In the arrangement shown in the given figure, gas is thermally insulated. An ideal gas is filled in the cylinder having pressure P₀ (> atmospheric pressure ${\mathrm{P}}_{\mathrm{a}}$). The spring of force constant K is initially unstretched. The piston of mass m and area S is frictionless. In equilibrium, the piston rises up by distance x₀, then

Equilibrium of piston gives

$\begin{array}{l}\mathrm{PS}={\mathrm{P}}_{\mathrm{a}}\mathrm{S}+\mathrm{mg}+{\mathrm{Kx}}_0\\ \mathrm{P}={\mathrm{P}}_{\mathrm{a}}+\frac{\mathrm{mg}}{\mathrm{S}}+\frac{{\mathrm{Kx}}_0}{\mathrm{S}}\\ (\mathrm{P}=\text{ final pressure of gas })\end{array}$

Work done by the gas: work done against atmospheric pressure + elastic potential energy stored in the spring + increase in gravitational potential energy of the piston

$={\mathrm{P}}_{\mathrm{a}}\mathrm{\Delta V}+\frac{1}{2}{\mathrm{Kx}}_0^2+{\mathrm{mgx}}_0={\mathrm{P}}_{\mathrm{a}}{\mathrm{Sx}}_0+\frac{1}{2}{\mathrm{Kx}}_0^2+{\mathrm{mgx}}_0$

There occurs decrease in internal energy of the gas, because the gas is thermally insulated and hence, work is done at the expense of internal energy of the gas.