Question

Calculate $w$ and $\Delta U$ for the conversion of $0.5$ mole of water at ${100}^{o}C$ to steak at $1$ atm pressure. Heat of vaporisation of water at ${100}^{o}C$ is $40670J{mol}^{-1}$.

Answer 1

Volume of $0.5$ mole of steam at $1$ atm pressure
$=\frac{nRT}{P}=\frac{0.5\times 0.0821\times 373}{1.0}=15.3L$
Change in volume $=$ Vol. of steam $-$ Vol. of water
$=15.3-$ negligible $=15.3L$
Work done by the system,
$w=P_{ext}\times volumechange$
$=1\times 15.3=15.3litre-atm$
$=15.3\times 101.3J=1549.89J$
$w$ should be negative as the work has been done by the system on the surroundings
$w=-1549.89J$
Heat required to convert $0.5$ mole of water at ${100}^{o}C$ to steam
$=0.5\times 40670J=20335J$
According to the first law of thermodynamics
$\Delta U=q+w=20335-1549.89=18785.11J$