Question:

The CORRECT Maxwell relation(s) derived from the fundamental equations of thermodynamics is (are)

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Use the equality of mixed partial derivatives on thermodynamic potentials ($U$, $H$, $A$, $G$) to derive Maxwell relations.
Updated On: Dec 5, 2025
  • $\left( \dfrac{\partial S}{\partial P} \right)_T = - \left( \dfrac{\partial V}{\partial T} \right)_P$
  • $\left( \dfrac{\partial S}{\partial V} \right)_T = \left( \dfrac{\partial P}{\partial T} \right)_V$
  • $\left( \dfrac{\partial T}{\partial V} \right)_S = \left( \dfrac{\partial P}{\partial S} \right)_V$
  • $\left( \dfrac{\partial T}{\partial P} \right)_S = \left( \dfrac{\partial V}{\partial S} \right)_P$
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The Correct Option is A, B, D

Solution and Explanation

Step 1: Recall the four thermodynamic potentials.
From the total differentials: \[ dU = T\,dS - P\,dV, \quad dH = T\,dS + V\,dP, \quad dA = -S\,dT - P\,dV, \quad dG = -S\,dT + V\,dP \] Step 2: Derive Maxwell relations.
From equality of mixed second derivatives: \[ \left( \frac{\partial T}{\partial V} \right)_S = -\left( \frac{\partial P}{\partial S} \right)_V \] \[ \left( \frac{\partial S}{\partial P} \right)_T = -\left( \frac{\partial V}{\partial T} \right)_P \] \[ \left( \frac{\partial S}{\partial V} \right)_T = \left( \frac{\partial P}{\partial T} \right)_V \] \[ \left( \frac{\partial T}{\partial P} \right)_S = \left( \frac{\partial V}{\partial S} \right)_P \] Step 3: Compare with options.
Options (A) and (B) match the standard Maxwell relations directly.
Step 4: Conclusion.
Correct Maxwell relations are (A) and (B).
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