If the atmospheric pressure is 535 mm of mercury, find the temperature at which water will boil. Latent heat of vaporisation of water is 545.5 cal/g.

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[latex]\begin{array}{l}P_{1}=535 \mathrm{~mm} \mathrm{~Hg} \quad ; \quad P_{2}=1 \mathrm{~atm}=760 \mathrm{~mm} \mathrm{~Hg} \ \ \mathrm{T}_{1}=? \quad ; \quad \mathrm{T}_{2}=100+273=373 \mathrm{~K}\ \ \Delta \mathrm{H}_{\mathrm{v}}=545.5 \mathrm{~cal} / \mathrm{g}=545.5 imes 18 \mathrm{~cal} / \mathrm{mol}=9819 \mathrm{~cal} \mathrm{~mol}^{-1} \ \ \mathrm{R}=1.987 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \ \ \log \frac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\frac{\Delta \mathrm{H}_{\mathrm{V}}}{2.303 \mathrm{R}}\left[\frac{\mathrm{T}_{2}-\mathrm{T}_{1}}{\mathrm{~T}_{2} \mathrm{~T}_{1}} ight] \ \ \log \frac{760}{535}=\frac{9819 \mathrm{~cal} \mathrm{~mol}^{-1}}{2.303 imes 1.987 \mathrm{~cal} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}}\left[\frac{1}{\mathrm{~T}_{1}}-\frac{1}{373} ight] \ \ \frac{1}{\mathrm{~T}_{1}}-\frac{1}{373}=\frac{0.1524 imes 2.303 imes 1.987}{9819}=0.00007102\ \ ext { or } \frac{1}{\mathrm{~T}_{1}}-0.002681 \mathrm{~K}^{-1}=0.00007102 \mathrm{~K}^{-1} \ \ \mathrm{~T}_{1}=\frac{1}{0.00275202}=363.37 \mathrm{~K} \end{array}[/latex]

Question 15.10.2: If the atmospheric pressure is 535 mm of mercury, find the t......

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