Question

For the relationship, $dH=TdS+VdP$, the Maxwell equation is:

• A. $\frac{\partial T}{\partial P}|_{S}=\frac{\partial V}{\partial S}|_{p}$
• B. $\frac{\partial T}{\partial P}|_{S}=-\frac{\partial V}{\partial S}|_{P}$
• C. $\frac{\partial U}{\partial T}|_{P}=\frac{\partial U}{\partial T}|_{P}$
• D. $\frac{\partial P}{\partial V}|_{T}=\frac{\partial S}{\partial T}|_{T}$

Hint:

To remember Maxwell relations, use the "Thermodynamic Square" or simply look at the differential equation: the variables in the partials come from the differentials ($dP$ and $dS$), and the "coefficients" ($T$ and $V$) become the numerators. If the original signs in the $dH$ equation are both positive, the Maxwell relation is positive.

Answer 1

The Correct Option is A

Concept: Maxwell relations are a set of equations in thermodynamics derived from the definitions of thermodynamic potentials and the property of exact differentials. For any exact differential $dz = Mdx + Ndy$, the condition $\frac{\partial M}{\partial y}|_x = \frac{\partial N}{\partial x}|_y$ must hold.

Step 1: Apply the exact differential rule to the enthalpy equation.
The given fundamental property relation for Enthalpy ($H$) is: \[ dH = T dS + V dP \] Comparing this to the general form $dz = M dx + N dy$:
• $z = H, x = S, y = P$
• $M = T = \frac{\partial H}{\partial S}|_P$
• $N = V = \frac{\partial H}{\partial P}|_S$

Step 2: Equate the mixed partial derivatives.
Since $H$ is a state function, its second mixed partial derivatives are equal: \[ \frac{\partial}{\partial P} \left( \frac{\partial H}{\partial S} \right) = \frac{\partial}{\partial S} \left( \frac{\partial H}{\partial P} \right) \] Substituting $T$ and $V$: \[ \left( \frac{\partial T}{\partial P} \right)_S = \left( \frac{\partial V}{\partial S} \right)_P \]