Question

One mole of a gas is put under a weightless piston of a vertical cylinder at temperature T. The space over the piston opens into atmosphere. Initially piston was in equilibrium. How much should be performed by some force to increase isothermally the volume under the piston to twice the volume ? (Neglect friction of piston).The answer is RT (1 - In x) then x =

Answer 1

Let the pressure of the gas under the piston be $P$ and $A$ be the cross-section area of the piston.

$\therefore$ $F+PA=P_{o}A$

$⟹F=(P_o-P)A$

Also $V_1=V$ and $V_2=2V$ (given)

$\therefore$ $W={\int }_{V_1}^{V_2}Fdx$

OR $W={\int }_{V_1}^{V_2}(P_o-P)Adx=$ ${\int }_{V_1}^{V_2}(P_o-P)dV$

Using $PV=nRT$ $⟹P=\frac{RT}{V}$ $(\because n=1)$

$⟹$ $W={\int }_{V_1}^{V_2}{P}_{o}dV-$$RT{\int }_{V_1}^{V_2}\frac{dV}{V}$

OR $W=P_o(V_2-V_1)-RT$ $ln\frac{V_2}{V_1}$

OR $W=P_{o}V(2-1)-RT$ $ln\left(2\right)$

$⟹$ $W=RT[1-$ $ln\left(2\right)]$ $(\because P_{o}V=RT)$