A vertical cylinder with heat-conducting wall | vedclass

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(D) Since the walls are heat-conducting and the process is slow,the temperature remains constant at $T_0$. This is an isothermal process.$1$. For an i

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The final pressure of the gas is $\frac{1}{\eta}$ times its initial pressure.

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(D) Since the walls are heat-conducting and the process is slow,the temperature remains constant at $T_0$. This is an isothermal process.$1$. For an isothermal process,$\Delta U = 0$ (Internal energy change is zero).$2$. Work done by the gas $W = \int P \, dV = \int_{V_0}^{\eta V_0} \frac{RT_0}{V} \, dV = RT_0 \ln \eta$.$3$. The piston is light,so the pressure of the gas is always equal to the atmospheric pressure $P_{atm}$. Since $PV = RT_0$,$P = \frac{RT_0}{V}$. Initially $P_i = \frac{RT_0}{V_0}$ and finally $P_f = \frac{RT_0}{\eta V_0} = \frac{P_i}{\eta}$. Thus,the final pressure is $\frac{1}{\eta}$ times the initial pressure.$4$. Work done against the atmosphere $W_{atm} = P_{atm} \Delta V = P_i (\eta V_0 - V_0) = P_i V_0 (\eta - 1) = RT_0(\eta - 1)$.Comparing these with the options,option $D$ states the final pressure is $\frac{1}{(\eta - 1)}$ times the initial pressure,which is incorrect.

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