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 m s ⁻¹, then the heat capacity at constant pressure is (Take gas constant R = 8.3 J K ⁻¹ mol ⁻¹) (a) 7.0 J K ⁻¹ mol ⁻¹ (b) 8.5 J K ⁻¹ mol ⁻¹ (c) 8.0 J K ⁻¹ mol ⁻¹ (d) 7.5 J K ⁻¹ mol ⁻¹

Answer: (c). 22.4 L at NTP is 1 mol, so M = 4 g/mol = 4×10⁻³ kg/mol. γ = (v² M)/(RT) = 952² × 4×10⁻³8.3 × 273 ≈ 1.6 Cp = γ Cv = 1.6 × 5.0 = 8.0 J K⁻¹mol⁻¹ So the answer is (c).

4.0 g of a gas occupies 22.4 litres at NTP. The specific heat… | NEET 2015 (AIPMT)

Q37NEET 2015 · AIPMTMedium

Speed of Waves

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 m s ${}^{-1}$ , then the heat capacity at constant pressure is

(Take gas constant R $=$ 8.3 J K ${}^{-1}$ mol ${}^{-1}$ )

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Answer: (c)

22.4 L at NTP is 1 mol, so $M = 4$ g/mol $= 4\times10^{-3}$ kg/mol.
$\gamma = \dfrac{v^2 M}{RT} = \dfrac{952^2 \times 4\times10^{-3}}{8.3 \times 273} \approx 1.6$
$C_p = \gamma C_v = 1.6 \times 5.0 = 8.0\ \text{J K}^{-1}\text{mol}^{-1}$
So the answer is (c).

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