Q12.127CP - Chemistry: Molecular Nature Of Matter And Change

Question

A greenhouse contains  $\mathbf{256}{\mathbf{m}}^3$ of air at a temperature of  $\mathbf{26}\mathbf{°}\mathbf{C}$, and a humidifier in it vaporizes 4.20 L of water. 

(a) What is the pressure of water vapor in the greenhouse, assuming that none escapes and that the air was originally completely dry (d of  ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$ = 1.00 g/mL)? 

(b) What total volume of liquid water would have to be vaporized to saturate the air (i.e., achieve $\mathbf{100}\mathbf{\%}$   relative humidity)? (See Table 5.4, p. 221.)

Step-by-Step Solution

Answer

1. The pressure of water vapour in the greenhouse is 0.22atm.
2. Total volume of liquid water that would have to be vaporized to saturate the air is $\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{-7}$.

Step 1: Subpart (a) The pressure of water vapor in the greenhouse.

Given,

 

The temperature ${\text{=26}}^{\text{o}}\text{C = 299k}$ .

 $\begin{array}{c}\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{water}=4.2\mathrm{L}\\ 1\mathrm{L}=1000\mathrm{mL}\\ \mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{water}=4.20\mathrm{L}\times \frac{1000\mathrm{mL}}{1\mathrm{L}}\end{array}$

  

 

Density of  ${\text{H}}_{\text{2}}\text{O = 1.00 g/mL}$

As,

 

 $\text{Number of Moles =}\frac{\text{Mass}}{\text{Molar Mass}}$

 

And,

 

 $\text{Density =}\frac{\text{Mass}}{\text{Volume}}$

 

 $\text{Mass = Volume × Density}$

 

Therefore,

 

 $\text{Number\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}Moles =}\frac{\text{4.20L × }\frac{\text{1000mL}}{\text{1L}}\text{×1.00g/mL}}{\text{18g/mole}}$

As,

Gas Constant,  $\text{R = 0.0821 Latm/k/mole}$

Then, the pressure at temperature  ${\text{=26}}^{\text{o}}\text{C = 299k}$.

$\mathbf{PV}\mathbf{=}\mathbf{nRT}$

Therefore,

 

 $\begin{array}{c}\mathrm{P}=\frac{\mathrm{n}\times \mathrm{R}\times \mathrm{T}}{}\\ \mathrm{P}=\frac{\left[\frac{4.20\times \frac{1000\mathrm{mL}}{1\mathrm{L}}\times 1.00\mathrm{g}/\mathrm{mL}}{18.02\mathrm{g}/\mathrm{mL}}\right]}{256{\mathrm{m}}^3\times \frac{1000\mathrm{L}}{1{\mathrm{m}}^3}}\times (0.0821\mathrm{Latm}/\mathrm{mol}/\mathrm{k})\times (299\mathrm{k})\\ \mathrm{P}=\frac{233.07}{}\times 24.55\\ \mathrm{P}=0.022\mathrm{atm}\end{array}$

 

 

Step 2: Subpart (b) The total volume of liquid water would have to be vaporized to saturate the air.

The Absolute Humidity as the saturated air with water vapour.

 

 $\begin{array}{c}\text{Absolute Humidity =}\frac{\text{Pressure}}{\text{Gas\hspace{0.17em}Cosntant\hspace{0.17em}of\hspace{0.17em}Water × Temperature}}\\ \text{Absolute Humidity =}\frac{\text{0.022atm}}{\text{461.5 × 299k}}\\ {\text{Absolute Humidity =1.6 × 10}}^{\text{-7}}\end{array}$