QJEEAdv23Ph0304: JEE Advanced 2023 Physics - Class 11 JEEAdv23Ph0304 Solution: A closed container contains a homogeneous mixture of two moles…

First, we identify the heat capacities at constant volume for monatomic and diatomic gases.

For a monatomic ideal gas, the heat capacity at constant volume, $C_{v1}$ , is $\left(\frac{3}{2}\right)R$ . This comes from the degrees of freedom for a monatomic gas, which are three (x, y, and z motions).

For a diatomic ideal gas, the heat capacity at constant volume, $C_{v2}$ , is $\left(\frac{5}{2}\right)R$ . This comes from the degrees of freedom for a diatomic gas, which are five (x, y, and z motions and rotation about two axes).

Next, we find the average heat capacity at constant volume for the gas mixture, $C_{v_{\text{mix}}}$ , which is a weighted average based on the number of moles of each gas :

$C_{v_{\text{mix}}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2 \times \left(\frac{3}{2}\right)R + 1 \times \left(\frac{5}{2}\right)R}{2 + 1} = \frac{11R}{6}$

The change in internal energy $\Delta U$ for a given number of moles $n$ and change in temperature $\Delta T$ is given by :

$\Delta U = n C_{v} \Delta T$

Given that the work done by the system at constant pressure is :

$W = n R \Delta T$

We can replace $\Delta T = \frac{W}{nR}$ in the equation for $\Delta U$ to get :

$\Delta U = n C_{v_{\text{mix}}} \times \frac{W}{n R} = C_{v_{\text{mix}}} \times \frac{W}{ R} = \frac{11R}{6} \times \frac{66}{ R} = 121 \, \text{Joule}$

So the change in internal energy is indeed 121 Joule.