Question

A vertical cylinder with heat-conducting wall is closed at the bottom and is filled with a smooth light piston. It contain one mole of an ideal gas. The temperature of the gas is always equal to the surrounding;s temperature $T_o$. The piston is moved up slowly to increase the volume of the gas to $\eta$ timers. Which of the following is incorrect?

• A. Work done by the gas is $R{T}_o$ in $\eta$.
• B. Work done against the atmosphere is $R{T}_o(\eta -1)$
• C. There is no change in the internal energy of the gas.
• D. The final pressure of the gas is $\frac{1}{(\eta -1)}$ times its initial pressure.

Answer 1

The correct option is D The final pressure of the gas is $\frac{1}{(\eta -1)}$ times its initial pressure.

Work Done by gas (isothermal process) :

$W=nRT\frac{V_2}{V_1}=R{T}_{0}ln\left(n\right)$

Work Done against Atmosphere (Pressure of atmosphere is constant P & Volume is decreased):

$W=P\Delta V=P(nV-V)=PV(n-1)=R{T}_0(n-1)$

As there is no change in temperature hence no change is Internal Energy (Always in synchronization with atmosphere).

Ideal Gas Equation:

$PV=P^{\prime }{V}^{\prime }$

$P^{\prime }\left(nV\right)=PV$

$P^{\prime }=\frac{P}{n}$