The molar heat capacity of a gas at constant volume is found to be `5 cal mol^(-1) K^(-1)`. Find the ratio `gamma = C_p/C_v` for the gas. The gas constant `R=2 cal mol^(-1) K^(-1)`.

Explanation

Using `pV = nRT`, the volume of 1 mole of air at STP is
V=nRT/p = (1mol) xx 8.3 J K^(-1) xx (273 K) / 1.0 xx 10^(5) N m^(-2)
= 0.0224 m^(3)
The mass of 1 mole is, therefore, (1.29 kg m^(-3) xx (0.0224 m^(3)) = 0.029 kg.
The number of moles in 1 kg is `1/0.029`. The molar heat capacity at constant volume is
`Cv = 170 cal / (1/0.029) mol K^(-1)`
= 4.93 cal K^(-1) mol^(-1)`.
Hence, Cp = gamma Cv = 1.4 xx 4.93 cal K^(-1) mol^(-1)
or, Cp - Cv =0.4 xx 4.93 cal K^(-1) mol^(-1)`.
Also Cp - Cv = R = 8.3 J K^(-1)mol^(-1).
Thus ` 8.3 J = 1.97 cal.`
The mechanical equivalent of heat is
`8.3 J/ 1.97 cal = 4.2 J cal^(-1).