0.50 mol of an ideal gas initially at a temperature of 300 K and at a pressure of 2 atm is expanded isothermally in three steps. In each step, the pressure is dropped suddenly and held constant until equilibrium is reestablished. The pressure at each of the three stages of expansion are 1.6, 1.2 and 1 atm. Calculate the work done ( |w| ) (in atm-litre) in this
process. [Use R = 0.08 atm-litre/mol.K]

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2.6 atm litre

7.4 atm litre

6.2 atm litre

4.7 atm litre

answer is A.

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Detailed Solution

$P_{i} V_{i} = n R T \Rightarrow V_{i} = \frac{n R T}{P_{i}} = \frac{(0.5) (0.08) 300}{(2)} = 6 L i n i t i a l$
As T is constant, $P V$ is also constant at the end of each step.
$P_{i} V_{i} = P_{1} V_{1} = P_{2} V_{2} = P_{3} V_{3} \Rightarrow V_{1} = \frac{P_{i} V_{i}}{P_{1}}, V_{2} = \frac{P_{i} V_{i}}{P_{2}}, V_{3} = \frac{P_{i} V_{i}}{P_{3}}$
$∴ W_{1} = - P_{1} (V_{1} - V_{i}) = - (P_{i} - P_{1}) V_{i}$
Similarly,
$∴ W_{2} = - P_{2} (V_{2} - V_{1}) = - (P_{1} - P_{2}) V_{1} ∴ W_{3} = - P_{3} (V_{3} - V_{2}) = - (P_{2} - P_{3}) V_{2} \begin{array}{l} ∴ W = (2 - 1.6) (6) + (1.6 - 1.2) (\frac{12}{1.6}) + (1.2 - 1) (\frac{12}{1.2}) \\ = (0.4) (6) + (0.4) (7.5) + (0.2) (10) \\ = 2.4 + 3 + 2 = 7.4 a t m - L \end{array}$