Consider the vaporization of 1g of water at `100^(@)C` at one atmosphere pressure. Compute the work done by the water system in the vaporization and change internal energy of the system.

Explanation

To change a system of mass m of liquid to vapour heat required is `Q=ml_("vapour")`
The process takes place at constant pressure so the work done by the system is the work
Where `DeltaV=(V_("vapour")-V_("liquid"))`
From first law of thermodynamics
`DeltaU=Q-W=mL_(v)-P(V_("vapour"-V_("liquid")))`
Latent heat of vaporization of water
`L_("vapour")=22.57xx10^(5)J//kg`
`Q=(1.00xx10^(-3))(22.57xx10^(5))=2.26xx10^(3)`
`because` No. of moles=weight /gram molecular weight
`therefore` Moles of water in `1g=(1)/(18)=0.0556` mole
`V_("vapour")=(nRT)/(P)=((0.0556)(8.315)(373))/(1.013xx10^(5))=1.70xx10^(-3)m`
The density of water is
`1.00xx10^(3) kg//m^(3)=1.00 g//cm^(3)`
`V_("liquid")=1.00xx10^(-6) m^(3)`
Thus the work done by the water system
Vaporization is `W=P(V_("vapour")-V_("liquid"))`
`=(1.013xx10^(5))(1.70xx10^(-3)-1.00xx10^(-6))=172J`
The work done by the system is positive since the volume of the system has increased. From first law,
`DeltaU=Q-QrArrDeltaU=2.26xx10^(3)-172=2.09xx10^(3)J`