`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^(-1)`. If the speed of sound in this gas at NTP is `952 ms^(-1)`. Then the heat capacity at constant pressure is

Options

A. `7.0 JK^(-1)mol^(-1)`
B. `8.5 JK^(-1)mol^(-1)`
C. `8.0 JK^(-1)mol^(-1)`
D. `7.5 JK^(-1)mol^(-1)`

Explanation

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 the sound in the gas is
`v=sqrt((gammaRT)/(M))`
where `gamma` is the ratio of two specific heats, R is the universal gas constant and T is the temperature of the gas.
`therefore gamma=(Mv^(2))/(RT)`
Here, `M=4.0g mol^(-1)=4.0xx10^(-3)kgmol^(-1)`
`v=952ms^(-1),R=8.3JK^(-1)mol^(-1)`
and `T=273K` (at NTP)
`therefore gamma=((4.0xx10^(-3)kgmol^(-1))(952ms^(-1))^(2))/((8.3JK^(-1)mol^(-1))(273K))=1.6`
By definition, `gamma=(C_(p))/(C_(v))orC_(p)=gammaC_(v)`
But `gamma=1.6andC_(v)=5.0JK^(-1)mol^(-1)`
`therefore C_(p)=(1.6)(5.0JK^(-1)mol^(-1))=8.0JK^(-1)mol^(-1)`