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QUESTION: 1

**Direction (Q. Nos. 1-10) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONLY ONE option is correct.**

**Q. Entropy change for the following reversible process is 1 mole H _{2}O**

**(l, 1 atm, 100°C ) 1 mole H _{2}O (g , 1 atm, 100°C)(ΔH_{vap} = 40850 J mol^{-1})**

Solution:

∆S=Change in energy/Absolute temp =∆H/(100+273)K =40850/373 = 109.52 J/K/mol

QUESTION: 2

Entropy change when 2 moles of an ideal gas expands reversibly from an initial volume of 1 dm^{3} to a final volume of 10 dm^{3} at a constant temperature of 298 K is

Solution:

∆S= 2.303nR × log(V_{2}/V_{1})Here n=2, R=8.314, V2= 10, V_{1}= 1

QUESTION: 3

3 moles of a diatomic gas are heated from 127° C to 727° C at a constant pressure of 1 atm. Entropy change is (log 2.5 = 0 .4)

Solution:

∆S = nCplnT_{2}/T_{1} + nRlnP_{1}/P_{2}

Since pressure is constant, so the second term will be zero.

Or ∆S = 3×7/2×8.314×2.303×log(1000/400)

= 80.42 JK^{-1}

QUESTION: 4

10 dm^{3} of an ideal monoatomic gas at 27° C and 1.01 x 10^{5} Nm^{-2} pressure are heated at constant pressure to 127°C. Thus entropy change is

Solution:

For isobaric process, we have ∆S =nC_{p}ln(T_{2}/T_{1})

T2 = 273+127 = 400K and T_{1} = 273+227 = 300K

Applying pV = nRT at initial condition,

1×10 = n×0.0821×300

n = 0.40

Applying ∆S =nC_{p}ln(T_{2}/T_{1})

∆S =0.40×5/2R×ln(400/300) = 2.38 JK^{-1}

QUESTION: 5

Exactly 100 J of heat was transferred reversibly to a block of gold at 25.00° C from a thermal reservoir at 25.01 °C, and then exactly 100 J of heat was absorbed reversibled from the block of gold by a thermal reservoir at 24.99° C. Thus entropy change of the system is

Solution:

QUESTION: 6

Given

I. C (diamond) + O_{2}(g) → CO_{2}(g) ; ΔH° = - 91.0 kcdl mol^{-1}

II. C(graphite) + O_{2}(g) → CO_{2}(g) ; ΔH° = - 94.0 kcal mol^{-1}

**Q. At 298 K, 2.4 kg of carbon (diamond) is converted into graphite form. Thus, entropy change is**

Solution:

The reaction is

C(diamond) → C(graphite) ∆H = (94-91) = 3 kcal mol-1

∆S = ∆H/T

∆H = (94-91)×2.4×103/12

= 600 kcal

∆S = 600/298 = 2.013 kcal K^{-1}

QUESTION: 7

Consider a reversible isentropic expansion of 1.0 mole of an ideal monoatomic gas from 25°C to 75°C. If the initial pressure was 1.0 bar, final pressure is

Solution:

Isentropic process means that entropy is constant. This is true only for reversible adiabatic process.

Applying P_{1}^{1-γ} T_{1}^{γ} = P_{2}^{1-γ} T_{2}^{γ} (for monatomic species, γ = 5/3)

(1/P)-⅔ = (75+273/25+273)^{5/3}

Or (1/P)-2 = (348/298)^{5}

Or P = 1.474 bar

QUESTION: 8

Consider the following figure representing the increase in entropy of a substance from absolute zero to its gaseous state at some temperature

**Q. ΔS° (fusion) and ΔS° (vaporisation) are respectively indicated by **

Solution:

You can co-relate both graph and the result will be option c.

QUESTION: 9

ΔH_{vap} = 30 kJ mol^{-1} and ΔS_{vap} = 75 J mol^{-1} K^{-1}. Thus, temperature of the vapour at 1 atm is

**[IIT JEE 2004]**

Solution:

Vapour pressure is equal to atmospheric pressure ,it means the substance is at boiling point

At boiling point, liquid and Gas are in equilibrium. Therefore dG=0

dG = H - TdS

dG = 0

⇒H = TdS

⇒T = H/dS

⇒ T = 30 103/75 = 400K

QUESTION: 10

For the process, and 1 atmosphere pressure, the correct choice is

**[JEE Advanced 2014]**

Solution:

At 100°C and 1 atmosphere pressure H_{2}O (l) ⇋ H_{2}O(g) is at equilibrium. For equilibrium

*Multiple options can be correct

QUESTION: 11

**Direction (Q. Nos. 11-14) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c) and (d), out of which ONE or MORE THANT ONE is correct.**

**Q. For an ideal gas, consider only (p -V) work in going from initial state X to the final state Z. The final state Z can be reached either of the two paths shown in the figure. Which of the following choice (s) is (are) correct?**

**(Take ΔS as change in entropy and W as work done)**

**[IIT JEE 2012]**

Solution:

∆S_{X→Z} = ∆S_{X→Y} + ∆S_{Y→Z}(Entropy is a state function, so it is additive)

W_{X→Y→Z} = WX→Y (work done in _{y→z} is zero as the process is isochoric)

*Multiple options can be correct

QUESTION: 12

Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s) is (are)

**[JEE Advanced 2013]**

Solution:

When an ideal solution is formed process is spontaneous

According to Raoult's law, for an ideal solution

ΔH=0

ΔVmix=0

Since there is no exchange of heat energy between system and surroundings

ΔSsurroundings=0

ΔSsys.=+ve

∴ From the relation

ΔG=ΔH–TΔS

ΔG=−ve

Hence, the correct options are B, C and D

*Multiple options can be correct

QUESTION: 13

Consider the following process

**Select correct choices(s)**

Solution:

∆S_{X→Y} = ∆S_{X→Z} + ∆S_{Z→Y}(Entropy is a state function, so it is additive)

However, ∆S_{X→Z} is zero. As for adiabatic process. There is no change in entropy.

∆S_{X→Y} = ∆S_{Z→Y} = 2.303 5 8.314 log(10/1) = 95.7 JK^{-1}

*Multiple options can be correct

QUESTION: 14

In which of the following cases, entropy of I is larger than that of II?

Solution:

a) More molar mass, more entropy. So ∆_{SN2O4} > ∆S_{NO2}

b) CO_{2} has more entropy than dry ice at -78°C

c) Pure alumina iss crystalline solid while ruby is amorphous. And ∆S_{amorphous} > ∆S_{Crystalline}. So alumina has less entropy than ruby.

d) At lower pressure, entropy be higher as gas particles are far from each other. So (∆S_{N2})1_{ bar} > (∆S_{N2})_{5 bar}

QUESTION: 15

Direction (Q. Nos. 15 and 18) This section contains 2 paragraphs, each describing theory, experiments, data etc. Four questions related to the paragraphs have been given. Each question has only one correct answer among the four given options (a), (b), (c) and (d)

**Passage I **

**Consider a series of isotherms and adiabates as shown**

AB, CD and EF are isotherms.

AC, CE, BD and D F are adiabates.

**Q. **

Solution:

AB, CD and EF are isotherms.

AC, CE, BD and D F are adiabates.

For adiabatic process, we have change n entropy = 0

So Entropy_{A→B} = Entropy_{C→D} = Entropy_{E→F} and Entropy_{A→C} = Entropy_{A→C} = Entropy_{C→E} = Entropy_{F→D} = Entropy_{D→B}

By this, all options are correct.

QUESTION: 16

**Passage I **

**Consider a series of isotherms and adiabates as shown**

AB, CD and EF are isotherms.

AC, CE, BD and D F are adiabates.

**Q. Select incorrect relationship**

Solution:

QUESTION: 17

Passage II

The stopcock connecting A and B is of negligible volume. Stopcock is opened and gases are allowed to mix isothermally.

**Q. Final pressure set up is**

Solution:

Volume in (I) = nRT/P_{1} = 1× 0.0821× 298/4 = 6.11 L

Volume in (II) = nRT/P_{2} = 1× 0.0821× 298/2 = 12.23 L

Total volume = 18.34 L

Applying PV = nRT at final condition,

P = 2× 0.0821× 298/18.34 = 2.66

QUESTION: 18

Passage II

The stopcock connecting A and B is of negligible volume. Stopcock is opened and gases are allowed to mix isothermally.

**Q. Entropy change for the system is **

Solution:

*Answer can only contain numeric values

QUESTION: 19

Direction (Q. Nos. 19 and 20) This section contains 2 questions. Each question, when worked out will result in an integer from 0 fo 9 (both inclusive).

**Q. Consider two Carnot engines (1) and (2)**

**Efficiency η _{2}
Derive the value of η_{2} η_{1}**

Solution:

η_{1} = 1-273/298 = 25/298

η_{2} = 1-248/298 = 50/298

η_{2}/η_{1} = (50/298)/25/298 = 2

QUESTION: 20

**A cyclic heat engine operates between a source temperature of 927 ^{o}C and a sink temperature of 27 ^{o}C. What will be the maximum efficiency of the heat engine?**

Solution:

Heat engine is operated between the temperatures

T1 = 927 0C = 927 + 273 = 1200 K and

T2 = 27 0C = 27 + 273 = 300 K

The maximum efficiency of a heat engine is given by,

η_{max} = 1 – (T2/T1)

η_{max} = 1 – (300/1200)

η_{max} = 0.75

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