91Ó°ÊÓ

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

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

The table shows the temperature and pressure

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

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

The partial pressure of water vapour is

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

Substituting the values

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

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,

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

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

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

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