(A) According to the first law of thermodynamics,$\Delta E = q + w$. Given: Heat absorbed $(q)$ = $+600 \ J$. Work done $(w)$ = $-P_{ext} \times \Delt

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

(A) According to the first law of thermodynamics,$\Delta E = q + w$. Given: Heat absorbed $(q)$ = $+600 \ J$. Work done $(w)$ = $-P_{ext} \times \Delta V$. $P_{ext} = 1 \ atm$,$\Delta V = 12 \ L - 2 \ L = 10 \ L$. $w = -1 \ atm \times 10 \ L = -10 \ L \cdot atm$. Conversion factor: $1 \ L \cdot atm = 101.3 \ J$. So,$w = -10 \times 101.3 \ J = -1013 \ J$. Now,$\Delta E = 600 \ J + (-1013 \ J) = -413 \ J$.

Class 11 Chemistry Thermodynamics First law of thermodynamics

Medium

$A$ sample of gas expands against a constant external pressure of $1 \ atm$ from a volume of $2 \ L$ to $12 \ L$. During this process,it absorbs $600 \ J$ of heat. Calculate $\Delta E$ of the system.

A
$-413 \ J$
B
$413 \ J$
C
zero
D
$-1613 \ J$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

Thermodynamics

Subtopic

First law of thermodynamics

At $298 \ K$,what is the value of $\Delta H - \Delta E$ in $J \ mol^{-1}$ when $1 \ mol$ of carbon monoxide gas is formed from its elements in their standard states? $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$

Medium

View Solution

Mark the correct statement.

Easy

View Solution

In a closed insulated container,a liquid is stirred with a paddle to increase its temperature. In this process,which of the following is true?

MediumAIPMT 2002

View Solution

For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ at $298 \ K$,the enthalpy change $\Delta H$ is $-92.38 \ kJ$. What is the value of $\Delta U$ in $kJ$ at $298 \ K$?

Medium

View Solution

$A$ gas occupies a volume of $2 \ L$ at $S.T.P.$ It absorbs $300 \ J$ of heat. If the pressure is $1 \ atm$,the volume becomes $2.5 \ L$. The change in internal energy $(\Delta U)$ for the process in $Joule$ is:

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

$A$ sample of gas expands against a constant external pressure of $1 \ atm$ from a volume of $2 \ L$ to $12 \ L$. During this process,it absorbs $600 \ J$ of heat. Calculate $\Delta E$ of the system.

Similar Questions

At $298 \ K$,what is the value of $\Delta H - \Delta E$ in $J \ mol^{-1}$ when $1 \ mol$ of carbon monoxide gas is formed from its elements in their standard states? $[R = 8.314 \ J \ K^{-1} \ mol^{-1}]$

Mark the correct statement.

In a closed insulated container,a liquid is stirred with a paddle to increase its temperature. In this process,which of the following is true?

For the reaction $N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)$ at $298 \ K$,the enthalpy change $\Delta H$ is $-92.38 \ kJ$. What is the value of $\Delta U$ in $kJ$ at $298 \ K$?

$A$ gas occupies a volume of $2 \ L$ at $S.T.P.$ It absorbs $300 \ J$ of heat. If the pressure is $1 \ atm$,the volume becomes $2.5 \ L$. The change in internal energy $(\Delta U)$ for the process in $Joule$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module