(A) For an isothermal process of an ideal gas,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = q

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

Both Statement-$I$ and Statement-$II$ are correct

(A) For an isothermal process of an ideal gas,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = q + w$.Substituting $\Delta U = 0$,we get $q = -w$.For an irreversible process,$w = -P_{\text{ext}} \Delta V = -P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$.Thus,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$. Statement-$I$ is correct.For an adiabatic process,the heat exchange $q = 0$.From the first law,$\Delta U = q + w = 0 + w_{\text{adiabatic}} = w_{\text{adiabatic}}$. Statement-$II$ is correct.

Class 11 Chemistry Thermodynamics Mix Examples-Thermodynamics and Thermochemistry

Easy

Given below are two statements:
Statement-$I$: For isothermal irreversible change of an ideal gas,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$
Statement-$II$: For adiabatic change,$\Delta U = W_{\text{adiabatic}}$
The correct answer is

AP EAMCET 2024 All AP EAMCET

A
Both Statement-$I$ and Statement-$II$ are correct
B
Both Statement-$I$ and Statement-$II$ are not correct
C
Statement-$I$ is correct but Statement-$II$ is not correct
D
Statement-$I$ is not correct but Statement-$II$ is correct

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

AP EAMCET 2024 Paper All AP EAMCET years →

Match the following items in List-$I$ with their corresponding expressions in List-$II$.
List-$I$ List-$II$
$A$. At constant volume the change in internal energy of a system $I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$
$B$. Isothermal irreversible change $II$. $W_{adiabatic} = \Delta U$
$C$. Isothermal reversible change $III$. $q_V = \Delta U$
$D$. Adiabatic change $IV$. $W = -p_{ex} (V_f - V_i)$
$V$. $\Delta U = \Delta H - \Delta nRT$

MediumAP EAMCET 2018

View Solution

For a reaction,$X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)}$. If $\Delta G_r^o = 20 \ kJ \ mol^{-1}$ and $\Delta S_r^o = -20 \ J \ K^{-1} \ mol^{-1}$ at $200 \ K$. Calculate $\Delta H_r^o$ at $400 \ K$ (Given $\Delta_r C_P = 20 \ J \ K^{-1} \ mol^{-1}$).

Medium

View Solution

Assertion $(A)$: If heat of combustion of $C_2H_6$ is $X \ kJ \ mol^{-1}$,heat liberated on combustion of $150 \ g$ of $C_2H_6$ is $5X \ kJ$.
Reason $(R)$: Enthalpy is an extensive property.

MediumAP EAMCET 2017

View Solution

When $1.0 \ g$ of solid oxalic acid $(H_2C_2O_4)$ is burned in a bomb calorimeter whose heat capacity is $8.75 \ kJ/K$,the temperature increases by $0.312 \ K$. The enthalpy of combustion of oxalic acid at $27 \ ^oC$ is

Difficult

View Solution

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 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-$I$	List-$II$
$A$. At constant volume the change in internal energy of a system	$I$. $W = -2.303 nRT \log \frac{V_f}{V_i}$
$B$. Isothermal irreversible change	$II$. $W_{adiabatic} = \Delta U$
$C$. Isothermal reversible change	$III$. $q_V = \Delta U$
$D$. Adiabatic change	$IV$. $W = -p_{ex} (V_f - V_i)$
	$V$. $\Delta U = \Delta H - \Delta nRT$

Given below are two statements:Statement-$I$: For isothermal irreversible change of an ideal gas,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$Statement-$II$: For adiabatic change,$\Delta U = W_{\text{adiabatic}}$The correct answer is

Similar Questions

For a reaction,$X_{2(g)} + Y_{2(g)} \rightleftharpoons 2XY_{(g)}$. If $\Delta G_r^o = 20 \ kJ \ mol^{-1}$ and $\Delta S_r^o = -20 \ J \ K^{-1} \ mol^{-1}$ at $200 \ K$. Calculate $\Delta H_r^o$ at $400 \ K$ (Given $\Delta_r C_P = 20 \ J \ K^{-1} \ mol^{-1}$).

Assertion $(A)$: If heat of combustion of $C_2H_6$ is $X \ kJ \ mol^{-1}$,heat liberated on combustion of $150 \ g$ of $C_2H_6$ is $5X \ kJ$.Reason $(R)$: Enthalpy is an extensive property.

When $1.0 \ g$ of solid oxalic acid $(H_2C_2O_4)$ is burned in a bomb calorimeter whose heat capacity is $8.75 \ kJ/K$,the temperature increases by $0.312 \ K$. The enthalpy of combustion of oxalic acid at $27 \ ^oC$ is

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 $kcal$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Given below are two statements:
Statement-$I$: For isothermal irreversible change of an ideal gas,$q = -w = P_{\text{ext}} (V_{\text{final}} - V_{\text{initial}})$
Statement-$II$: For adiabatic change,$\Delta U = W_{\text{adiabatic}}$
The correct answer is

Assertion $(A)$: If heat of combustion of $C_2H_6$ is $X \ kJ \ mol^{-1}$,heat liberated on combustion of $150 \ g$ of $C_2H_6$ is $5X \ kJ$.
Reason $(R)$: Enthalpy is an extensive property.