$$1\,gm$$  of water at a pressure of $$1.01 \times {10^5}\,Pa$$   is converted into steam without any change of temperature. The volume of $$1\,g$$  of steam is $$1671\,cc$$   and the latent heat of evaporation is $$540\,cal.$$  The change in internal energy due to evaporation of $$1\,gm$$  of water is

Solution :
$$\eqalign{ & dW = P\Delta V = 1.01 \times {10^5}\left[ {1671 - 1} \right] \times {10^{ - 6}}\,Joule \cr & = \frac{{1.01 \times 167}}{{4.2}}cal. = 40\,cal.\,{\text{nearly}} \cr & \Delta Q = mL = 1 \times 540, \cr & \Delta Q = \Delta W + \Delta Q \cr & {\text{or}}\,\,\Delta U = 540 - 40 = 500\,cal. \cr} $$