Q12.118CP - Chemistry: Molecular Nature Of Matter And Change

Question

In an experiment, 5.00 L of  ${\mathbf{N}}_2$  is saturated with water vapour at  $\mathbf{22}\mathbf{°}\mathbf{C}$  and then compressed to half its volume at constant T. 

(a) What is the partial pressure of  ${\mathbf{H}}_2\mathbf{O}$ in the compressed gas mixture? 

(b) What mass of water vapor condenses to liquid?

Step-by-Step Solution

Answer

1. The partial pressure of the molecule ${\text{H}}_{\text{2}}\text{O}$  is 19.8torr at  ${\text{22}}^{\text{o}}\text{C}$.
2. The mass of water vapor condenses to liquid is 720g.

Step 1: Formulas

The equation for the ideal gas is

 

 $\mathbf{PV}\mathbf{=}\mathbf{nRT}$

 

The moles are calculated as,

 

 $\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{Moles}\mathbf{=}\frac{\mathrm{Mass}}{\mathrm{Molar} \mathrm{Mass}}$

Step 2: Subpart (a) The partial pressure of H 2 O in the compressed gas mixture.

At temperature $22°\mathrm{C}$  , the vapor pressure is 19.8torr.

The vapor pressure depends upon the temperature. Here, as the temperature is the same for the reaction, therefore the vapor pressure also remains the same.

Step 2: Subpart (b) The mass of water vapor condenses to liquid.

Given,

Temperature  $22°\mathrm{C}$  is equal to 295K.

Volume of the water 5L is 5000mL.

Gas Constant, R = 8.314 J/k/mole

Pressure at temperature   $22°\mathrm{C}$ is 19.8torr.

 

Therefore, the number of moles of water at  $22°\mathrm{C}$:

 

 $\begin{array}{c}\mathrm{PV}=\mathrm{nRT}\\ \mathbf{19}\mathbf{.}\mathbf{8}\mathbf{\times }\mathbf{5000}\mathbf{mL}\mathbf{=}\mathbf{n}\mathbf{\times }\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{295}\end{array}$

And,

 $\begin{array}{c}\mathbf{n}\mathbf{=}\frac{\mathbf{19}\mathbf{.}\mathbf{8}\mathbf{\times }\mathbf{5000}\mathbf{mL}}{\mathbf{8}\mathbf{.}\mathbf{314}\mathbf{\times }\mathbf{295}}\\ n=40\mathrm{mole}\end{array}$

 

The mass of the water molecule at  ${\text{22}}^{\text{o}}\text{C}$:

 

As,

 $\begin{array}{c}\text{Number\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}Moles = }\frac{\text{Mass}}{\text{Molar Mass}}\\ \text{ 40 = }\frac{\text{Mass}}{\text{18}}\end{array}$

Hence,

 

 $\begin{array}{c}\text{Mass = 18 × 40}\\ \text{Mass = 720g}\end{array}$