One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the $PV$-diagram below. Among these four processes,one is isobaric,one is isochoric,one is isothermal and one is adiabatic. Match the processes mentioned in List-$I$ with the corresponding statements in List-$II$.
List-$I$ List-$II$ $P$. In process $I$ $1$. Work done by the gas is zero $Q$. In process $II$ $2$. Temperature of the gas remains unchanged $R$. In process $III$ $3$. No heat is exchanged between the gas and its surroundings $S$. In process $IV$ $4$. Work done by the gas is $6 P_0 V_0$
| List-$I$ | List-$II$ |
| $P$. In process $I$ | $1$. Work done by the gas is zero |
| $Q$. In process $II$ | $2$. Temperature of the gas remains unchanged |
| $R$. In process $III$ | $3$. No heat is exchanged between the gas and its surroundings |
| $S$. In process $IV$ | $4$. Work done by the gas is $6 P_0 V_0$ |
- A$P$ $\rightarrow 4, Q$ $\rightarrow 3, R$ $\rightarrow 1, S$ $\rightarrow 2$
- B$P$ $\rightarrow 1, Q$ $\rightarrow 3, R$ $\rightarrow 2, S$ $\rightarrow 4$
- C$P$ $\rightarrow 3, Q$ $\rightarrow 4, R$ $\rightarrow 1, S$ $\rightarrow 2$
- D$P$ $\rightarrow 3, Q$ $\rightarrow 4, R$ $\rightarrow 2, S$ $\rightarrow 1$
Explore More
Similar Questions
The heat of reaction for $C_6H_{12}O_{6(s)} + 6O_{2(g)} \to 6CO_{2(g)} + 6H_2O_{(l)}$ at constant pressure is $-651 \, kcal$ at $17 \, ^oC$. Calculate the heat of reaction at constant volume at $17 \, ^oC$ in $kcal$.
Medium
View SolutionFor the reaction,$X_2O_{4(l)} \to 2XO_{2(g)}$,$\Delta U = 2.1 \ kcal$ and $\Delta S = 20 \ cal \ K^{-1}$ at $300 \ K$. Calculate $\Delta G$ in $kcal$.
DifficultNEET 2014
View SolutionReversible expansion of an ideal gas under isothermal and adiabatic conditions are as shown in the figure.
$AB \to$ Isothermal expansion
$AC \to$ Adiabatic expansion
Which of the following options is not correct?
$AB \to$ Isothermal expansion
$AC \to$ Adiabatic expansion
Which of the following options is not correct?
DifficultNEET 2019
View SolutionThe value of $\Delta H_{transition}$ for $C(\text{graphite}) \rightarrow C(\text{diamond})$ is $1.9 \ kJ/mol$ at $25^{\circ}C$. The entropy of graphite is higher than the entropy of diamond. This implies that which of the following is incorrect $:-$
Easy
View SolutionMatch the thermodynamic processes given under Column $I$ with the expression given under Column $II$:
Column $I$
Column $II$
$A$. Freezing of water at $273 \ K$ and $1 \ atm$
$P$. $q=0$
$B$. Expansion of $1 \ mol$ of an ideal gas into a vacuum under isolated conditions
$Q$. $w=0$
$C$. Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container
$R$. $\Delta S_{sys} < 0$
$D$. Reversible heating of $H_{2(g)}$ at $1 \ atm$ from $300 \ K$ to $600 \ K$,followed by reversible cooling to $300 \ K$ at $1 \ atm$
$S$. $\Delta U=0$
$T$. $\Delta G=0$
| Column $I$ | Column $II$ |
| $A$. Freezing of water at $273 \ K$ and $1 \ atm$ | $P$. $q=0$ |
| $B$. Expansion of $1 \ mol$ of an ideal gas into a vacuum under isolated conditions | $Q$. $w=0$ |
| $C$. Mixing of equal volumes of two ideal gases at constant temperature and pressure in an isolated container | $R$. $\Delta S_{sys} < 0$ |
| $D$. Reversible heating of $H_{2(g)}$ at $1 \ atm$ from $300 \ K$ to $600 \ K$,followed by reversible cooling to $300 \ K$ at $1 \ atm$ | $S$. $\Delta U=0$ |
| $T$. $\Delta G=0$ |
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo