When 1gm of water changes from liquid to vapour phase at constant pressure of 1 atmosphere, the volume increase from 1 $c{m}^3$ to 1671 $c{m}^3$. The heat of vaporisation at this pressure is 540 cal/gm. Find (a) the work done (in J) in change of phase and (b) increase in internal energy of water.(1 atmosphere = $1.01\times {10}^{6}dyne/c{m}^2)$

The correct option is B 168.67J, 2099.33J
(a) As the process is isobaric
$\Delta W=\int PdV=P[V_F-V_I]i.e.=1.01\times {10}^6[1671-1]=1686.7\times {10}^{6}ergs$
or, = 168.67J (= 40 cal)
[as 1 erg = ${10}^{-7}$J]
(b) According to first law of thermodynamics
$\Delta Q=\Delta U+\Delta W$
Here =mL=1 $\times$ 540cal=2268J
[as 1 cal = 4.2J]
So, $\Delta U=\Delta Q+\Delta W$
= 2268 - 168.67
= 2099.33J