MCQ
Match List-I with List-II.
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
Choose the correct answer from the options given below :
  • A
    (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
  • (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  • C
    (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
  • D
    (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Answer

Correct option: B.
(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
(B) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Sol. (A) $dG = VdP - SdT$
Constant pressure
$
\begin{array}{l}
dG=-SdT \\
\left(\frac{\partial G}{\partial T}\right)_{p}=-S\end{array}$

(B) $dH =( dq )_{ p }= nCpdT$
$
\left(\frac{\partial H}{\partial T}\right)_{P}=C_{p}
$

(C) $dG = VdP - SdT$
At constant temperature
$
\begin{array}{l}
dG=VdP \\
\left(\frac{\partial G}{\partial P}\right)_{T}=V
\end{array}
$
(D)$
\begin{array}{l}\\dU=nC_{V} dT=(q)_{V} \\
\left(\frac{\partial U}{\partial T}\right)_{V}=C_{V}
\end{array}
$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free