Q. Match List - I with List - II. [table] List - I (Partial Derivatives) | List - II (Thermodynamic Quantity) ; (A) $\left(\frac{\partial G}{\partial T}\right)_P$ | (I) $C_p$ ; (B) $\left(\frac{\partial H}{\partial T}\right)_P$ | (II) $-S$ ; (C) $\left(\frac{\partial G}{\partial P}\right)_T$ | (III) $C_v$ ; (D) $\left(\frac{\partial U}{\partial T}\right)_V$ | (IV) $V$ [/table] Choose the correct answer from the options given below :
A) (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
B) (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
C) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
D) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Explanation
[Solution by Dr. Abubacker] From $dG=-S\,dT+V\,dP$: $\left(\frac{\partial G}{\partial T}\right)_P=-S$ and $\left(\frac{\partial G}{\partial P}\right)_T=V$. By definition $\left(\frac{\partial H}{\partial T}\right)_P=C_p$ and $\left(\frac{\partial U}{\partial T}\right)_V=C_v$. So A–II, B–I, C–IV, D–III. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.