Ordinarily, the partial pressure of water vapour in the air is less than the equilibrium vapour pressure at the ambient temperature; this is why a cup of water will spontaneously evaporate. The ratio of the partial pressure of water vapour to the equilibrium vapour pressure is called the relative humidity. When the relative humidity is 100%, so that water vapour in the atmosphere would be in diffusive equilibrium with a cup of liquid water, we say that the air is saturated. The dew point is the temperature at which the relative humidity would be 100%, for a given partial pressure of water vapour.

(a) Use the vapour pressure equation (Problem 5.35) and the data in Figure 5.11 to plot a graph of the vapour pressure of water from 0°C to 40°C. Notice that the vapour pressure approximately doubles for every 10° increase in temperature.

(b) Suppose that the temperature on a certain summer day is 30° C. What is the dew point if the relative humidity is 90%? What if the relative humidity is 40%?

Using the result of problem 5.35, the vapour pressure equation is as follows:

$$\begin{gathered} P=\left(\text{constant}\right)e^{\frac{-L}{RT}} \\ =D{e}^{\frac{-L}{RT}} \end{gathered}$$

Where,

D is the constant

L is the latent heat of vaporisation

R is universal gas constant

T is the temperature

Rearranging for D and substituting the values

Substituting the value of D and solving for P

$P=\left(1.626\times {10}^6\text{bar}\right)e^{\frac{L}{RT}}$

The table shows the temperature and pressure

$\begin{array}{|ll|}\hline T\text{in}{}^{\circ }\mathrm{C}& P=\left(1.626\times {10}^6\mathrm{bar}\right)e^{\frac{L}{RT}}\\ 0& 0.006237\\ 10& 0.01237\\ 20& 0.023413\\ 30& 0.042487\\ 40& 0.07422\\ \hline\end{array}$

The graph shows the variation of vapour pressure of water from 0°C to 40°C

The partial pressure of water vapour is

$P=\left(1.626\times {10}^6\text{bar}\right)e^{\frac{L}{RT}}$

Substituting the values

role="math" localid="1646942082761" $$\begin{gathered} P=\left(1.626\times {10}^6\mathrm{bar}\right)e^{\frac{(43990\mathrm{J}/\mathrm{mol})}{(8.315\mathrm{J}/\mathrm{mol}·\mathrm{K})\left(\right(30+273\left)\mathrm{K}\right)}} \\ P=0.043bar \end{gathered}$$

The pressure on the day when relative humidity is 90% is

$$\begin{gathered} P=(0.043\text{bar})\left(\frac{90}{100}\right) \\ P=0.038bar \end{gathered}$$

Substituting the values,

$$\begin{gathered} P=\left(1.626\times {10}^6\text{bar}\right)e^{\frac{L}{R{T}_{dw}}}\text{to solve for}{T}_{\text{dew}} \\ 0.038\text{bar}=\left(1.626\times {10}^6\text{bar}\right)e^{\frac{43990\mathrm{J}/\mathrm{mol}}{(8.315\mathrm{J}/\mathrm{molK}){\tau }_{\mathrm{de}w}}} \\ {e}^{\frac{43990\mathrm{J}\mathrm{mol}}{(8315\mathrm{J}/\mathrm{mol}.\mathrm{K}){\mathrm{Tdec}}_{\mathrm{de}}}}=\frac{0.038\mathrm{bar}}{1.626\times {10}^6\mathrm{bar}} \\ {T}_{\text{dew}}=301.1\mathrm{K} \\ {\mathrm{T}}_{\mathrm{dew}}=(301.1-273)\mathrm{K} \\ {\mathrm{T}}_{\mathrm{dew}}=28.1°\mathrm{C} \end{gathered}$$

As a result, when the relative humidity is 90%, the dew point is 28°C.

Water vapour has a partial pressure of 0.043 bar. The pressure on that day is as follows: Because the relative humidity is 40%, the pressure is as follows:

$$\begin{gathered} P=(0.043\text{bar})\left(\frac{40}{100}\right) \\ P=0.017bar \end{gathered}$$

Substitute the values in

$$\begin{gathered} P=\left(1.626\times {10}^6\text{bar}\right)e^{\frac{L}{R{T}_{dx}}} \\ 0.017\text{bar}=\left(1.626\times {10}^6\text{bar}\right)e^{\frac{43990/\mathrm{mol}}{(8.315\mathrm{J}/\mathrm{mol}.\mathrm{K})T_{\mathrm{dec}}}} \\ {e}^{\frac{43990\mathrm{J}/\mathrm{mol}}{(8.315\mathrm{Jmol}.\mathrm{K})r_{\mathrm{doc}}}}=\frac{0.017\mathrm{bar}}{1.626\times {10}^6\mathrm{bar}} \\ {T}_{\text{dew}}=287.9\mathrm{K} \\ {\mathrm{T}}_{\mathrm{dew}}=(287.9-273)\mathrm{K} \\ {\mathrm{T}}_{\mathrm{dew}}=14.9°\mathrm{C} \end{gathered}$$

As a result, when the relative humidity is 40%, the dew point is 14.9°C.