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Real Gas Equation

n moles of a ...

Question

$n$ moles of a perfect gas undergoes a cyclic process $A B C A$ (see figure) consisting of the following processes -

$A \to B$ : Isothermal expansion at temperature $T$ so that the volume is doubled from $V_{1}$ to $V_{2}$ and pressure changes from $P_{1}$ to $P_{2}$ .

$B \to C$ : Isobaric compression at pressure $P_{2}$ to initial volume $V_{1}$ .

$C \to A$ : Isochoric change, leading to change of pressure from $P_{2}$ to $P_{1}$ .

Total work done in the complete cycle $A B C A$ is :

0

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n R T (ln 2 + \frac{1}{2})

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n R T (ln 2)

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n R T (ln 2 - \frac{1}{2})

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Solution

The correct option is D $n R T (ln 2 - \frac{1}{2})$
Given:
$A \to B$ : Isothermal process
$B \to C$ : Isobaric process
$C \to A$ : Isochoric process

Also, $V_{2} = 2 V_{1}$
So, $P_{2} = \frac{P_{1}}{2}$ as $A B$ is an Isothermal process.

Work done by gas in the complete cycle $A B C A$ is

$W = W_{A B} + W_{B C} + W_{C A} . . . . . (1)$

Now,
$W_{C A} = 0$ as Isochoric Process.

$W_{A B} = P_{1} V_{1} ln (\frac{V_{2}}{V_{1}}) = n R T ln 2$

$W_{B C} = P_{2} (V_{1} - V_{2}) = P_{2} (V_{1} - 2 V_{1}) = - P_{2} V_{1} = - \frac{n R T}{2}$

Now, putting the values of $W_{A B}, W_{B C}$ and $W_{C A}$ in equation $(1)$ , we get :

$W = n R T ln 2 - \frac{n R T}{2} + 0$
$\Rightarrow W = n R T (ln 2 - \frac{1}{2})$

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