A closed container contains a homogeneous mixture of two moles of...

Change in internal energy for the mixture can be written as,$∆ U = {n}_{1}{\left({C}_{v}\right)}_{1} ∆ T + {n}_{2}{\left({C}_{v}\right)}_{2} ∆ T$

= \left({n}_{1}{\left({C}_{v}\right)}_{1} + {n}_{2}{\left({C}_{v}\right)}_{2}\right) ∆ T . . . $\left(i\right)$

For isobaric process, work done $= P ∆ V$

$= \left({n}_{1} + {n}_{2}\right)R ∆ T . . . \left(ii\right)$

Divide \left(i\right)$by \left(ii\right)$, we get

$\dfrac{∆ U}{W} = \dfrac{\left({n}_{1}{\left({C}_{v}\right)}_{1} + {n}_{2}{\left({C}_{v}\right)}_{2}\right) ∆ T}{\left({n}_{1} + {n}_{2}\right)R ∆ T}$

$⇒ ∆ U = \dfrac{W}{R}\left(\dfrac{\left({n}_{1}{\left({C}_{v}\right)}_{1} + {n}_{2}{\left({C}_{v}\right)}_{2}\right)}{{n}_{1} + {n}_{2}}\right)$

$= \dfrac{66}{R}\dfrac{\left[2 \times \dfrac{3R}{2} + 1 \times \dfrac{5R}{2}\right]}{2 + 1}$

$= 121J$