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$1\,\, gm$ water at $100\,^oC$and $10^5\,\,Pa$ pressure converts into $1841\,\,cm^3$ of steam at constant temperature and pressure. If latent heat of vapourization of water is $2250 \,\,J/gm$. The change in internal energy of water in this process is ..... $J$
$0$
$2250$
$2066$
none
Solution
mass of water $=1 gm =10^{-3}\,kg$
Pressure $=10^5 Pa =10^5\,N – m ^2$
volume of steam $=1841\,cm ^3=1841 \times 10^{-6}\,m ^3$
Latent heat of $=2250\,J / gm$ vahorization
$\because \quad$ volume of water $=\frac{\text { mass }}{\text { density }} \quad$ (Swater $=10^3 \frac{ g }{ m }$ )
$=\frac{10^{-3}}{10^3}=10^{-6} m ^3$
Change in internal energy is given by,
$d u =d \theta-p d v$
$d u =m L-P\left(V_b-t_i\right)$
$=10^{-3}\left(2250 \times 10^3\right)-10^5\left(1841 \times 10^{-6}-1 \times 10^{-6}\right.$
$=2250-184$
$=2066\,J$
So change in internal energy, is $2066\,J$.
