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Q.
The volume of a gas expands by $0.25\, m ^{3}$ at a constant pressure of $10^{3} N m ^{-2}$. The work done is equal to
Thermodynamics
Solution:
We know, work done, $w =- P _{ ex }+\Delta V$
Given, pressure is $10^{3}\, N m ^{-2}$
and, $\Delta V =0.25 \,m ^{3}$
$\therefore w=-10^{3} Nm ^{-2} \times 0.25 \,m ^{3}=-250\, J$
So, work done by the gas is $250\, J$.