Using `pV = nRT`, the volume of `1` mole of air at `STP` is
`V = (nRT)/(p) = ((1mol)xx(8.3J//mol-K)xx(273K))/(1.01xx10^(5)N//m^(2))= 0.024 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
`C_(v) = (170 cal)/((1//0.029)mol-K) = 4.94 cal//mol-K`.
Hence, `C_(p) = gamma C_(v) = 1.4 xx 4.93 cal//mol-K`
or, `C_(p) - C_(v) = 0.4 xx 4.93 cal//mol-K`
`=1.97 cal//mol-K`.
Also, `C_(p) - C_(v) = R = 8.3 J//mol-K`.
Thus, `8.3J = 1.97 cal`.
The mechanical equivalent of heat is
`(8.3J)/(1.97 cal) = 4.2 J//cal`.