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Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IICalculation of DSEasy2 minPYQ_2023
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Which of the following relations are correct?
(A)ΔU=q+pΔV
(B)ΔG=ΔH-TΔS
(C)ΔS=qrevT
(D)ΔH=ΔU-ΔnRT
Choose the most appropriate answer from the options given below :

Options:

Answer:
B
Solution:

The Gibbs free energy can be written as a function of enthalpy and entropy as follows,

(B) G=H-TS

At constant T 

ΔG=ΔH-TΔS

(A) According to first law of thermodynamics, ΔU=Q+W

If we apply constant P and reversible work.

W=-PV

ΔU=Q-PΔV

(C) By definition of entropy change dS=dqrevT

At constant T

ΔS=qrevT

Enthalpy can be written as a function of internal energy, pressure and volume as follows,

(D) H=U+PV

For ideal gas equation, PV=nRT

H=U+nRT

At constant T

ΔH=ΔU+ΔnRT

Stream:JEESubject:ChemistryTopic:Thermodynamics - IISubtopic:Calculation of DS
2mℹ️ Source: PYQ_2023

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