91Ó°ÊÓ

The variation of vapour pressure of water between $50°\mathrm{C}$ and $100°\mathrm{C}$.

Write the expression for the vapor pressure.

$P=K{e}^{-LJRT}$

Here, $P$is the vapour pressure, $K$is a constant, $L$is the latent heat, $R$is universal gas constant and $T$is the absolute temperature.

The values of latent heat of water at $50°\mathrm{C}$and $100°\mathrm{C}$are $42.92×{10}^3\mathrm{J}/\mathrm{mol}$and $40.66×{10}^3\mathrm{J}/\mathrm{mol}$. The value of latent heat for a temperature range between $50°\mathrm{C}$and $100°\mathrm{C}$is not constant.

Calculate the average value of latent heat.

$L=\frac{1}{2}(42.92+40.66)×{10}^3\mathrm{J}/\mathrm{mol}$

$=41.79×{10}^3\mathrm{J}/\mathrm{mol}$

$K=P{e}^{IIRT}$

Substitute $\left(0.1234×{10}^5\mathrm{Pa}\right)$for $P,\left(41.79×{10}^3\mathrm{J}/\mathrm{mol}\right)$for $L,(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K})$for $R$and $323\mathrm{K}$for $T$in the above expression.

$K=\left(0.1234×{10}^5\mathrm{Pa}\right)e^{\left(41.79×{10}^3\mathrm{J}/\mathrm{mol}\right)/\left(\right(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K}\left)\right(323\mathrm{K}\left)\right)}$

$=7.075×{10}^{10}\mathrm{Pa}$

A graph to show the variation of pressure with respect to temperature.

The estimated values of the boiling temperature of water for different altitudes.

Write the expression for the vapour pressure.

$P=K{e}^{-LKT}$

Here, $P$is the vapour pressure, $K$is a constant, $L$is the latent heat, $R$is universal gas constant and $T$is the absolute temperature.

Rearrange the above expression.

$T=\frac{L}{R\ln \left(\frac{K}{P}\right)}\mathrm{K}$

Convert the above temperature in centigrade unit.

$T_c=\left(\frac{L}{R\ln \left(\frac{K}{P}\right)}-273\right){}^{∘}\mathrm{C}$

Substitute $\left(41.79×{10}^3\mathrm{J}/\mathrm{mol}\right)$for $$L,(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K})$ for $K$and $0.844×{10}^5\mathrm{Pa}$for $P$for an altitude of $1430\mathrm{m}$in the above expression.

$T_c=\left(\frac{\left(41.79×{10}^3\mathrm{J}/\mathrm{mol}\right)}{(8.314\mathrm{J}/\mathrm{mol}·\mathrm{K})\ln \left(\frac{7.005×{10}^{10}{\mathrm{p}}_{\mathrm{a}}}{0.844×105\mathrm{P}2}\right)}-273\right)°\mathrm{C}$

localid="1651135640596" $=73.5^{o}C$

The variation of the dependency of boiling temperature of water on altitude.

Write the expression for the Clausius-Clapeyron equation.

$\frac{dP}{P}=\frac{L}{R}\frac{dT}{T^2}……...\text{(2)}$

Here, $\mathit{P}$is the pressure, $L$is the latent heat, $R$is the universal gas constant and $T$is the temperature.

Write the barometric equation.

$dP=-\frac{M_{B}P}{R{T}_b}dz……\left(3\right)$

Here, $M$ is the molar mass, $z$ is the value of the altitude and $g$ is the acceleration due to gravity.

Substitute $\left(-\frac{M_R}{K{T}_b}dz\right)$for $\left(\frac{dP}{P}\right)$in expression (2).

$-\frac{M_R}{R{T}_b}dz=\frac{L}{R}\frac{dT}{T^2}$

Rearrange the above expression.

$\frac{dT}{T^2}=-\frac{M_g}{T_{b}L}dz$

Integrate both sides of the above expression.

$\frac{1}{T}=-\frac{Mgz}{T_{b}L}+C_{…….….}\left(4\right)$

Here, $C$is the integrating constant.

At $z=0$the value of the boiling point is $373\mathrm{K}$.

Substitute $373\mathrm{K}$for $T$and 0 for $z$in expression (4).

$C=\frac{1}{373K}$

Substitute $0.029\mathrm{Kg}$for $M,9.8\mathrm{m}·{\mathrm{s}}^{-2}$for $T_b$and $\left(41.79×{10}^3\mathrm{J}/\mathrm{mol}\right)$for $L$and $2.681×{10}^{-3}{\mathrm{K}}^{-1}$for $C$in expression (4).

$T=\frac{1}{-\frac{\left(0.09{\mathrm{K}}_8\right)\left(9.8\mathrm{ms}{\mathrm{s}}^{-}\right)}{(2×3\mathrm{K})\left(41.79×{10}^3\mathrm{Nmol}\right)}z+\left(2.681×{10}^{-3}{\mathrm{K}}^{-1}\right)}$

$=-\frac{1}{\left(2.403×{10}^{-8}{\mathrm{m}}^{-1}{\mathrm{K}}^{-1}\right)z+\left(2.681×{10}^{-3}{\mathrm{K}}^{-1}\right)}$

A graph: