Animated Solution for Physics - Thermodynamics: Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? [Take, $R=8.3\text{ J/mol-K}$]

The problem of finding the specific heat of a gas mixture is a classic application of the Kinetic Theory of Gases and the principle of conservation of energy. Let's break down the physics and the math behind this beautiful concept.

Imagine a closed container where we are mixing two different gases. On one side, we have $2$ moles of Helium ($\text{He}$). Helium is a noble gas, meaning it exists as single, isolated atoms—it is monoatomic. On the other side, we have $3$ moles of Hydrogen (${\text{H}}_2$). Hydrogen atoms pair up to form molecules, making it a diatomic gas.

The problem explicitly states that the hydrogen molecules are "taken to be rigid." This is a crucial detail! A rigid diatomic molecule can translate in 3 directions and rotate in 2 independent axes, giving it exactly $5$ degrees of freedom ($f=5$). It does not vibrate. Helium, being monoatomic, only has translational kinetic energy, giving it $3$ degrees of freedom ($f=3$).

When we mix these gases, the total internal energy of the mixture is simply the sum of the internal energies of the individual gases. There is no energy lost or gained from the outside world.

We know that the internal energy $U$ of an ideal gas is given by $n{C}_{V}T$. If we consider a small change in temperature $\Delta T$ for the mixture, the energy equation becomes:

Canceling out $\Delta T$ from both sides, we get the standard weighted average formula for the molar specific heat of a mixture:

From the degrees of freedom, we can write the specific heats at constant volume ($C_V=\frac{f}{2}R$): Helium (Monoatomic): $C_{V_1}=\frac{3R}{2}$ Hydrogen (Diatomic, Rigid): $C_{V_2}=\frac{5R}{2}$

Now, we substitute these along with the number of moles ($n_1=2$, $n_2=3$) into our master equation:

Let's simplify the numerator carefully. The $2$ in the first term cancels out, leaving us with $3R$. The second term becomes $\frac{15R}{2}$. The denominator is simply $5$.

To add the terms in the numerator, we can write $3R$ as $\frac{6R}{2}$:

Finally, we substitute the value of the universal gas constant, $R=8.3\text{ J/mol-K}$:

And there we have it! The molar specific heat of the mixture at constant volume is $17.4\text{ J/mol-K}$. Always remember to check the atomicity and rigidity of the gases involved, as they completely dictate the degrees of freedom and the resulting specific heat.