300 mol of perfect gas occupies 13 L at 320 K. Calculate the work done in joules when the gas expands isothermally against a constant external pressure of 0.20 atm to 16 L.

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Solution

Given that
Number of moles of gas, $n = 300 m o l$
Initial volume, $V_{1} = 13 L$
Change in volume, $Δ V = 16 - 13 = 3 L$
Pressure, $P_{e x t} = 0.2 a t m$
Temperature, $T = 320 K .$

In an isothermal process against a constant external pressure of 0.20 atm, the process is irreversible.

$∴ P_{e x t} = 0.20 a t m, Δ V = 3 L$

$W = - P_{e x t} Δ V$
$= - 0.20 \times 3 = 0.6 L a t m$ ,
$= - 0.6 \times 101.3 J$
$= - 60.78 J$

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