Question 1: Thermodynamics & KTG

Question

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.

Accepted Answer

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 $\left(\frac{3}{2}ight)R$. 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 $\left(\frac{5}{2}ight)R$. 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, $C_{v_{	ext{mix}}}$, which is a weighted average based on the number of moles of each gas :
$C_{v_{	ext{mix}}} = \frac{n_1 C_{v1} + n_2 C_{v2}}{n_1 + n_2} = \frac{2 	imes \left(\frac{3}{2}ight)R + 1 	imes \left(\frac{5}{2}ight)R}{2 + 1} = \frac{11R}{6}$
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 :
$\Delta U = n C_{v_{	ext{mix}}} 	imes \frac{W}{n R} = C_{v_{	ext{mix}}} 	imes \frac{W}{ R} = \frac{11R}{6} 	imes \frac{66}{ R} = 121 \, 	ext{Joule}$
So the change in internal energy is indeed 121 Joule.

Thermodynamics & KTG

Integer / Numeric Answer Type