Animated Solution for Physics - Thermodynamics: If one mole of a monoatomic gas ($\gamma =5/3$) is mixed with one mole of a diatomic gas ($\gamma =7/5$), the value of $\gamma$ for the mixture is

When dealing with a mixture of gases, one of the most common mistakes students make is trying to find the adiabatic exponent ($\gamma$) of the mixture by simply taking the average of the individual $\gamma$ values. This is fundamentally incorrect because $\gamma$ is a ratio of heat capacities ($C_P/C_V$), and mathematically, the average of ratios is not equal to the ratio of averages.

To find the correct $\gamma$ for the mixture, we must first determine the total molar heat capacity at constant volume ($C_V$) and the total molar heat capacity at constant pressure ($C_P$) for the entire mixture.

Let's write down what we are given: - Gas 1 is monoatomic: $n_1=1\text{ mole}$, ${\gamma }_1=\frac{5}{3}$ - Gas 2 is diatomic: $n_2=1\text{ mole}$, ${\gamma }_2=\frac{7}{5}$

We know that the molar heat capacity at constant volume is related to $\gamma$ by the formula $C_V=\frac{R}{\gamma -1}$. Using this, we can find the individual heat capacities:

For the monoatomic gas:

For the diatomic gas:

The molar heat capacity at constant volume for a mixture of non-reacting ideal gases is simply the weighted average of their individual heat capacities:

Let's substitute our values into this master equation:

Adding the numerators gives $\frac{8}{2}R=4R$. Dividing by the total number of moles ($2$) gives:

Now that we have $C_{V_{\text{mix}}}$, finding $C_{P_{extmix}}$ is a breeze thanks to Mayer's relation, which holds perfectly true for ideal gas mixtures:

Finally, the adiabatic exponent for the mixture is the ratio of its total $C_P$ to its total $C_V$:

Substituting the values we just calculated:

And there we have it! The $\gamma$ for the mixture is exactly 1.50.