1. Use the ideal-gas relation at constant pressure

For any ideal gas mixture heated at constant pressure, $W = P\Delta V = nR\Delta T$ where $n$ is the total number of moles.

Here,

• monatomic gas: $2$ moles
• diatomic gas: $1$ mole

So total moles, $n=2+1=3$

Given work done, $W=66\text{ J}$ Hence, $nR\Delta T = 66$ $3R\Delta T = 66$

2. Find total heat capacity at constant volume of the mixture

For an ideal gas, $\gamma = \frac{C_p}{C_v}, \qquad C_p-C_v=R$ Thus, $C_v = \frac{R}{\gamma-1}$

(a) Monatomic gas

Given $\gamma=\frac{5}{3}$, $C_{v,m} = \frac{R}{\frac{5}{3}-1} = \frac{R}{\frac{2}{3}} = \frac{3R}{2}$ For $2$ moles, $C_{V,\text{mono total}} = 2\cdot \frac{3R}{2} = 3R$

(b) Diatomic gas

Given $\gamma=\frac{7}{5}$, $C_{v,d} = \frac{R}{\frac{7}{5}-1} = \frac{R}{\frac{2}{5}} = \frac{5R}{2}$ For $1$ mole, $C_{V,\text{dia total}} = \frac{5R}{2}$

Therefore total heat capacity at constant volume of mixture is $C_{V,\text{total}} = 3R + \frac{5R}{2} = \frac{11R}{2}$

3. Compute change in internal energy

For an ideal gas mixture, $\Delta U = C_{V,\text{total}}\Delta T$ So, $\Delta U = \frac{11R}{2}\Delta T$

From step 1, $3R\Delta T=66 \implies R\Delta T=22$

Hence, $\Delta U = \frac{11}{2}(R\Delta T)=\frac{11}{2}\times 22=121\text{ J}$

4. Final answer

$\boxed{121\text{ J}}$