$C_v$ and $C_p$ Are Independent of Volume and Pressure for an Ideal Gas: 21-Mat-A1 Thermodynamics, December 2018, Q3

Question 3: $C_v$ and $C_p$ Are Independent of Volume and Pressure for an Ideal Gas (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. For an ideal gas, $(\partial U/\partial V)_T=0$ and $(\partial H/\partial P)_T=0$.

Find. (a) $(\partial C_v/\partial V)_T=0$; (b) $(\partial C_p/\partial P)_T=0$.

(a) $C_v$ independent of $V$ at constant $T$. By definition $C_v=(\partial U/\partial T)_V$. Because $U(T,V)$ is a state function, its mixed second partial derivatives are equal regardless of the order of differentiation (Euler's reciprocity/Clairaut's theorem): $$\left(\frac{\partial C_v}{\partial V}\right)_T=\left[\frac{\partial}{\partial V}\left(\frac{\partial U}{\partial T}\right)_V\right]_T=\left[\frac{\partial}{\partial T}\left(\frac{\partial U}{\partial V}\right)_T\right]_V.$$ The given condition $(\partial U/\partial V)_T=0$ holds at every temperature for an ideal gas, so the quantity being differentiated with respect to $T$ on the right-hand side is identically zero for all $T$; its derivative with respect to $T$ is therefore also zero: $$\left(\frac{\partial C_v}{\partial V}\right)_T=\left[\frac{\partial}{\partial T}(0)\right]_V=\boxed{0}.$$ Hence $C_v$ does not depend on volume at constant temperature — it can only be a function of $T$ alone, $C_v=C_v(T)$.

(b) $C_p$ independent of $P$ at constant $T$. By definition $C_p=(\partial H/\partial T)_P$. By the same mixed-partial-derivative argument applied to $H(T,P)$: $$\left(\frac{\partial C_p}{\partial P}\right)_T=\left[\frac{\partial}{\partial P}\left(\frac{\partial H}{\partial T}\right)_P\right]_T=\left[\frac{\partial}{\partial T}\left(\frac{\partial H}{\partial P}\right)_T\right]_P.$$ Since $(\partial H/\partial P)_T=0$ for an ideal gas at every temperature, differentiating this identically-zero quantity with respect to $T$ gives $$\left(\frac{\partial C_p}{\partial P}\right)_T=\left[\frac{\partial}{\partial T}(0)\right]_P=\boxed{0}.$$ Hence $C_p$ does not depend on pressure at constant temperature — it can only be a function of $T$ alone, $C_p=C_p(T)$.

Result	Conclusion
(a) $(\partial C_v/\partial V)_T$	0 — proven, $C_v=C_v(T)$ only
(b) $(\partial C_p/\partial P)_T$	0 — proven, $C_p=C_p(T)$ only

Concept & Theory

Topic. Using the equality of mixed second partial derivatives of a state function to relate a heat capacity's pressure/volume dependence to the energy/enthalpy equation of state.

Key relations. $C_v=(\partial U/\partial T)_V$; $C_p=(\partial H/\partial T)_P$; for any state function $Z(x,y)$, $\partial^2Z/\partial x\partial y=\partial^2Z/\partial y\partial x$.

Why this works. $U$ and $H$ are exact differentials (state functions), so the order of partial differentiation never matters. That single fact converts a question about how a heat capacity varies with $V$ or $P$ into a question about how $(\partial U/\partial V)_T$ or $(\partial H/\partial P)_T$ varies with $T$ — and for an ideal gas both of those are identically zero at every temperature, so their $T$-derivatives vanish too.

Common pitfall. Trying to prove the result by assuming $C_v$ or $C_p$ is constant from the outset — the proof must start only from the two given zero-derivative conditions and the reciprocity of mixed partials; assuming the conclusion (constant $C_v$/$C_p$) to prove itself is circular.

Source. Gaskell, Introduction to the Thermodynamics of Materials, Ch. 3 (energy equation, exact differentials, ideal-gas internal energy and enthalpy).

Question 3 of 7: $C_v$ and $C_p$ Are Independent of Volume and Pressure for an Ideal Gas

Question 3: $C_v$ and $C_p$ Are Independent of Volume and Pressure for an Ideal Gas (20 marks)