Question

The state of an ideal gas is changed through an isothermal process at temperature $T_o$ as shown in figure. The work done by the gas going from state $B$ to $C$ is double the work done by gas in going from state $A$ to $B$. If the pressure in the state $B$ is $\frac{P_o}{2}$, then the pressure of the gas in state $C$ is

• A. $\frac{P_o}{3}$
• B. $\frac{P_o}{4}$
• C. $\frac{P_o}{6}$
• D. $\frac{P_o}{8}$

Answer 1

The correct option is D $\frac{P_o}{8}$

From indicator diagram ($P-V$ diagram), For an isothermal process, both pressure and volume vary.

In the diagram, pressure variation is given
$P_i=P_o;P_f=\frac{P_0}{2}$
$\therefore$ Workdone $\left(W_{AB}\right)=nR{T}_o\ln \left(\frac{P_i}{P_f}\right)$
$=nR{T}_o\ln \left(2\right)$

To find $W_{BC};P_i=\frac{P_o}{2},P_f=P_c$
$W_{BC}=nR{T}_o\ln \left(\frac{P_o}{2P_c}\right)$
given that,
$W_{BC}=2W_{AB}$
$nR{T}_o\ln \left(\frac{P_o}{2P_c}\right)=2nR{T}_o\ln \left(2\right)$
$\Rightarrow \frac{P_o}{2P_c}=4$
$\therefore P_c=\frac{P_o}{8}$