A paramagnetic salt of mass m is held at temperature T in a magnetic field H. If S
is the entropy of the salt and M is its magnetization, then dG = -SdT - MdH,
where G is the Gibbs free energy. If the magnetic field is changed adiabatically by
ΔH → 0 and the corresponding infinitesimal changes in entropy and temperature
are ΔS and ΔT, then which of the following statements are correct
Options:
(A) ΔS = - 1
T \left(𝜕G
𝜕T\right)
ΔT
(B) ΔS = 0
(C) ΔT = -
(𝜕M
𝜕T )
(𝜕S
𝜕T)
ΔH
(D) ΔT = 0
Official Answer Key:
B,C
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2022 Q.47)**
**1. Thermodynamics Principles (Magnetic Systems & Adiabatic Demagnetization)**:
- First Law of Thermodynamics for magnetic system:
$T dS = dU - H dM$
$dG = -S dT - M dH \implies \left(\frac{\partial S}{\partial H}\right)_T = \left(\frac{\partial M}{\partial T}\right)_H \quad \text{(Maxwell relation)}$
**2. Detailed Step-by-Step Derivation**:
- For an adiabatic process, $dS = 0 \implies \Delta S = 0$. **Statement (B) is TRUE**.
- Expressing $S$ as a function of $T$ and $H$:
$dS = \left(\frac{\partial S}{\partial T}\right)_H dT + \left(\frac{\partial S}{\partial H}\right)_T dH = 0$
$\implies \left(\frac{\partial S}{\partial T}\right)_H dT = -\left(\frac{\partial S}{\partial H}\right)_T dH$
$dT = -\frac{(\partial S / \partial H)_T}{(\partial S / \partial T)_H} dH$
- Substituting Maxwell relation $\left(\frac{\partial S}{\partial H}\right)_T = \left(\frac{\partial M}{\partial T}\right)_H$:
$\Delta T = -\frac{\left(\frac{\partial M}{\partial T}\right)_H}{\left(\frac{\partial S}{\partial T}\right)_H} \Delta H$
**Statement (C) is TRUE**.
**3. Official Verification**:
The verified official IIT Master Answer is **Options (B, C)**.
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