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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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The heat of transition (ΔH_t) of graphite into diamond would be, where

$C (graphite) + O_{2} (g) \to C O_{2} (g); Δ H = x k J$

$C (diamond) + C_{2} (g) \to C O_{2} (g); Δ H = y k J$ [Pb. PET 1985]

$(x + y) k J m o l^{- 1}$

$(x - y) k J m o l^{- 1}$

$(y - x) k J m o l^{- 1}$

None of these

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Explanation

Graphite → diamond $Δ H_{t} = (x - y) k J m o l^{- 1}$ .

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The enthalpy of combustion at 25°C of H₂, cyclohexane (C₆H₁₂) and cyclohexene (C₆H₁₀) are –241, –3920 and –3800 KJ / mole respectively. The heat of hydrogenation of cyclohexene is [BHU 2005]

– 121 KJ / mole

+ 121 KJ / mole

– 242 KJ / mole

+ 242 KJ / mole

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Explanation

(i) $H_{2} + \frac{1}{2} O_{2} \to H_{2} O, Δ H = - 241 k J$

(ii) $C_{6} H_{10} + \frac{17}{2} O_{2} \to 6 C O_{2} + 5 H_{2} O, Δ H = - 3800 k J$

(iii) $C_{6} H_{12} + 9 O_{2} \to 6 C O_{2} + 6 H_{2} O, Δ H = - 3920 k J$

$C_{6} H_{10} + H_{2} \to C_{6} H_{12}$

Eq. (i) + Eq. (ii) – Eq. (iii)

$Δ H = - 241 - 3800 - (- 3920)$

= –4041 + 3920 = – 121 kJ

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