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Thermodynamics question

2025 · 22 Jan · Shift 2 · Q12
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Thermodynamics question

2025 · 22 Jan · Shift 2 · Q12

JEE MainChemistryThermodynamicsMCQ+4 / −1

Match List - I with List - II.

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

Choose the correct answer from the options given below :

  1. A
    (A)-(I), (B)-(II), (C)-(IV), (D)-(III)
  2. B
    (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
  3. C
    (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
  4. D
    (A)-(II), (B)-(I), (C)-(III), (D)-(IV)
View written solutionFree

Correct answer: C

  1. We use the standard thermodynamic differential relations.

  2. For Gibbs free energy: dG=V dP−S dTdG = V\,dP - S\,dTdG=VdP−SdT Hence, (∂G∂T)P=−S\left(\frac{\partial G}{\partial T}\right)_P = -S(∂T∂G​)P​=−S So, (A)→(II)\text{(A)} \to \text{(II)}(A)→(II)

  3. For enthalpy, the heat capacity at constant pressure is defined as: CP=(∂H∂T)PC_P = \left(\frac{\partial H}{\partial T}\right)_PCP​=(∂T∂H​)P​ So, (B)→(I)\text{(B)} \to \text{(I)}(B)→(I)

  4. Again from dG=V dP−S dTdG = V\,dP - S\,dTdG=VdP−SdT at constant temperature, (∂G∂P)T=V\left(\frac{\partial G}{\partial P}\right)_T = V(∂P∂G​)T​=V So, (C)→(IV)\text{(C)} \to \text{(IV)}(C)→(IV)

  5. For internal energy, the heat capacity at constant volume is defined as: CV=(∂U∂T)VC_V = \left(\frac{\partial U}{\partial T}\right)_VCV​=(∂T∂U​)V​ So, (D)→(III)\text{(D)} \to \text{(III)}(D)→(III)

  6. Therefore the complete matching is: (A)-(II), (B)-(I), (C)-(IV), (D)-(III)\text{(A)-(II), (B)-(I), (C)-(IV), (D)-(III)}(A)-(II), (B)-(I), (C)-(IV), (D)-(III)

  7. Checking the options, this corresponds to Option C.

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