Chapter 13, Problem 69P

To determine

To calculate: The mass of the water evaporate from the open pan on the basis of information provided.

Answer to Problem 69P

Mass of the water evaporate in the air is $3.1\text{kg}$ .

Explanation of Solution

Given:

  $V(\text{Volume}\text{ofthe}\text{gas})=680{\text{m}}^{\text{3}}$

  $T(\text{Temperature}\text{of}\text{the}\text{gas})={25}^0\text{C}$

Formula used:

Ideal gas equation:

  $\begin{array}{l}PV=nRT\\ n=\frac{PV}{RT}\end{array}$

Here, $P$ is the pressure,

  $V$ is the volume,

  $T$ is the temperature,

  $R$ is the universal gas constant,

  $n$ is the no of moles.

Calculation:

Saturated vapour pressure of the water at ${25}^0\text{C}$ is

  $P=3.17\times {10}^3\text{Pa}$

Ideal gas equation:

  $\begin{array}{l}PV=nRT\\ n=\frac{PV}{RT}=\frac{\left(3.17\times {10}^3\text{Pa}\right)\left(680{\text{m}}^3\right)}{\left(8.31\text{J/mol}\text{.K}\right)\left(273+25\right)\text{K}}\\ =8.70\times {10}^2\text{moles}\end{array}$

According to the question

Relative humidity is only $80\%$ means $80\%$ water is present in air,

Remaining $20\%$ is evaporate in the air.

  $\begin{array}{l}{m}_{\text{evaporate}}=0.20(8.70\times {10}^2\text{moles})\left(\frac{18\times {10}^{-3}\text{kg}}{1\text{mole}}\right)\\ =3.1\text{kg}\end{array}$

Conclusion:

Mass of the water evaporate in the air is $3.1\text{kg}$ .