Question

$3$ moles of an ideal gas are contained within a cylinder with a frictionless piston and are initially at temperature $T$. The pressure of the gas remains constant, while its volume gets doubled. If $R$ is the molar gas constant, the work done by the gas in this process is

• A. $\frac{3}{2}RT\ln 2$
• B. $3RT\ln 2$
• C. $\frac{3}{2}RT$
• D. $3RT$

Answer 1

The correct option is D $3RT$

Given :
Number of moles of an ideal gas $n=3$
Final volume $V_2=2V_1$

Also given that, pressure is constant i.e isobaric process
Work done in isobaric process is
$W=P(V_2-V_1)$
$W=P(2V_1-V_1)=P{V}_1$

From ideal gas equation,
$P{V}_1=nRT$
Hence, $W=P{V}_1=nRT=3RT$