The state of an ideal gas is changed through an isothermal process at temperature $${T_0}$$ as shown in figure. The work done by gas in 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 $$\frac{{{P_0}}}{2}$$ then the pressure of the gas in state $$C$$ is
A.
$$\frac{{{P_0}}}{2}$$
B.
$$\frac{{{P_0}}}{4}$$
C.
$$\frac{{{P_0}}}{6}$$
D.
$$\frac{{{P_0}}}{8}$$
Answer :
$$\frac{{{P_0}}}{8}$$
Solution :
Work done by gas in going isothermally from state $$A$$ to $$B$$ is
$$\Delta {W_{AB}} = nRT\ln \frac{{{P_A}}}{{{P_B}}} = nRT\ln 2\,.....\left( {\text{i}} \right)$$
Work done by gas in going isothermally from state $$B$$ to $$C$$ is
$$\Delta {W_{BC}} = nRT\ln \frac{{{P_B}}}{{{P_C}}} = nRT\frac{{{P_0}}}{{2{P_C}}}\,.....\left( {{\text{ii}}} \right)$$
It is given that $$\Delta {W_{BC}} = 2\Delta {W_{AB}}$$
$$\eqalign{
& \ln \frac{{{P_0}}}{{2{P_C}}} = \ln {\left( 2 \right)^2} \cr
& \therefore {P_C} = \frac{{{P_0}}}{8} \cr} $$
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