Question

An ideal gas, contained in a cylinder by a frictionless piston, is allowed to expand from volume $V_1$ at pressure $p_1$ to volume $V_2$ at pressure $p_2$. its temperature is kept constant throughout . the work done by the gas is :

• A. zero, because it obeys Boyles's law and therefore $P_2{V}_2-P_1{V}_1=0$
• B. negative, because the pressure has decreased and so the force on the piston has been diminishing
• C. zero, because it has been kept at constant temperature and so its internal energy is unchanged
• D. positive, because the volume has increased

Answer 1

The correct option is D positive, because the volume has increased

The expression for work done is given by,

$W=P\Delta V$

Given that

P= Constant

And volume is doubled from the initial position.

$W=P(2V-V)$

$=PV$

But, by Ideal gas equation $PV=nRT$

For 1 mole gas, $PV=RT$

$=2RT-RT=RT$.

$\Rightarrow W=RT$

$Thus\; workdone\; is\; positive\; because\; volume\; increased.$

$Hence\; Option\; D\; is\; right\; answer$