Topic: Thermodynamics
The state of a real gas if changed from pressure $P_{1}$, temperature $T_{1}$ to pressure $P_{2}$ temperature $T_{2}.$ The change in enthalpy, $h_{2}-h_{1},$ is given by
- A. $\int_{T_{1}}^{T_{2}}C_{p}dT$
- B. $\int_{T_{1}}^{T_{2}}C_{p}dT+\int_{T_{1}}^{T_{2}}(\frac{\partial V}{\partial p})_{T}dp$
- C. $\int_{T_{1}}^{T_{2}}C_{v}dT+\oint_{P_{1}}^{P_{2}}[V-T(\frac{\partial V}{\partial p})_{p}]dp$
- D. $\int_{T_{1}}^{T_{2}}C_{p}dT+\oint_{P_{1}}^{P_{2}}[V-T(\frac{dV}{dT})_{p}]dp$
Answer: C. $\int_{T_{1}}^{T_{2}}C_{v}dT+\oint_{P_{1}}^{P_{2}}[V-T(\frac{\partial V}{\partial p})_{p}]dp$
Explanation: From fundamental thermodynamic potential equation, $dH=Tds+Vdp$. Expression of entropy as a function of P (Pressure) and T (Temperature), $Tds=T(\frac{\partial s}{\partial P})_{T}dp+T(\frac{\partial s}{\partial T})_{p}dT$. From a Maxwell relation, $(\frac{\partial s}{\partial P})_{T}=-(\frac{\partial v}{\partial T})_{P}$. Using $T(\frac{\partial s}{\partial T})_{p}=(\frac{\partial H}{\partial T})_{p}=C_{p}$, substituting yields the change in enthalpy. (Note: The provided solution in the text points to option (c) structurally as printed).