Three moles of an inert monoatomic gas is mixed with two moles of a diatomic gas and the mixture is

Question

Three moles of an inert monoatomic gas is mixed with two moles of a diatomic gas and the mixture is heated by a process $V=\frac{k}{\sqrt{T}}$. Where $V$ and $T$ are volume and temeprature of the mixture and $k$ is a constant. The molar heat capacity of this process for this mixture is.....

Professor PC Thomas Classes · Answer

$\begin{aligned} & \mathrm{V}=\frac{\mathrm{K}}{\sqrt{\mathrm{T}}}, \mathrm{T}=\frac{\mathrm{PV}}{\mathrm{nR}} 	ext { is put then } \mathrm{PV}^3=	ext { constant }\left(	ext { say } \mathrm{k}_1ight) \ & 	herefore \Delta \mathrm{W}=\mathrm{PdV}=\int \frac{\mathrm{k}_1}{\mathrm{v}^3} \mathrm{dv}=\frac{-\mathrm{nR} \Delta \mathrm{T}}{2}=\frac{-5 \mathrm{R} \Delta \mathrm{T}}{2} \ & \Delta U=n C v \Delta T=\left(n_1 C v_1+n_2 C v_2ight) \Delta T=\left(3 	imes \frac{3 R}{2}+2 	imes \frac{5 R}{2}ight) \Delta T=\frac{19}{2} R \Delta T \ & \Delta Q=\Delta U+\Delta W=\frac{14 R \Delta T}{2}=7 R \Delta T \ & C_{	ext {(process) }}=\frac{\Delta Q}{n \Delta T}=\frac{7 R \Delta T}{5 \Delta T}=\frac{7 R}{5}\end{aligned}$