$10\mathrm{.}4$ Examine the following experimental data for water in the light of equations $(10\mathrm{.}12)$ and $(10\mathrm{.}16).$

$\backslash$table$[[$temperature $\frac{?}{?}C,0,5,10,15,20],[$vapour pressure $\frac{?}{mmHg},4\mathrm{.}581,6\mathrm{.}536,9\mathrm{.}198,12\mathrm{.}77,17\mathrm{.}51],[$temperature $\frac{?}{?}C,25,30,35,40,],[$vapour pressure $\frac{?}{mmHg},23\mathrm{.}73,31\mathrm{.}79,42\mathrm{.}14,55\mathrm{.}29,],[$For water vapour, $c_p=1\mathrm{.}9kJ{K}^{-1}k{g}^{-1}.,,,,,]]$

$10\mathrm{.}4.$ Integration of the Clausius$-$Clapeyron equation

It is sometimes useful to have an explicit functional form for the relationship between vapour pressure and temperature. The Clausius$-$Clapeyron equation itself is exact. By making approximations which are reasonable under certain conditions it is possible to integrate it to obtain an explicit approximate expression for the vapour pressure.

If the pressure is not too high and we are not too near to the critical point, it is reasonable to assume that the vapour obeys the perfect gas law and that the specific volume of the condensed phase is negligible in comparison with that of the vapour. With these assumptions the Clausius$-$Clapeyron equation becomes

$\frac{dp}{(d)T}=\frac{Lp}{R{T}^2}$

Applying again the condition that $V_v{V}_c,$ using the perfect gas law and substituting from $(10\mathrm{.}12),$

$\frac{d}{dT}(\frac{L}{T})=\frac{C_{pv}-C_{pc}}{T}-\frac{L}{T^2}$

$dL=(C_{pv}-C_{pc})dT.$