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Chemical Thermodynamics MCQs for NEET — Chemistry Questions with Answers

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Which of the following is true for the reaction $H_{2} O (l) ⇌_{}^{} H_{2} O (g)$ at 100°C and 1 atmosphere [KCET 1991; AIIMS 1996]

ΔE = 0

ΔH = 0

ΔH = ΔE

ΔH = TΔS

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Explanation

At equilibrium, ΔG = 0

Hence 0 = ΔH – TΔS or ΔH = TΔS.

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The enthalpy of vapourization water is 386 kJ. What is the entropy of water [BHU 1997]

0.5 kJ

1.03 kJ

1.5 kJ

22.05 kJ

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Explanation

$∆ S = ∆ H_{v a p \div} T ∆ S = 386 \times 1000 \div 373 = 1.03 K J$

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Identify the correct statement regarding entropy [CBSE PMT 1998; BHU 2001]

At 0°C, the entropy of a perfectly crystalline substance is taken to be zero

At absolute zero of temperature, the entropy of a perfectly crystalline substance is +ve

At absolute zero of temperature, the entropy of all crystalline substances is taken to be zero

At absolute zero of temperature, the entropy of a perfectly crystalline substance is taken to be zero

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Explanation

This is the statement of third law of thermodynamics.

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An engine operating between 150°C and 25°C takes 500 J heat from a higher temperature reservoir if there are no frictional losses, then work done by engine is [MH CET 1999]

147.7 J

157.75 J

165.85 J

169.95 J

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Explanation

$T_{2} = 150 + 273 = 423 K$

$T_{1} = 25 + 273 = 298 K$

Q = 500 K

$\frac{W}{Q} = \frac{T_{2} - T_{1}}{T_{2}}$ ; $W = 500 (\frac{423 - 298}{423}) = 147 .7 J$ .

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The standard entropies of CO₂(g), C(s) and O₂(g) are 213.5, 5.690 and 205 JK^–1 respectively. The standard entropy of formation of CO₂(g) is [CPMT 2001]

1.86 JK^–1

1.96 JK^–1

2.81 JK^–1

2.86 JK^–1

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Explanation

Formation of CO₂ is,

$C_{(s)} + O_{2 (g)} \to C O_{2 (g)}$

$Δ S^{o} = 213 .5 - 5 .690 - 205 = 2 .81 J K^{- 1}$ .

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Equal volumes of monoatomic and diatomic gases at same initial temperature and pressure are mixed. The ratio of specific heats of the mixture (C_p/C_v) will be [AFMC 2002]

1.67

1.5

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Explanation

$C_{v} = \frac{3}{2} R T; C_{p} = \frac{5}{2} R T$ for monoatomic gas

$C_{v} = \frac{5}{2} RT; C_{p} = \frac{7}{2} R T$ for diatomic gas

Thus for mixture of 1 mole each, $C_{v} = \frac{\frac{3}{2} R T + \frac{5}{2} R T}{2}$ and $C_{p} = \frac{\frac{5}{2} R T + \frac{7}{2} R T}{2}$

Therefore, $C_{p} / C_{v} = \frac{3 R T}{2 R T} = 1 .5$ .

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The unit of entropy is [CBSE PMT 2002]

J mol^–1

JK mol^–1

J mol^–1 K^–1

J^–1 K^–1 mol^–1

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Explanation

$Δ S = \frac{q_{r e v}}{T}$ ∴ unit of S is JK^–1 mol^–1.

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The entropy changed involved in the conversion of 1 mole of liquid water at 373 K to vapour at the same temperature will be $[Δ H_{vap} = 2 .257 k J / g m]$ [MP PET 2002]

0.119 kJ

0.109 kJ

0.129 kJ

0.120 kJ

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Explanation

$H_{2} O_{(l)} ⇌_{}^{} H_{2} O_{(g)}, Δ S = \frac{Δ H_{v a p}}{T},$

$Δ H_{v a p .} = 2 .257 K J / g$

or $Δ H_{v a p} = 2 .257 \times 18 k J / m o l . = 40 .7 K J / m o l$

hence, $Δ S = \frac{40 .7}{373} = 0 .109 k J / m o l / K .$

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When a liquid boils, there is [JIPMER 2002]

An increase in entropy

A decrease in entropy

An increase in heat of vaporization

An increase in free energy

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Explanation

Liquid → Vapour, entropy increases.

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The work done to contract a gas in a cylinder, is 462 joules. 128 joule energy is evolved in the process. What will be the internal energy change in the process [MP PMT 2003]

+ 590 joules

– 334 joules

+ 334 joules

– 590 joules

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Explanation

As the work is done on system, it will be positive i.e. W = +462 joule, E = –128 joule (heat is evolving)

From the I^st law of thermodynamics

$Δ E = q + w = (- 128) + (+ 462) = + 334 J o u l e s .$

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