21-Mat-A1 Thermodynamics · May 2014
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — May 2014 — 10-Met-A1 Metallurgical Thermodynamics. Three-hour, closed-book exam using an approved (Casio or Sharp) calculator; candidates were told to state any interpretive assumptions. Any five of the seven questions constitute a complete paper — all seven are solved below for completeness. All questions are of equal value (20 marks each out of 100).
Reference texts: Gaskell, D. R., Introduction to the Thermodynamics of Materials (2nd–5th ed.) — the exam's own Ellingham diagram (Fig. 9-3) is reproduced from this text, which also supplies the standard-state enthalpy/entropy data used below; supporting 298 K entropies for the elements, CO, H₂, H₂O(g) and CO₂ from the NIST-JANAF Thermochemical Tables.
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. An ideal gas for which $(\partial U/\partial V)_T=0$ and $(\partial H/\partial P)_T=0$ (Joule's law and its enthalpy analogue).
Find. Show $(\partial C_v/\partial V)_T=0$ and $(\partial C_p/\partial P)_T=0$.
Approach. Both proofs use the same idea: a heat capacity is itself a partial derivative of a state function, and for any state function the order in which two partial derivatives are taken does not matter (the mixed second derivatives are equal). Differentiating the definition of $C_v$ with respect to $V$, and swapping the order of differentiation, turns the question directly into the derivative of the given quantity $(\partial U/\partial V)_T$ — which is zero.
(a) By definition $C_v=(\partial U/\partial T)_V$. Differentiating $C_v$ with respect to $V$ at constant $T$, $$\left(\frac{\partial C_v}{\partial V}\right)_T=\frac{\partial}{\partial V}\left[\left(\frac{\partial U}{\partial T}\right)_V\right]_T=\frac{\partial}{\partial T}\left[\left(\frac{\partial U}{\partial V}\right)_T\right]_V,$$ where the last step exchanges the order of differentiation — valid because $U$ is a state function with a continuous, exact differential. Substituting the given condition $(\partial U/\partial V)_T=0$ for every $T$ makes the right-hand side the $T$-derivative of a function that is identically zero: $$\left(\frac{\partial C_v}{\partial V}\right)_T=\frac{\partial}{\partial T}(0)=\boxed{0}.$$ Hence $C_v$ does not depend on $V$ at constant $T$.
(b) The argument mirrors (a) with $H$ and $P$ in place of $U$ and $V$. By definition $C_p=(\partial H/\partial T)_P$, so $$\left(\frac{\partial C_p}{\partial P}\right)_T=\frac{\partial}{\partial P}\left[\left(\frac{\partial H}{\partial T}\right)_P\right]_T=\frac{\partial}{\partial T}\left[\left(\frac{\partial H}{\partial P}\right)_T\right]_P=\frac{\partial}{\partial T}(0)=\boxed{0},$$ using the given $(\partial H/\partial P)_T=0$ and the same mixed-partial exchange. Hence $C_p$ does not depend on $P$ at constant $T$.
| Quantity | Result |
|---|---|
| (a) $(\partial C_v/\partial V)_T$ | 0 — $C_v$ independent of $V$ |
| (b) $(\partial C_p/\partial P)_T$ | 0 — $C_p$ independent of $P$ |
