4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is . If the speed of sound in this gas at NTP is , then the heat capacity at constant pressure is (Take gas constant R = 8.3 ) [2015]

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Solution

Q.
4.0 g of a gas occupies 22.4 litres at NTP. The specific heat capacity of the gas at constant volume is $5.0 J K^{- 1} {mol}^{- 1}$ . If the speed of sound in this gas at NTP is $952 {ms}^{- 1}$ , then the heat capacity at constant pressure is (Take gas constant R = 8.3 $J K^{- 1} {mol}^{- 1}$ ) [2015]

1 $7.0 J K^{- 1} {mol}^{- 1}$

2 $8.5 J K^{- 1} {mol}^{- 1}$

3 $8.0 J K^{- 1} {mol}^{- 1}$

4 $7.5 J K^{- 1} {mol}^{- 1}$

Ans.
(3)

Since 4.0 g of a gas occupies 22.4 litres at NTP, so the molecular mass of the gas is $M = 4.0 g$ ${mol}^{- 1}$

As the speed of sound in the gas is $v = \sqrt{\frac{γ R T}{M}}$

where $γ$ is the ratio of two specific heats, $R$ is the universal gas constant and $T$ is the temperature of the gas.

$∴$ $γ = \frac{M v^{2}}{R T}$

Here, $M = 4.0$ $g$ ${mol}^{- 1} = 4.0 \times 10^{- 3}$ $kg$ ${mol}^{- 1}$ , $v = 952$ $m$ $s^{- 1}$

$R = 8.3 J$ $K^{- 1} {mol}^{- 1}$ and

$T = 273$ $K$ (at NTP)

$∴$ $γ = \frac{(4.0 \times 10^{- 3} {kg mol}^{- 1}) {(952 {m s}^{- 1})}^{2}}{(8.3 {J K}^{- 1} {mol}^{- 1}) (273 K)} = 1.6$

By definition, $γ = \frac{C_{P}}{C_{V}}$ or $C_{P} = γ C_{V}$

But $γ = 1.6$ and $C_{V} = 5.0$ $J$ $K^{- 1}$ ${mol}^{- 1}$

$∴$ $C_{P} = (1.6) (5.0 {J K}^{- 1} {mol}^{- 1}) = 8.0$ $J$ $K^{- 1} {mol}^{- 1}$

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