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Q.

The state of an ideal gas is changed through an isothermal process at temperature $T_{0}$ as shown in the 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

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a

$\frac{P_{0}}{2}$

b

$\frac{P_{0}}{4}$

c

$\frac{P_{0}}{6}$

d

$\frac{P_{0}}{8}$

answer is D.

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Detailed Solution

$n R T_{0} \ln \frac{V_{C}}{V_{B}} = 2 n R T_{0} \ln \frac{V_{B}}{V_{A}} \frac{V_{C}}{V_{B}} = \frac{V_{B}^{2}}{V_{A}^{2}} \Rightarrow \frac{P_{B}}{P_{C}} = {(\frac{P_{A}}{P_{B}})}^{2} = P_{C} = \frac{P_{B}^{3}}{P_{A}^{2}} = \frac{{(\frac{P_{0}}{2})}^{3}}{P_{0}^{2}} = \frac{P_{0}}{8}$

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