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Chemistry MCQs for NEET — Practice Questions with Answers

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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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For a carnot engine, the source is at 500 K and the sink at 300 K. What is efficiency of this engine [BHU 2004]

0.2

0.4

0.6

0.3

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Explanation

Given that, $T_{1} = 500 K, T_{2} = 300 K$

By using, $η = \frac{T_{1} - T_{2}}{T_{1}} = \frac{500 - 300}{500} = \frac{200}{500} = 0 .4$ .

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From Kirchhoff's equation which factor affects the heat of reaction [MP PMT 1990]

Pressure

Temperature

Volume

Molecularity

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Explanation

Effect of temperature in heat of reaction is given by Kirchoff’s equation.

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The absolute enthalphy of neutralisation of the reaction $M g O (s) + 2 H C l (a q) \to M g C l_{2} (a q) + H_{2} O (l)$ will be [CBSE PMT 2005]

Less than –57.33 kJ mol^–1

–57.33 kJ mol^–1

Greater than –57.33 kJ mol^–1

57.33 kJ mol^–1

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Explanation

Heat of neutralisation will be less than –57.33 kJ/mole because some amount of this energy will be required for the dissociation of weak base (MgO)

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