A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas $\gamma =\frac{5}{3}$ and one mole of an ideal diatomic gas $\gamma =\frac{7}{5}$. The gas mixture does a work of 66 Joule when heated at constant pressure. The change in its internal energy is .

Correct Answer: The change in internal energy of the mixture is 121 J (or 121J).

Let us solve the problem step-by-step using fundamental concepts of thermodynamics.

Step 1: Identify the given information
- Number of moles of ideal monatomic gas ($n_1$): 2 moles
- Ratio of specific heats for monatomic gas (${\gamma }_1$): $\frac{5}{3}$
- Degrees of freedom for monatomic gas ($f_1$): 3
- Number of moles of ideal diatomic gas ($n_2$): 1 mole
- Ratio of specific heats for diatomic gas (${\gamma }_2$): $\frac{7}{5}$
- Degrees of freedom for diatomic gas ($f_2$): 5
- Work done by the gas mixture at constant pressure ($W$): 66 J

Step 2: Express work done at constant pressure
For a mixture of ideal gases undergoing an isobaric (constant pressure) expansion, the work done $W$ is given by:

$W=P\Delta V=(n_1+n_2)R\Delta T$

Substituting the given values of $n_1=2$ and $n_2=1$:

$66=(2+1)R\Delta T=3R\Delta T$

$R\Delta T=\frac{66}{3}=22\text{J}$

Step 3: Calculate the change in internal energy ($\Delta U$)
The change in internal energy of a mixture of ideal gases depends on the individual degrees of freedom and change in temperature:

$\Delta U=\Delta U_1+\Delta U_2$

$\Delta U=n_1\frac{f_1}{2}R\Delta T+n_2\frac{f_2}{2}R\Delta T$

Substituting the known values into the equation:

$\Delta U=[2\times \frac{3}{2}+1\times \frac{5}{2}]R\Delta T$

$\Delta U=[3+2.5]R\Delta T=5.5R\Delta T=\frac{11}{2}R\Delta T$

Now, substitute $R\Delta T=22\text{J}$:

$\Delta U=\frac{11}{2}\times 22=11\times 11=121\text{J}$

Thus, the change in internal energy of the mixture is 121 J.