Let, CV= molar heat capacity of the first gas, CV′′= molar heat capacity of the second gas, CV= molar heat capacity of the mixture, and similar symbols for other quantities. Then γ=
CP′
CV′
=1.67 and C′P=CV′+R This gives CV′=3∕2R and CP′=5∕2R Similarly, γ=1.4 gives CV′′=5∕2R and CP′′=7∕2R Suppose the temperature of the mixture is increased by dT. The increase in the internal energy of the first gas =n1CV 'dT. The increase in internal energy of the second gas =n2CV′′dT Thus, (n1+n2)CVdT=n1CV⋅dT+n2CV′′dT CV=