Question

Boiling water: Suppose 1.0 g of water vaporizes isobarically at atmospheric pressure `(1.01xx10^5 Pa)`. Its volume in the liquid state is `V_i=V_(liqu i d)=1.0cm^3` and its volume in vapour state is `V_f=V_(vapour)=1671 cm^3`. Find the work done in the expansion and the change in internal energy of the system. Ignore any mixing of the stream and the surrounding air. Take latent heat of vaporization `L_v=2.26xx10^6J//kg`.

Answer 1

Correct Answer - A::B
Because the expansion takes place at constant pressure, the work done is
`W=int_(V_i)^(V_f)p_0dV=p_0int_(V_i)^(V^f)dV=p_0(V_f-V_i)`
`=(1.01xx10^5)(1671xx10^-6-1.0xx10^-6)`
`=169J`
`Q=mL_v=(1.0xx10^-3)(2.26xx10^6)`
`=2260J`
Hence, from the first law, the change in internal energy
`DeltaU=Q-W=2260-169`
`=2091J`