Question

In the arrangement shown in figure, gas is thermally insulated. An ideal gas is filled in the cylinder having pressure P greater than atmospheric pressure $P_0$. The spring of force constant K is initially unstreteched. The piston of mass m and area s is frictionless. In equilibrium, the piston rises up a 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 $\frac{1}{2}K{x}_0^2+mg{x}_0+P_{0}s{x}_0$
• 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 $\frac{1}{2}K{x}_0^2+mg{x}_0+P_{0}s{x}_0$

Equilibrium of piston gives
$Ps=P_{0}S+K{x}_0+mg$
$P=P_0+\frac{K{x}_0}{s}+\frac{mg}{s}$
Work done by the gas
$W=P_0\Delta V+\frac{1}{2}K{x}_0^2+mg{x}_0=P_{0}s{x}_0+\frac{1}{2}K{x}_0^2+mg{x}_0$
Also, $W=-\Delta U$
Decrease in internal energy of the gas $=P_{0}s{x}_0+\frac{1}{2}K{x}_0^2+mg{x}_0$