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.

For an ideal gas kept at constant pressure, the work done is related to the temperature rise as
$$W = P\Delta V = nR\Delta T$$

The mixture contains
- 2 moles of monatomic gas with $$\gamma_1 = \dfrac{5}{3}$$,
- 1 mole of diatomic gas with $$\gamma_2 = \dfrac{7}{5}$$.
Total moles
$$n = 2 + 1 = 3$$

Given work done
$$W = 66 \text{ J}$$, hence from the relation above
$$\Delta T = \frac{W}{nR} = \frac{66}{3R} = \frac{22}{R}$$ $$-(1)$$

For each component we first evaluate its molar specific heat at constant volume $$C_v$$.
The relation between $$\gamma$$, $$C_p$$ and $$C_v$$ is
$$\gamma = \frac{C_p}{C_v}, \qquad C_p - C_v = R$$ $$-(2)$$

Monatomic gas: $$\gamma_1 = \dfrac{5}{3}$$.
Using $$(2)$$,
$$C_{v1} = \frac{R}{\gamma_1 - 1} = \frac{R}{\frac{5}{3} - 1} = \frac{R}{\frac{2}{3}} = \frac{3}{2}R$$

Diatomic gas: $$\gamma_2 = \dfrac{7}{5}$$.
Similarly,
$$C_{v2} = \frac{R}{\gamma_2 - 1} = \frac{R}{\frac{7}{5} - 1} = \frac{R}{\frac{2}{5}} = \frac{5}{2}R$$

The change in internal energy of the mixture is the sum over all moles:
$$\Delta U = n_1 C_{v1}\Delta T + n_2 C_{v2}\Delta T$$
$$\Delta U = 2\left(\frac{3}{2}R\right)\Delta T \;+\, 1\left(\frac{5}{2}R\right)\Delta T$$
$$\Delta U = \left(3R + \frac{5}{2}R\right)\Delta T = \frac{11}{2}R\Delta T$$ $$-(3)$$

Substitute $$\Delta T$$ from $$(1)$$ into $$(3)$$:
$$\Delta U = \frac{11}{2}R \left(\frac{22}{R}\right) = \frac{11 \times 22}{2} = 121 \text{ J}$$

Hence, the change in internal energy of the gas mixture is 121 Joule.