Chapter 13, Problem 66P

To determine

To Calculate:The temperature of a day on relative humidity $40\%$ and partial pressure $530\text{Pa}$ .

Answer to Problem 66P

Temperature of the waterat $40\%$ relative humidity is ${11}^{\text{0}}\text{C}$ .

Explanation of Solution

Given:

Partial pressure= $530\text{Pa}$

Relative humidity = $40\%$

Formula used:

  $\begin{array}{l}\text{Relative}\text{humidity}\text{=}\frac{\text{Partial}\text{pressure}\text{of}\text{water}}{\text{Saturated}\text{pressure}\text{of}\text{water}}\text{×100\%}\\ \text{Saturated}\text{pressure}\text{of}\text{water=}\frac{\text{Partial}\text{pressure}\text{of}\text{water}}{\text{Relative}\text{humidity}}\end{array}$

Calculation:

  $\begin{array}{l}\text{Relative}\text{humidity}\text{=}\frac{\text{Partial}\text{pressure}\text{of}\text{water}}{\text{Saturated}\text{pressure}\text{of}\text{water}}\text{×100\%}\\ \text{Saturated}\text{pressure}\text{of}\text{water=}\frac{\text{Partial}\text{pressure}\text{of}\text{water}}{\text{Relative}\text{humidity}}\\ P_{\text{saturated}}=\frac{530\text{Pa}}{0.40}=1325\text{Pa}\end{array}$

  $1325\text{Pa}$ is the saturated vapour which lies between the ${10}^0\text{C}$ to ${15}^0\text{C}$

But here is the temperature is given ${40}^0\text{C}$ which is not acceptable

At temp ${10}^0{\text{C-15}}^{\text{0}}\text{C}$

  $\frac{{\left(15-10\right)}^0\text{C}}{\left(1.71-1.23\right)\times {10}^3\text{Pa}}=1.402\times {10}^{-2}{}^0\text{C/Pa}$

Temperature corresponding to $1325\text{Pa}$ is

  $\begin{array}{l}={10}^0\text{C}+[(1325-1230)\text{Pa}]\left(1.402\times {10}^{-2}{}^0{}^0\text{C/Pa}\right)\\ ={10.99}^{\text{0}}\text{C}\approx {11}^{\text{0}}\text{C}\end{array}$

Conclusion:

Temperature of the waterat $40\%$ relative humidity is ${11}^{\text{0}}\text{C}$ .