Animated Solution for Physics - Thermodynamics: Two moles of an ideal gas with $\frac{C_p}{C_V}=\frac{5}{3}$ are mixed with 3 mol of another ideal gas with $\frac{C_p}{C_V}=\frac{4}{3}$. The value of $\frac{C_p}{C_V}$ for the mixture is

Calculating the Effective Gamma

Imagine you are in a laboratory, holding two separate containers of ideal gases. Gas 1 is a monatomic gas (like Helium) with $2$ moles and an adiabatic index ${\gamma }_1=\frac{5}{3}$. Gas 2 is a polyatomic gas with $3$ moles and an adiabatic index ${\gamma }_2=\frac{4}{3}$.

Now, you open a valve and let them mix into a single, larger container. The question is: what is the effective adiabatic index (${\gamma }_{mix}$) of this new, blended gas?

Why We Can't Just Average Gammas

It is incredibly tempting to just take a weighted average of the gammas: $\frac{n_1{\gamma }_1+n_2{\gamma }_2}{n_1+n_2}$. Do not fall for this trap!

Gamma ($\gamma$) is defined as the ratio of two specific heats: $\gamma =\frac{C_p}{C_V}$. It is an intensive property, much like temperature or density. When you mix two gases, their intensive properties do not simply add up. Instead, it is their extensive properties—like total internal energy and total heat capacity—that are strictly additive.

To find the true ${\gamma }_{mix}$, we must first find the total molar heat capacity at constant pressure ($C_{p(mix)}$) and the total molar heat capacity at constant volume ($C_{V(mix)}$) for the entire mixture.

The heat capacity of a mixture is the weighted average of the individual heat capacities:

When we take the ratio to find ${\gamma }_{mix}$, the total moles $(n_1+n_2)$ in the denominators beautifully cancel out:

We aren't given $C_p$ or $C_V$ directly, but we have Mayer's relation ($C_p-C_V=R$) and the definition of gamma ($\gamma =\frac{C_p}{C_V}$). By combining these, we can express the heat capacities purely in terms of $\gamma$ and the universal gas constant $R$:

Substituting these into our mixture ratio, we get a powerful, universal formula. Notice how the gas constant $R$ appears in every single term and completely factors out:

Now, we breathe, stay calm, and substitute our given values: $n_1=2$, ${\gamma }_1=\frac{5}{3}$, $n_2=3$, and ${\gamma }_2=\frac{4}{3}$.

Let's simplify the denominators inside the brackets first. $\frac{5}{3}-1=\frac{2}{3}$, and $\frac{4}{3}-1=\frac{1}{3}$.

The fractions simplify elegantly as the $3$s cancel out:

Evaluating this fraction gives us our final answer:

And there you have it! By respecting the fundamental laws of thermodynamics and averaging the extensive properties rather than the intensive ones, we arrive at the perfect solution.