First slide

Work

Question

At 298K a gas in state ‘A’ (P = 20 atm, V = 1 l) changes to state ‘B’ (P = 2atm, V = 10 l). Workdone if carried out reversibly is

Moderate

Solution

At constant temperature $P_{1} V_{1} = P_{2} V_{2}$
$PV = Constant$
$PV = nRT$
$W_{rev} = - 2 .303 nRTlog \frac{V_{2}}{V_{1}}$
=- $2.303 PV$ $\log \frac{V_{2}}{V_{1}}$
= $- 2.303 (20)$ $\log \frac{10}{1}$
$= - 46.06 L . atm$

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