Physics - Heat & Thermo Question with Solution | TestHub

In the curve, $V\propto \frac{1}{T}$ $$\begin{array}{rl}& \Rightarrow V=m\frac{1}{T}\text{ where, }m=\text{ slope of curve. }\\ & \Rightarrow VT=\text{ constant }\\ & \therefore TdV+VdT=0\\ & \Rightarrow dV=-\frac{V}{T}dT\\ & \text{ As, }Q=\Delta U+W\\ & CdT=C_{V}dT+pdV\\ & CdT=C_{V}dT-\frac{p}{T}VdT\\ & \Rightarrow C=C_V-\frac{pV}{T}\\ & =C_V-R\left(\because \frac{pV}{T}=R\right)\\ & \end{array}$$ For diatomic gas, $C_V=\frac{5}{2}R$ $$\therefore C=C_V-R=\frac{5}{2}R-R=\frac{3R}{2}$$