JEE Advance - Physics (2023 - Paper 1 Online - No. 11)
A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas $(\gamma=5 / 3)$ and one mole of an ideal diatomic gas $(\gamma=7 / 5)$. Here, $\gamma$ is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of 66 Joule when heated at constant pressure. The change in its internal energy is _______ Joule.
Answer
121
Explanation
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.
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.
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