Which of the following relations are correct? (A) $\Delta U=q+p\Delta V$ (B) $\Delta G=\Delta H-T\Delta S$ (C) $\Delta S=\frac{{q}_{\mathrm{rev}}}{T}$ (D) $\Delta H=\Delta U-\Delta nRT$ Choose the most appropriate answer from the options given below :

The Gibbs free energy can be written as a function of enthalpy and entropy as follows, (B) $G=H-\mathrm{TS}$ At constant $T$ $\Delta G=\Delta H-T\Delta S$ (A) According to first law of thermodynamics, $\Delta U=Q+W$ If we apply constant $P$ and reversible work. $W=-P\Delta V$ $\Delta U=Q-P\Delta V$ (C) By definition of entropy change $\mathrm{dS}=\frac{{\mathrm{dq}}_{\mathrm{rev}}}{T}$ At constant $T$ $\Delta S=\frac{{q}_{\mathrm{rev}}}{T}$ Enthalpy can be written as a function of internal energy, pressure and volume as follows, (D) $H=U+\mathrm{PV}$ For ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$ $H=U+\mathrm{nRT}$ At constant $T$ $\Delta H=\Delta U+\Delta nRT$

JEE Main 2023 — Chemistry Physical Chemistry

2023mcqeasy

Which of the following relations are correct?

(A) $\Delta U=q+p\Delta V$

(B) $\Delta G=\Delta H-T\Delta S$

(D) $\Delta H=\Delta U-\Delta nRT$

Choose the most appropriate answer from the options given below :

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

Options

A
C and D only
B
B and C only
C
A and B only
D
B and D only

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Solution

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

(B) $G=H-\mathrm{TS}$

At constant $T$

$\Delta G=\Delta H-T\Delta S$

(A) According to first law of thermodynamics, $\Delta U=Q+W$

If we apply constant $P$ and reversible work.

$W=-P\Delta V$

$\Delta U=Q-P\Delta V$

At constant $T$

$\Delta S=\frac{{q}_{\mathrm{rev}}}{T}$

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

(D) $H=U+\mathrm{PV}$

For ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$

$H=U+\mathrm{nRT}$

At constant $T$

$\Delta H=\Delta U+\Delta nRT$

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Solution

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

(B) $G=H-\mathrm{TS}$

At constant $T$

$\Delta G=\Delta H-T\Delta S$

(A) According to first law of thermodynamics, $\Delta U=Q+W$

If we apply constant $P$ and reversible work.

$W=-P\Delta V$

$\Delta U=Q-P\Delta V$

At constant $T$

$\Delta S=\frac{{q}_{\mathrm{rev}}}{T}$

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

(D) $H=U+\mathrm{PV}$

For ideal gas equation, $\mathrm{PV}=\mathrm{nRT}$

$H=U+\mathrm{nRT}$

At constant $T$

$\Delta H=\Delta U+\Delta nRT$

JEE Main 2023 — Chemistry Physical Chemistry

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Solution

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JEE Main Physical Chemistry in other years

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JEE Main 2023 — Chemistry Physical Chemistry

Options

Did you get this right?

Solution

Your note

JEE Main Physical Chemistry in other years

More JEE Main Physical Chemistry questions

Other Chemistry topics for JEE Main

Organic Chemistry

Inorganic Chemistry

Keep practicing Physical Chemistry