• # question_answer Unit mass of a liquid with volume ${{V}_{1}}$ is completely changed into a gas of volume ${{V}_{2}}$ at a constant external pressure P and temperature T. If the latent heat of evaporation for the given mass is L, then the increase in the internal energy of the system is     A) Zero      B) $P({{V}_{2}}-{{V}_{1}})$ C) $L-P({{V}_{2}}-{{V}_{1}})$ D) L

 [c] $\Delta Q=\Delta V+P\Delta V$Þ mL = DU + P(V2  V1) Þ DU = L  P (V2  V1)             ($\because$ m = 1)