Question

A thermally insulated chamber of volume $2V_0$ is divided by light and frictionless piston of area S into two equal parts A and B. Part A has an ideal gas at pressure $P_0$ and temperature $T_0$ and in part B is the vacuum. A massless spring of force constant K is connected with the piston and the wall of the container as shown. Initially, the spring is unstretched. The gas in chamber A is allowed to expand. Let in equilibrium the spring be compressed by $z_0$ Then.

• A. final pressure of the gas is$\frac{K{x}_0}{S}$
• B. work done by the gas is $\frac{1}{2}K{x}_0^2$
• C. change in internal energy of the gas is $-\frac{1}{2}K{x}_0^2$
• D. temperature of the gas is decreased

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

final pressure of the gas is$\frac{K{x}_0}{S}$

B

work done by the gas is $\frac{1}{2}K{x}_0^2$

C

change in internal energy of the gas is $-\frac{1}{2}K{x}_0^2$

D

temperature of the gas is decreased

Equilibrium of piston gives

PS = $K{x}_0$ or P = $\frac{K{X}_0}{S}$

Since the chamber is thermally insulated. $\Delta Q=0$; $\therefore$ Elastic P.E. of springs = Work done by gas or Work done by gas = $\frac{1}{2}K{x}_0^2$

This work is done at the expense of internal energy of gas. Therefore, internal energy of the gas is decreased by $\frac{1}{2}K{x}_0^2$

Internal energy of an ideal gas depends on its temperature only.

Internal energy of the gas is decreasing. Therefore, temperature of the gas will decrease.