(A) . Work done in reversible isothermal expansion: $W = -nRT \ln(\frac{V_2}{V_1}) = -2 \times 8.314 \times 300 \times \ln(10) \approx -11488 \ J = -1

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(A) . Work done in reversible isothermal expansion: $W = -nRT \ln(\frac{V_2}{V_1}) = -2 \times 8.314 \times 300 \times \ln(10) \approx -11488 \ J = -11.5 \ kJ$. Magnitude is $11.5 \ kJ$ $(II)$.$B$. Work done in irreversible isothermal expansion: $W = -P_{ext}(V_2 - V_1) = -3 \times 10^3 \ Pa \times (3 - 1) \ m^3 = -6000 \ J = -6 \ kJ$. Magnitude is $6 \ kJ$ $(III)$.$C$. Change in internal energy: $\Delta U = nC_V \Delta T = 1 \times \frac{3}{2} \times 8.314 \times 320 \approx 3990 \ J \approx 4 \ kJ$. Magnitude is $4 \ kJ$ $(I)$.$D$. Change in enthalpy: $\Delta H = nC_P \Delta T = 1 \times \frac{5}{2} \times 8.314 \times 337 \approx 7004 \ J \approx 7 \ kJ$. Magnitude is $7 \ kJ$ $(IV)$.Thus, the correct match is $A-II, B-III, C-I, D-IV$.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

Difficult

Match the List-$I$ with List-$II$

List-$I$ Thermodynamic Process List-$II$ Magnitude in $kJ$

$A$. Work done in reversible, isothermal expansion of $2 \ mol$ of ideal gas from $2 \ dm^3$ to $20 \ dm^3$ at $300 \ K$. $I$. $4$

$B$. Work done in irreversible isothermal expansion of $1 \ mol$ ideal gas from $1 \ m^3$ to $3 \ m^3$ at $300 \ K$ against a constant pressure of $3 \ kPa$. $II$. $11.5$

$C$. Change in internal energy for adiabatic expansion of a $1 \ mol$ ideal gas with change of temperature $= 320 \ K$ and $\overline{C}_V = \frac{3}{2} R$. $III$. $6$

$D$. Change in enthalpy at constant pressure of $1 \ mole$ ideal gas with change of temperature $= 337 \ K$ and $\overline{C}_P = \frac{5}{2} R$. $IV$. $7$

Choose the correct answer from the option given below:

JEE MAIN 2026 All JEE MAIN

A
$A-II, B-III, C-I, D-IV$
B
$A-II, B-III, C-IV, D-I$
C
$A-I, B-II, C-III, D-IV$
D
$A-III, B-II, C-I, D-IV$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

Thermodynamics

Subtopic

Mix Examples-Thermodynamics and Thermochemistry

Previous Year

JEE MAIN 2026 Paper All JEE MAIN years →

Calculate the enthalpy change when $6 \text{ g}$ of $CO(g)$ reacts with sufficient $NO_2(g)$ according to the following reaction: $4 \text{ CO}(g) + 2 \text{ NO}_2(g) \rightarrow 4 \text{ CO}_2(g) + 2 \text{ N}_2(g)$; $\Delta_r H^0 = -1200 \text{ kJ}$ (in $\text{ kJ}$)

DifficultMHT CET 2026

View Solution

Assertion : The increase in internal energy $(\Delta E)$ for the vaporization of one mole of water at $1 \ atm$ and $373 \ K$ is zero.
Reason : For all isothermal processes,$\Delta E = 0$.

EasyAIIMS 2003

View Solution

For a reaction $2 CO_{(g)} + O_{2(g)} \rightleftharpoons 2 CO_{2(g)}$,$\Delta_{r} G^0 = -128 \ kJ$ at $300 \ K$. If $\Delta_{r} S^0$ of the reaction is $-40 \ J \ K^{-1}$,calculate $\Delta_{r} U$ of the reaction. (in $kJ$)

MediumAP EAMCET 2022

View Solution

State $1 \longleftarrow$ State $2 \longleftarrow$ State $3$
$\left(\begin{array}{c} T=300 \ K \\ P=15 \ bar \\ 1 \ mole \end{array}\right) \left(\begin{array}{c} T=300 \ K \\ P=10 \ bar \\ 1 \ mole \end{array}\right) \left(\begin{array}{c} T=300 \ K \\ P=5 \ bar \\ 1 \ mole \end{array}\right)$
The above shows a cyclic process. Calculate the total work done during one complete cycle. (Assume a single step to reach the next state).

MediumAP EAMCET 2022

View Solution

Match List-$I$ with List-$II$. Given $V_1$ and $V_2$ are initial and final volumes respectively.

List-$I$ (Isothermal process) List-$II$ (Expression)

$A$. Reversible expansion $I$. $q = 0$

$B$. Free expansion $II$. $q = nRT \ln \frac{V_2}{V_1}$

$C$. Irreversible Compression $III$. $w = -P_{ext}(V_1 - V_2)$

$D$. Cyclic reversible $IV$. $\frac{q_{rev}}{T} = 0$

DifficultJEE MAIN 2026

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

List-$I$ Thermodynamic Process	List-$II$ Magnitude in $kJ$
$A$. Work done in reversible, isothermal expansion of $2 \ mol$ of ideal gas from $2 \ dm^3$ to $20 \ dm^3$ at $300 \ K$.	$I$. $4$
$B$. Work done in irreversible isothermal expansion of $1 \ mol$ ideal gas from $1 \ m^3$ to $3 \ m^3$ at $300 \ K$ against a constant pressure of $3 \ kPa$.	$II$. $11.5$
$C$. Change in internal energy for adiabatic expansion of a $1 \ mol$ ideal gas with change of temperature $= 320 \ K$ and $\overline{C}_V = \frac{3}{2} R$.	$III$. $6$
$D$. Change in enthalpy at constant pressure of $1 \ mole$ ideal gas with change of temperature $= 337 \ K$ and $\overline{C}_P = \frac{5}{2} R$.	$IV$. $7$

List-$I$ (Isothermal process)	List-$II$ (Expression)
$A$. Reversible expansion	$I$. $q = 0$
$B$. Free expansion	$II$. $q = nRT \ln \frac{V_2}{V_1}$
$C$. Irreversible Compression	$III$. $w = -P_{ext}(V_1 - V_2)$
$D$. Cyclic reversible	$IV$. $\frac{q_{rev}}{T} = 0$

Similar Questions

Calculate the enthalpy change when $6 \text{ g}$ of $CO(g)$ reacts with sufficient $NO_2(g)$ according to the following reaction: $4 \text{ CO}(g) + 2 \text{ NO}_2(g) \rightarrow 4 \text{ CO}_2(g) + 2 \text{ N}_2(g)$; $\Delta_r H^0 = -1200 \text{ kJ}$ (in $\text{ kJ}$)

Assertion : The increase in internal energy $(\Delta E)$ for the vaporization of one mole of water at $1 \ atm$ and $373 \ K$ is zero.Reason : For all isothermal processes,$\Delta E = 0$.

For a reaction $2 CO_{(g)} + O_{2(g)} \rightleftharpoons 2 CO_{2(g)}$,$\Delta_{r} G^0 = -128 \ kJ$ at $300 \ K$. If $\Delta_{r} S^0$ of the reaction is $-40 \ J \ K^{-1}$,calculate $\Delta_{r} U$ of the reaction. (in $kJ$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion : The increase in internal energy $(\Delta E)$ for the vaporization of one mole of water at $1 \ atm$ and $373 \ K$ is zero.
Reason : For all isothermal processes,$\Delta E = 0$.