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GATE 2022 PH – Question 47

Thermodynamics and Statistical Mechanics · Laws of thermodynamics · 2 marks · Multiple select

A paramagnet obeys $dG=-S\,dT-M\,dH$. For an infinitesimal adiabatic field change, which statements hold?

  1. $\Delta S=-T^{-1}(\partial G/\partial T)_H\Delta T$
  2. $\Delta S=0$
  3. $\Delta T=-[(\partial M/\partial T)_H/(\partial S/\partial T)_H]\Delta H$
  4. $\Delta T=0$

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Correct answer: (B) $\Delta S=0$; (C) $\Delta T=-[(\partial M/\partial T)_H/(\partial S/\partial T)_H]\Delta H$

Explanation

For a paramagnet, $dG=-S\,dT-M\,dH$. Since $G$ is a state function, the cross derivatives are equal (a Maxwell relation):
$$\left(\frac{\partial S}{\partial H}\right)_T=\left(\frac{\partial M}{\partial T}\right)_H .$$

**Reversible adiabatic change of the field:** the entropy is constant, so $\Delta S=0$. ✓ (B)

Write $dS=0$ in terms of $T$ and $H$:
$$0=\left(\frac{\partial S}{\partial T}\right)_HdT+\left(\frac{\partial S}{\partial H}\right)_TdH=\left(\frac{\partial S}{\partial T}\right)_HdT+\left(\frac{\partial M}{\partial T}\right)_HdH .$$

$$\Delta T=-\frac{(\partial M/\partial T)_H}{(\partial S/\partial T)_H}\,\Delta H\quad(\text{option C}),$$
which is the magnetocaloric effect. So $\Delta T\neq0$ in general (D is false), and A is not correct.

Answer **B and C**.