Question

In the arrangement shown in figure, gas is thermally insulated. An ideal gas is filled in the cylinder having pressure $P_0(>$ atmospheric pressure $P_a)$. 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 $x_0$, then

• A. final pressure of the gas is $P_0+\frac{K{x}_0}{S}+\frac{mg}{S}$
• B. work done by the gas is $\frac{1}{2}K{x}_0^2+mg{x}_0$
• C. decrease in internal energy of the gas is $\frac{1}{2}K{x}_0^2+mg{x}_0+P_{a}S{x}_0$
• D. all of the above

Answer 1

Answer: (A), (C)

The correct options are
A final pressure of the gas is $P_0+\frac{K{x}_0}{S}+\frac{mg}{S}$
C decrease in internal energy of the gas is $\frac{1}{2}K{x}_0^2+mg{x}_0+P_{a}S{x}_0$

Equilibrium of piston gives


$PS=P_{a}S+mg+K{x}_0$
$P=P_a+\frac{mg}{S}+\frac{K{x}_0}{S}$
$(P=\text{final pressure of gas})$
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
$=P_a\Delta V+\frac{1}{2}K{x}_0^2+mg{x}_0=P_{a}S{x}_0+\frac{1}{2}K{x}_0^2+mg{x}_0$
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.