Question

Effect of altitude on boiling water.

(a) Use the result of the previous problem and the data in Figure 5.11 to plot a graph of the vapor pressure of water between $50°\mathrm{C}$ and $100°\mathrm{C}$. How well can you match the data at the two endpoints?

(b) Reading the graph backward, estimate the boiling temperature of water at each of the locations for which you determined the pressure in Problem 1.16. Explain why it takes longer to cook noodles when you're camping in the mountains.

(c) Show that the dependence of boiling temperature on altitude is very nearly (though not exactly) a linear function, and calculate the slope in degrees Celsius per thousand feet (or in degrees Celsius per kilometer).

Short answer

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

b). The estimate values of the boiling temperature of water for different altitudes is $73.5^{o}C$.

c). A graph to show the variation of the altitude with pressure.

Step-by-step solution

Part (a) Step 1: Given Information

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

Part (a): Step 2: Explanation

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 $42.92\times {10}^3\mathrm{J}/\mathrm{mol}$and $40.66\times {10}^3\mathrm{J}/\mathrm{mol}$. 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.

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

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

Part (a) Step 3: Explanation

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

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

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

$=7.075\times {10}^{10}\mathrm{Pa}$

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

Part (b) Step 4: Given Information

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

Part (b) Step 5: Explanation

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.

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

Part (b) Step 6: Explanation 

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

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

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

Part (c) Step 7: Given Information

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

Part (c) Step 8: Explanation

Write the expression for the Clausius-Clapeyron equation.

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

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.

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

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).

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

Part (c) Step 9: Explanation

Rearrange the above expression.

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

Integrate both sides of the above expression.

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

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 $\left(41.79\times {10}^3\mathrm{J}/\mathrm{mol}\right)$for $L$and $2.681\times {10}^{-3}{\mathrm{K}}^{-1}$for $C$in expression (4).

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

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

A graph: