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Q.

Under isothermal condition for one mole of ideal gas what is the ratio of work done under reversible to irreversible process, initially held at 20 atm undergoes expansion from 1L to 2L, at 298K, under external pressure of 10 atm?

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a

1.7

b

1.4

c

1.0

d

2.0

answer is A.

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Detailed Solution

$\begin{aligned} - W_{irreversible} = P_{ext} (V_{2} - V_{1}) \\ = 10 atm (2 L - 1 L) \\ = 10 atm - L \end{aligned} - W_{reversible} = \int_{V_{1}}^{V_{2}} P_{ex} dv = 2 .303 nRT \log \frac{V_{2}}{V_{1}} = 1 \times 2 .303 \times 0 .0821 atm - L / K / mol \times \log \frac{2}{1} = 16 .96 atm-L \frac{W_{reversible}}{W_{irreversible}} = \frac{16 .96}{10 .00} = 1 .69 \approx 1 .7$

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