Question

Two moles of a gas at 8.21 bar and 300 K are expanded at constant temperature up to 2.73 bar against a constant pressure of 1 bar. How much work (in Latm) is done by the gas?

(neglect the sign)

Answer 1

$V_{1}=\frac{2 \times 0.0821 \times 300}{8.21}=6.0 \, L$

$V_{2}=\frac{2 \times 0.0821 \times 300}{2.73}=18.0 \, L$

$W_{i r r , i s o t h e r m a l}=-P_{e x t}\times \Delta V=-1\times \left(\right.18-6\left.\right)$

$=-12 \, L \, atm$