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Physics Question on 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 JK−1 mol−1JK^{-1} \, mol^{-1}. If the speed of sound in this gas at NTP is 952 ms−1ms^{-1}, then the heat capacity at constant pressure is (Take gas constant R = 8.3 JK−1 mol−1JK^{-1} \, mol^{-1} )

A

7.0 JK−1 mol−17.0 \, JK^{-1} \, mol^{-1}

B

8.5 JK−1 mol−18.5 \, JK^{-1} \, mol^{-1}

C

8.0 JK−1 mol−18.0 \, JK^{-1} \, mol^{-1}

D

7.5 JK−1 mol−17.5 \, JK^{-1} \, mol^{-1}

Answer

8.0 JK−1 mol−18.0 \, JK^{-1} \, mol^{-1}

Explanation

Solution

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−1M = 4.0 \, g\, mol^{-1}
As the speed of the sound in the gas is
v=?RTMv=\sqrt{\frac{?RT}{M}}
where ?? is the ratio of two specific heats, R is the universal gas constant and T is the temperature of the gas.
∴ γ=Mv2RT\therefore \, \gamma = \frac{Mv^2}{RT}
Here,    M=4.0 g mol−1=4.0×10−3kg mol−1 \, \, \, M = 4.0 \, g \, mol^{-1} = 4.0 \times 10^{-3}kg\, mol^{-1}
v=952 ms−1,R=8.3 JK−1 mol−1v=952 \, ms^{-1}, R = 8.3\, JK^{-1}\, mol^{-1},
v=952 mol−1,R=8.3 JK−1 mol−1v= 952\, mol^{-1}, R = 8.3\,JK^{-1}\, mol^{-1}
and T = 273 K (at NTP)
∴γ=(4.0×10−3 kg mol−1)(952 ms−1)2(8.3JK−1mol−1)(273 K)=1.6\therefore \gamma= \frac{(4.0 \times 10^{-3} \,kg\, mol^{-1})(952\,ms^{-1})^2}{(8.3 JK^{-1} mol^{-1})(273\,K)}=1.6
By definition,
γ=CPVV\gamma=\frac{C_P}{V_V} or CP=γCVC_P = \gamma C_V
But γ=1.6\gamma=1.6 and CV=5.0 JK−1 mol−1C_V=5.0 \,JK^{-1} \, mol^{-1}
∴  CP=(1.6)(5.0JK1 mol−1)\therefore \, \, C_P = (1.6)(5.0 JK^{1} \, mol^{-1})
    =8.0 JK−1mol−1\, \, \, \, =8.0 \,JK^{-1} mol^{-1}