Exercises — Maxwell relations — derivation from each potential

WHAT we look for: a relation with $S$ differentiated with respect to $P$ at constant $T$ on one side. Scan the toolkit — that is the fourth relation: ${\left(\frac{\partial S}{\partial P}\right)}_T=-{\left(\frac{\partial V}{\partial T}\right)}_P$ WHY this one: the left side matches exactly. It comes from Gibbs free energy $G$, whose natural variables are $(T,P)$ — precisely the two variables in play here ($T$ held, $P$ varied). WHY it's useful: the right side is $-V\alpha$, which you measure with a thermometer and a ruler. So ${\left(\frac{\partial S}{\partial P}\right)}_T=-V\alpha$.

WHAT: find where $T$ is differentiated by $V$. That is the first relation (from $U$): ${\left(\frac{\partial T}{\partial V}\right)}_S=-{\left(\frac{\partial P}{\partial S}\right)}_V$ WHY the subscript is $S$: $U$'s natural variables are $(S,V)$. When we differentiate with respect to $V$, the other natural variable, $S$, is the one held fixed. Dropping or changing that subscript makes the relation false — see the mistake box below.

Step 1 — pick the relation (WHY): we need $\partial S/\partial V$ at constant $T$. The Helmholtz relation gives it in measurable terms: ${\left(\frac{\partial S}{\partial V}\right)}_T={\left(\frac{\partial P}{\partial T}\right)}_V$ Step 2 — evaluate the RHS (WHAT): from $P=RT/V$, differentiate with respect to $T$ holding $V$ fixed: ${\left(\frac{\partial P}{\partial T}\right)}_V=\frac{R}{V}$ So ${\left(\frac{\partial S}{\partial V}\right)}_T=\frac{R}{V}$. Step 3 — integrate at fixed $T$: $S=\int \frac{R}{V}dV=R\ln V+\text{const}(T)$ Sanity check: this is exactly the volume-dependence of the Sackur–Tetrode entropy — the famous $R\ln V$ term — recovered from a Maxwell relation alone.

Step 1 (WHY): the Gibbs relation ${\left(\partial S/\partial P\right)}_T=-{\left(\partial V/\partial T\right)}_P$. By the definition of $\alpha$, ${\left(\partial V/\partial T\right)}_P=V\alpha$, so immediately ${\left(\frac{\partial S}{\partial P}\right)}_T=-V\alpha .$ Step 2 — ideal gas $\alpha$: with $V=RT/P$, ${\left(\partial V/\partial T\right)}_P=R/P$, so $\alpha =\frac{1}{V}\cdot \frac{R}{P}=\frac{1}{T}$ (since $PV=RT$). For an ideal gas $\alpha =1/T$. Step 3 — number: ${\left(\frac{\partial S}{\partial P}\right)}_T=-V\alpha =-\frac{RT}{P}\cdot \frac{1}{T}=-\frac{R}{P}$. $=-\frac{8.314}{1.00\times 10^5}=-8.314\times 10^{-5}{\text{J K}}^{-1}{\text{Pa}}^{-1}.$ Meaning: raising the pressure lowers the entropy (squeezing orders the gas) — the sign is negative, exactly as intuition demands.

Step 1 — start from entropy (WHY): heat capacities are $C_V=T{\left(\partial S/\partial T\right)}_V$ and $C_P=T{\left(\partial S/\partial T\right)}_P$. We want their difference, so express $S$ once as $S(T,V)$ and compare. Step 2 — expand $S(T,V)$: $dS={\left(\frac{\partial S}{\partial T}\right)}_{V}dT+{\left(\frac{\partial S}{\partial V}\right)}_{T}dV.$ Divide by $dT$ at constant $P$: ${\left(\frac{\partial S}{\partial T}\right)}_P={\left(\frac{\partial S}{\partial T}\right)}_V+{\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_P.$ Step 3 — multiply by $T$ to turn each $\partial S/\partial T$ into a heat capacity: $C_P=C_V+T{\left(\frac{\partial S}{\partial V}\right)}_T{\left(\frac{\partial V}{\partial T}\right)}_P.$ Step 4 — apply the Helmholtz Maxwell relation ${\left(\partial S/\partial V\right)}_T={\left(\partial P/\partial T\right)}_V$: $C_P-C_V=T{\left(\frac{\partial P}{\partial T}\right)}_V{\left(\frac{\partial V}{\partial T}\right)}_P.$ Step 5 — kill the unmeasurable ${\left(\partial P/\partial T\right)}_V$ with the cyclic (triple-product) rule. For any three variables $P,V,T$ tied by an equation of state, the general identity is ${\left(\frac{\partial P}{\partial T}\right)}_V{\left(\frac{\partial T}{\partial V}\right)}_P{\left(\frac{\partial V}{\partial P}\right)}_T=-1.$ Solve this for the piece we want by moving the other two factors to the right (note that ${\left(\partial T/\partial V\right)}_P=1/{\left(\partial V/\partial T\right)}_P$, the reciprocal): ${\left(\frac{\partial P}{\partial T}\right)}_V=\frac{-1}{{\left(\partial T/\partial V\right)}_P{\left(\partial V/\partial P\right)}_T}=-\frac{{\left(\partial V/\partial T\right)}_P}{{\left(\partial V/\partial P\right)}_T}.$ Now insert the two measurable coefficients, ${\left(\partial V/\partial T\right)}_P=V\alpha$ and ${\left(\partial V/\partial P\right)}_T=-V{\kappa }_T$: ${\left(\frac{\partial P}{\partial T}\right)}_V=-\frac{V\alpha }{-V{\kappa }_T}=\frac{\alpha }{{\kappa }_T}.$ Step 6 — assemble (using ${\left(\partial V/\partial T\right)}_P=V\alpha$ once more): $\boxed{C_P-C_V=T\frac{\alpha }{{\kappa }_T}\cdot V\alpha =\frac{TV{\alpha }^2}{{\kappa }_T}}$ Step 7 — ideal-gas check: $\alpha =1/T$, ${\kappa }_T=1/P$ (from $V=RT/P$: ${\left(\partial V/\partial P\right)}_T=-RT/P^2$, so ${\kappa }_T=-\frac{1}{V}\cdot (-RT/P^2)=RT/(P^{2}V)=1/P$). Then $C_P-C_V=\frac{T\cdot V\cdot (1/T{)}^2}{1/P}=\frac{VP}{T}=\frac{RT}{T}=R.✓$ > This is the celebrated result of Heat capacities $C_P - C_V$ — note it required a Maxwell relation to prove for a general substance, not just ideal gases. Because $T$, $V$, ${\alpha }^2$ and ${\kappa }_T$ are all $\ge 0$, this also proves $C_P\ge C_V$ always.

Step 1 — the energy equation (proven in the parent note): ${\left(\partial U/\partial V\right)}_T=T{\left(\partial P/\partial T\right)}_V-P$. Step 2 — solve for $P$: $P=\frac{RT}{V-b}-\frac{a}{V^2}$. Step 3 — differentiate at constant $V$: only the first term depends on $T$: ${\left(\frac{\partial P}{\partial T}\right)}_V=\frac{R}{V-b}.$ Step 4 — assemble: ${\left(\frac{\partial U}{\partial V}\right)}_T=T\cdot \frac{R}{V-b}-\left(\frac{RT}{V-b}-\frac{a}{V^2}\right)=\frac{a}{V^2}.$ $\boxed{{\left(\frac{\partial U}{\partial V}\right)}_T=\frac{a}{V^2}>0}$ Interpretation: because $a>0$ (attractions), internal energy rises as the gas expands — you must feed energy in to pull molecules apart against their mutual attraction. Notice the excluded-volume constant $b$ cancels out entirely: the energy's volume-dependence is set purely by the attractions. This is the microscopic origin of cooling in a Joule (free) expansion of a real gas.

Step 1 — the process is isenthalpic (WHY): throttling through a porous plug conserves enthalpy $H$, so we differentiate $T$ with respect to $P$ at constant $H$. Step 2 — cyclic rule on $(T,P,H)$: the three variables $T$, $P$, $H$ are linked (any two fix the state), so the same general cyclic identity used in Problem 3.1 applies to them: ${\left(\frac{\partial T}{\partial P}\right)}_H{\left(\frac{\partial P}{\partial H}\right)}_T{\left(\frac{\partial H}{\partial T}\right)}_P=-1.$ Solving for the piece we want (and using ${\left(\partial P/\partial H\right)}_T=1/{\left(\partial H/\partial P\right)}_T$): ${\left(\frac{\partial T}{\partial P}\right)}_H=-\frac{(\partial H/\partial P{)}_T}{(\partial H/\partial T{)}_P}.$ The denominator is ${\left(\partial H/\partial T\right)}_P=C_P$ by definition. Step 3 — evaluate ${\left(\partial H/\partial P\right)}_T$ from $dH=TdS+VdP$. Divide by $dP$ at constant $T$: ${\left(\frac{\partial H}{\partial P}\right)}_T=T{\left(\frac{\partial S}{\partial P}\right)}_T+V.$ Step 4 — kill the entropy term with the Gibbs Maxwell relation ${\left(\partial S/\partial P\right)}_T=-{\left(\partial V/\partial T\right)}_P=-V\alpha$: ${\left(\frac{\partial H}{\partial P}\right)}_T=-TV\alpha +V=V(1-T\alpha ).$ Step 5 — assemble: $\boxed{{\mu }_{JT}=\frac{V}{C_P}\left(T\alpha -1\right)}$ Step 6 — ideal gas: $\alpha =1/T$, so $T\alpha -1=0$ and ${\mu }_{JT}=0$. An ideal gas does not cool on throttling — consistent with ${\left(\partial U/\partial V\right)}_T=0$ and ${\left(\partial H/\partial P\right)}_T=0$. Real gases cool when $T\alpha >1$ (below the inversion temperature), which is how liquefiers work.

Step 1 — write $C_V$ via entropy: $C_V=T{\left(\partial S/\partial T\right)}_V$. Differentiate with respect to $V$ at constant $T$: ${\left(\frac{\partial C_V}{\partial V}\right)}_T=T\frac{\partial }{\partial V}{\left(\frac{\partial S}{\partial T}\right)}_V.$ Step 2 — swap the order of differentiation (WHY — Schwarz): since $S$ is a state function, mixed partials commute: $\frac{\partial }{\partial V}{\left(\frac{\partial S}{\partial T}\right)}_V=\frac{\partial }{\partial T}{\left(\frac{\partial S}{\partial V}\right)}_T.$ Step 3 — apply the Helmholtz Maxwell relation ${\left(\partial S/\partial V\right)}_T={\left(\partial P/\partial T\right)}_V$: $\frac{\partial }{\partial T}{\left(\frac{\partial P}{\partial T}\right)}_V={\left(\frac{{\partial }^{2}P}{\partial T^2}\right)}_V.$ Step 4 — assemble: $\boxed{{\left(\frac{\partial C_V}{\partial V}\right)}_T=T{\left(\frac{{\partial }^{2}P}{\partial T^2}\right)}_V}$ Step 5 — ideal gas: $P=RT/V$ is linear in $T$ at fixed $V$, so ${\left({\partial }^{2}P/\partial T^2\right)}_V=0$. Hence ${\left(\partial C_V/\partial V\right)}_T=0$: $C_V$ cannot depend on volume for an ideal gas — again derived, not assumed.

Step 1 — bring in a Maxwell relation (WHY): $\alpha =\frac{1}{V}{\left(\partial V/\partial T\right)}_P$, and the Gibbs relation converts this into an entropy derivative: ${\left(\frac{\partial V}{\partial T}\right)}_P=-{\left(\frac{\partial S}{\partial P}\right)}_T.$ So $\alpha =-\frac{1}{V}{\left(\frac{\partial S}{\partial P}\right)}_T$. Step 2 — invoke the third law (Nernst): as $T\to 0$, the entropy of every system approaches a constant $S_0$ that is independent of $P$ (and of $V$). Therefore ${\lim }_{T\to 0}{\left(\frac{\partial S}{\partial P}\right)}_T=0.$ Step 3 — conclude: $\boxed{\underset{T\to 0}{\lim }\alpha =-\frac{1}{V}\cdot 0=0}$ Meaning: this is a universal prediction — no material expands or contracts (to first order in $T$) as it approaches absolute zero. It has been confirmed for solids, liquid helium, everything. A pure thermodynamic theorem falling out of a Maxwell relation plus one law.

Step 1 — what "adiabatic reversible" means (WHY): no heat flow and reversible $\Rightarrow$ constant entropy, $S=\text{const}$. We want the sign of ${\left(\partial T/\partial P\right)}_S$ under compression ($dP>0$). Step 2 — cyclic rule on $(T,P,S)$: ${\left(\frac{\partial T}{\partial P}\right)}_S=-\frac{(\partial S/\partial P{)}_T}{(\partial S/\partial T{)}_P}.$ Step 3 — identify the pieces. Denominator: ${\left(\partial S/\partial T\right)}_P=C_P/T>0$ (heat capacity and temperature are positive). Numerator: use the Gibbs Maxwell relation ${\left(\frac{\partial S}{\partial P}\right)}_T=-{\left(\frac{\partial V}{\partial T}\right)}_P=-V\alpha .$ Step 4 — assemble: ${\left(\frac{\partial T}{\partial P}\right)}_S=-\frac{-V\alpha }{C_P/T}=\frac{TV\alpha }{C_P}.$ Step 5 — read off the sign. $T$, $V$, $C_P$ are all positive, so the sign of ${\left(\partial T/\partial P\right)}_S$ is the sign of $\alpha$.

• Normal material ($\alpha >0$): compression warms it ($dT>0$) — the familiar bicycle-pump heating.
• Negative-expansion material ($\alpha <0$): compression cools it ($dT<0$). $\boxed{\text{For }\alpha <0:{\left(\frac{\partial T}{\partial P}\right)}_S<0\Rightarrow \text{adiabatic compression lowers }T.}$ Meaning: the counter-intuitive expansion behaviour of water near freezing flips the sign of its adiabatic heating — a striking, testable consequence extracted from one Maxwell relation and the cyclic rule.

L1: recognise which potential owns which derivative (match the differentiated variable and the subscript). L2: apply a relation by pushing the calculation onto the measurable equation-of-state side. L3: combine relations with the cyclic rule to get $C_P-C_V$ and $(\partial U/\partial V{)}_T$. L4: synthesise brand-new results — Joule–Thomson, $\partial C_V/\partial V$ — using Maxwell + Schwarz together. L5: master universal theorems ($\alpha \to 0$ at $T=0$; the sign-flip for negative-expansion materials) where thermodynamics alone forecasts real experiments. Throughout: derive, never memorize signs.