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(D) For a mixture of gases at constant pressure,the work done is given by $W = n_{mix} R \Delta T = 66 \ J$.The molar heat capacity at constant volume

Vedclass · Accepted Answer

$121$

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(D) For a mixture of gases at constant pressure,the work done is given by $W = n_{mix} R \Delta T = 66 \ J$.The molar heat capacity at constant volume for the mixture is given by $(C_V)_{mix} = \frac{n_1 C_{V1} + n_2 C_{V2}}{n_1 + n_2}$.For a monatomic gas,$C_{V1} = \frac{3}{2} R$. For a diatomic gas,$C_{V2} = \frac{5}{2} R$.Substituting the values: $(C_V)_{mix} = \frac{2 	imes (3/2)R + 1 	imes (5/2)R}{2 + 1} = \frac{3R + 2.5R}{3} = \frac{5.5R}{3} = \frac{11}{6} R$.The change in internal energy is $\Delta U = n_{total} (C_V)_{mix} \Delta T$.Since $n_{total} = n_1 + n_2 = 3$,we have $\Delta U = 3 	imes (\frac{11}{6} R) \Delta T = \frac{11}{2} R \Delta T$.From the work equation,$n_{total} R \Delta T = 66$,so $3 R \Delta T = 66$,which means $R \Delta T = 22$.Substituting this into the internal energy equation: $\Delta U = \frac{11}{2} 	imes 22 = 11 	imes 11 = 121 \ J$.

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