Suppose I have $1$ mole of a monoatomic gas and $1$ mole of a diatomic gas. If I mix them, the ratio of their specific heats at constant pressure to that constant volume becomes:$$\gamma = \frac{3}{2}$$
I came up with this result be averaging $C_p$ and $C_v$ of both the gases over $1+1=2$ moles. My question is why this works? Why averaging out gives the right solution? How will I go about calculating the $\gamma$ of mixture of gases if there are more than $1$ mole in each gaseous system?
Best Answer
The change in internal energy and enthalpy of mixing ideal gases is zero. According to Gibbs' Theorem, the individual contribution of each species in an ideal gas mixture to the extensive thermodynamic properties of the mixture is the same as that of the pure species at the same temperature and at the partial pressure of the species in the mixture.