Consider $1 \,kg$ of liquid water undergoing change in phase to water vapour at $100^{\circ} C$. At $100^{\circ} C$, the vapour pressure is $1.01 \times 10^5 \,N - m ^2$ and the latent heat of vaporization is $22.6 \times 10^5 \,Jkg ^{-1}$. The density of liquid water is $10^3 \,kg m ^{-3}$ and that of vapour is $\frac{1}{1.8} \,kg m ^{-3}$. The change in internal energy in this phase change is nearly ............ $\,J kg ^{-1}$
KVPY 2009, Advanced
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(b)
Change in volume of $1 \,kg$ of water due to phase change is
$\Delta V=V_{\text {steam }}-V_{\text {water }}$
$=\frac{m_{\text {seam }}-m_{\text {water }}}{\rho_{\text {steam }}}-\rho_{\text {water }}$
$=\frac{1}{(1 / 18)}-\frac{1}{1000}$
$=1800-0.001=1799 \,m ^3$
Work done against atmospheric pressure during phase change is
$\Delta W =p \Delta V$
$=101 \times 10^5 \times 1799$
$=181699 \,J$
Heat absorbed during phase change is
$\Delta Q =m L$
$=1 \times 22.6 \times 10^5$
$=2260000 \,J$
So, change in internal energy is
$\Delta U =\Delta Q-\Delta W$
$=2260000-181699$
$=2078301 J \,kg$
$=208 \times 10^5 \,J kg ^{-1}$
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