An ideal gas is subjected to an isothermal expansion such that its volume changes from $V_i$ to $V_f$ and pressure from $P_i$ to $P_f$. The work done on the gas is:

  • A
    $W = nRT \ln(V_f/V_i)$
  • B
    $W = -nRT \ln(V_f/V_i)$
  • C
    $W = nRT \ln(P_f/P_i)$
  • D
    $W = -nRT \ln(P_f/P_i)$

Explore More

Similar Questions

If an ideal gas is compressed isothermally,then:

$A$ perfect gas of volume $5 \ L$ is compressed isothermally to a volume of $1 \ L$. The $r.m.s.$ speed of the molecules will

Two identical containers $A$ and $B$ with frictionless pistons contain the same ideal gas at the same temperature and the same volume $V$. The mass of the gas in $A$ is ${m_A}$ and that in $B$ is ${m_B}$. The gas in each cylinder is now allowed to expand isothermally to the same final volume $2V$. The changes in the pressure in $A$ and $B$ are found to be $\Delta P$ and $1.5 \Delta P$ respectively. Then:

In an isothermal process,which statement is wrong?

$Assertion :$ The isothermal curves intersect each other at a certain point.
$Reason :$ The isothermal change takes place slowly,so,the isothermal curves have very little slope.

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