting on the body. The buoyant force depends only on the density of the liquid and the weight of the liquid displaced by the body.

In the given case, as both the bodies undergo an equal weight loss, that means they displace the same amount of liquid when immersed. For that, both the ball must have the same volume.

Hence, the correct answer is option 3.

#### Question 29:

When two solids are put in two pans of a beam balance, they exactly balance each other in air. When the two solids are completely immersed in water alongwith  pans of balance, then they no longer balance each other. Which of the following is the incorrect statement about these balls ?
(1) they have equal masses in air
(2) they have equal weights in air
(3) they have equal volumes in air
(4) they have unequal densities

As both the solid balance each other on a beam balance in the air, so they have the same mass and weight. When they are immersed in water along with beam balance, they no longer balance each-other because both the solids experience different buoyant force. The buoyant force depends on the volume of the body immersed, that means the solids do not have equal volumes.

Hence, the correct answer is option 3.

#### Question 30:

The temperature-time graph obtained when hot water is allowed to cool resembles the graph given in one of the following figures. The correct figure is :

(1) A
(2) B
(3) C
(4) D


text:
# mean absolute deviation

1. ### How do I find percentage from two values which are coming from two different population sets that has large difference in population sizes?

I need to find percentage from two values which are coming from two different population sets that has large difference in population sizes. They are also different in nature. I set is teachers and another set is students. I have the performance score of 10 schools. Each schools have filled...
2. ### Higher standard deviation than the mean

HI ! Excuse me for my english, my writing is not very good. I want to know what implies or what explains in an excercise when the standard deviation (STD=67) is higher than the mean (M=38), and the histogram behaves like an exponential distribution. Thanks by the way ! :tup: Regards from...
3. ### Normal distribution with Mean Absolute Deviation What percent lies within 1,2 AbsDev

Roughly 95% of the normal distribution lies within 2 standard deviations. However, recently I have begun using the mean absolute deviation a lot. What percent of the normal distribution is within 2 mean absolute deviations? Is this possible to calculate?


text:
HP-25C Eumulator
01-05-2015, 12:56 PM
Post: #126
 Chris Chung Member Posts: 218 Joined: Aug 2014
RE: HP-25C Eumulator
Of the V3 1st batch, I received PM that one of the units have a digit gone blank. I believe it is a cold solder joint and if that is the case, it could be easily fixed w/ a soldering iron and some solder wires.

My 1st few builts had buttons stop working after a few weeks and I discovered that the soldering is weak. Afterwards the newer units I let my iron stay 1/2 second longer on each joint to make them more secure. Still it will happen and you can try and fix this.

If it is a lone tactile button, you could just re-heat the joint w/ a soldering iron, watch the solder melt and flow and sink / stick to the pad on PCB. You should "tin" your iron w/ a bit of fresh solder (new from your solder wire spool) before applying to the cold joint, otherwise it is difficult to melt the older solder on the joint. You can google "good / bad solder joints". This is a good visual..

Now if your failed tactile buttons are in a group of 3 or more, this means that the cold joint is on the IC side. And you need to follow the schematic / ic pinout (or ask me) to locate which IC pin corresponds to the failed buttons. And re-flow it to try and solve the problem.

The same applies to the LED display. If one digit / segment goes blank, and if at the same time you do not have button failures, it is likely that a joint on the LED modules went bad. And If it's digits not showing + some buttons fails, it is likely a joint on the MCU side, since a MCU I/O pin is shared.

The schematic w/ IC pinouts can be found in the source code (from my github repo). I am including the ascii schematic here; If one of the digits / segments failed, we can trace back which I/O port (digits are on MCU, segments are on 164 shift register) it is connected to, and find which pin is in question from the pinout diagrams.

It is easier for me as I would use a blank PCB and a DMM to trace the failed connections, and then re-flow the joints. In previous batches there are always a few cold solder joints and I need to re-flow after testing all the buttons and displays. For the latest batch, I had only 1 joint that I need to re-flow, so I guess I am getting better at them now.

Since the PCB design is components-on-both-sides, sometimes you cannot re-flow a join on the original size. In this case you should secure it on the other side (i.e. component side).

Code:
                MSP430G25x3              -----------------          /|\|                 |           | |                 |           --|RST  digit 0 P2.6|--\             |     digit 1 P1.0|--\       digits bus (12)             |     digit 2 P2.7|-----//--+---------+             |     digit 3 P1.1|--/      |         ~  3 units of 4 digits bubble led             |     digit 4 P1.2|--/      |         |             |     digit 5 P1.3|--/      |    +---------+ +---------+ +---------+             |     digit 6 P2.0|--/      |    | % % % % | | % % % % | | % % % % |----+             |     digit 7 P1.4|--/      ~    +---------+ +---------+ +---------+    |             |     digit 8 P1.5|--/      |                                           |             |     digit 9 P2.1|--/      | digits bus x 3 scan lines                 |             |     digit a P1.6|--/      +-----+-----+---/ ~~ /--+-----+-----+-----+ |             |     digit b P1.7|--/  _=_ | _=_ | _=_ | _/ ~~ /=_ | _=_ | _=_ | _=_ | |             |      scan X P2.5|-----o o-+-o o-+-o o-+-/ ~~ /o o-+-o o-+-o o-+-o o-+ |             |      scan Y P2.4|-----o o-+-o o-+-o o-+/ ~~ /-o o-+-o o-+-o o-+-o o-+ |             |      scan Z P2.3|-----o o-+-o o-+-o o-/ ~~ /+-o o-+-o o-+-o o-+-o o-+ |             |                 |                                                     ~             |                 |      (32 tactile button)                            |             |                 |                                                     |             |                 |        +--------+                                   |             |             P2.2|--+---->|CLK ~CLR|---+---- Vcc                       |             |                 |  |     |       B|---+                               |             |                 |  |     |  0(d)QA|-\                                 |             |                 |  |     |  1(f)QB|-\                segments bus (8) |             |                 |  -     |  2(b)QC|--------------//-------------------+             |                 | | |    |  3(a)QD|-/             |                 | | |1k  |  4(g)QE|-/  74hc164 segment driver             |                 |  -     |  5(e)QF|-/             |                 |  |     |  6(c)QG|-/             |                 |  +---->|A 7(h)QH|-/             |                 |  |     +--------+             |                 | === (102)    1 wire shift register via RC delay             |                 | ===  1nF     to shift 1, Hi-Low-Hi             +-----------------+  |           to shift 0, Hi-Looooooow-Hi (>20us Lo)                                 ___                                 ///                74hc164                o---------+                |A     Vcc|                |B      QH|                |QA     QG|                |QB     QF|                |QC     QE|                |QD   ~CLR|                |Gnd   CLK|                +---------+     4 digit bubble led from sparkfun     +--+--+--+--+--+    |a  b  2  f  d  g|    |1  e  c  3  h  4|     +--+--+--+--+--+     msp430g25x3 pinout (a=port1, b=port2)     +--+--+--+--+--+--+--+--+--+    |- b6 b7 CK IO a7 a6 b5 b4 b3|    |                            |    |+ a0 a1 a2 a3 a4 a5 b0 b1 b2|     o--+--+--+--+--+--+--+--+--+

For this failed unit's missing digit, it is the 4th digit (i.e. 1st LED module). From the pin-out, this would be pin 6 (i.e. bottom row, 6th pin). The solder side has been hidden, so the proper fix would be to solder on the component side.

From the schematic, this digit (digit 3) is connected to I/O port P1.1(a1) of the MCU, refer to the pinout this would be pin 3. A reflow can refresh the connection. If it is this joint that failed, it would have affected also 3 buttons. So my guess is that the cold joint is on the LED module.

There is also a chance that the LED module failed. I do not think it is likely as I had gone thru 100+ LED modules and never experience one that fails.

I hope this is helpful when you encounter a cold joint in your unit.
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text:
# 克爾效应

$\Delta n = \lambda K E^2,\$

## 来源

1. ^ Weinberger, P. John Kerr and his Effects Found in 1877 and 1878. Philosophical Magazine Letters. 2008, 88 (12): 897–907. doi:10.1080/09500830802526604.


text:
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