Calculate the maximum work when 24 g of oxygen are expanded isothermally and reversibly from a pressure of $1.6 \times 10^{5}$ Pa to 100 kPa at 298 K.

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Solution

Solution:-
Given weight $= 24 g$
Molecular weight $= 32 g$
No. of moles $(n) = \frac{24}{32} = 0.75 mol$
As we know that work done in an isothermal process is given as-
$W = - 2.303 n R T log \frac{P_{1}}{P_{2}}$
Given:-
$T = 298 K$
$P_{1} = 1.6 \times 10^{5} P a = 160 k P a$
$P_{2} = 100 k P a$
Therefore,
$W = - 2.303 \times 0.75 \times 0.0821 \times 298 \times log \frac{160}{100}$
$\Rightarrow W = - 42.26 \times log 1.6 = - 8.62 a t m - L$

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