Chemistry · Thermodynamics & Thermochemistry

JEE Main 2025 — 22 January, Evening Shift — Question 32

Match List-I with List-II.

List-I (Partial Derivatives)List-II (Thermodynamic Quantity)
(A)(∂G∂T)P{{\left( \frac{\partial \text{G}}{\partial \text{T}} \right)}_{\text{P}}}(I)CP{{\text{C}}_{\text{P}}}
(B)(∂H∂T)P{{\left( \frac{\partial \text{H}}{\partial \text{T}} \right)}_{\text{P}}}(II)-S
(C)(∂G∂P)T{{\left( \frac{\partial \text{G}}{\partial \text{P}} \right)}_{\text{T}}}(III)CV{{\text{C}}_{\text{V}}}
(D)(∂U∂T)V{{\left( \frac{\partial \text{U}}{\partial \text{T}} \right)}_{\text{V}}}(IV)V

Choose the correct answer from the options given below:

  1. Option A:

    (A)-(II), (B)-(I), (C)-(III), (D)-(IV)

  2. Option B:

    (A)-(II), (B)-(I), (C)-(IV), (D)-(III)

    Correct
  3. Option C:

    (A)-(I), (B)-(II), (C)-(IV), (D)-(III)

  4. Option D:

    (A)-(II), (B)-(III), (C)-(I), (D)-(IV)

Answer: B

Step-by-step solution

(A) dG=VdP−SdT\mathrm{dG}=\mathrm{VdP}-\mathrm{SdT}

Constant pressure

dG=−SdT\mathrm{dG}=-\mathrm{SdT}

(∂G∂ T)P=−S \left(\frac{\partial \mathrm{G}}{\partial \mathrm{~T}}\right)_{\mathrm{P}}=-\mathrm{S}

(B) dH=(dq)P=nCpdT\mathrm{dH}=(\mathrm{dq})_{\mathrm{P}}=\mathrm{nCpdT}

(∂H∂ T)P=CP\left(\frac{\partial \mathrm{H}}{\partial \mathrm{~T}}\right)_{\mathrm{P}}=\mathrm{C}_{\mathrm{P}}

(C) dG=VdP−SdT\mathrm{dG}=\mathrm{VdP}-\mathrm{SdT}

At constant temperature

dG=VdP\mathrm{dG}=\mathrm{VdP}

(∂G∂P)T=V\left(\frac{\partial \mathrm{G}}{\partial \mathrm{P}}\right)_{\mathrm{T}}=\mathrm{V}

(D) dU=nCVdT=(q)v\mathrm{dU}=\mathrm{nC}_{\mathrm{V}} \mathrm{dT}=(\mathrm{q})_{\mathrm{v}}

(∂U∂ T)V=CV\left(\frac{\partial \mathrm{U}}{\partial \mathrm{~T}}\right)_{\mathrm{V}}=\mathrm{C}_{\mathrm{V}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Chemistry
Chapter
Thermodynamics & Thermochemistry
Topic
Gibbs Free Energy and the Third Law of Thermodynamics
Match List-I with List-II. List-I (Partial Derivatives) List-II… | JEE Main 2025 PYQ with Solution · DhiX AI