1.00kg of liquid water at 100‌∘C undergoes a phase change into steam at 100‌∘C at 1.0atm (take it to be 1.00×105Pa ). The initial volume of the liquid water was 1.00×10−3m3 which is changed to 2.001m3 of steam. Find the change in the internal energy of the system. [Use heat of vaporization ≃2000kJ∕kg ]
We know, from first law of thermodynamics, ∆Q=∆U+W‌‌‌⋅⋅⋅⋅⋅⋅⋅(i) Given, volume of liquid, Vl=10−3m3 And volume of gas or volume of steam, Vg=200mm3 Now, Vg≫Vl ∴‌‌(Vg−Vl)≈Vg W=pVg=(105×2.001)J=2.001×102‌kJ=200‌kJ Now, heat required for phase change to convert the liquid into steam, ∆Q=mL=1×2000‌kJ. Put in Eq (i), ∆U=∆Q−W ⇒‌‌∆U=(2000−200)‌kJ‌‌ [from Eq. (i)] ⇒‌‌∆U=1800‌kJ