Chapter 9, Problem 61QAP

Consider a sealed flask with a movable piston that contains 5.25 L of O₂ saturated with water vapor at 25°C. The piston is depressed at constant temperature so that the gas is compressed to a volume of 2.00 L. (Use the table in Appendix 1 for the vapor pressure of water at various temperatures.)

(a) What is the vapor pressure of water in the compressed gas mixture?

(b) How many grams of water condense when the gas mixture is compressed?

Interpretation Introduction

Interpretation:

The amount of O₂ saturated water vapor in a sealed flask with movable piston is 5.25 L at temperature 25°C. The gas is compressed to a volume of 2.00 L by depressing the piston at a constant temperature. The vapor pressure of water in the compressed gas mixture is to be determined

Concept Introduction :

The vapor pressure of water is calculated by the following formula-

  $R{H}_1=\frac{P}{P^{*}}$

The concept of Ideal gas law will also be used which is given as-

  $PV=nRT$

Answer to Problem 61QAP

The vapor pressure of water is 23.8 mm Hg or 3.17×10-³Pa in the compressed gas mixture.

Explanation of Solution

The relative humidity is a method by which the amount of water vapor present in the air is to be measured. Mathematically, it is defined as the ratio of the partial pressure of water vapor to the vapor pressure at equilibrium at the given temperature.

  $R{H}_1=\frac{P}{P^{*}}$.......(1)

Where,

RH = relative humidity

P = partial pressure of water vapor

P* = vapor pressure at equilibrium at the given temperature

Ideal gas law is used to the relation between volume, pressure, and temperature. The ideal gas law equation is given as-

  $PV=nRT$.......(2)

Where,

P = pressure

V =volume

n = number of moles

R = gas constant

T = temperature

Conversion of pressure into Pascal is done by the following formula-

  $\mathrm{P}(Pa)=P(mmHg)\times \left(\frac{101.3kPa}{760mmHg}\right)$

  $\text{P(Pa)=P(mmHg)×}\left(\frac{\text{101}\text{.3kPa}}{\text{760mmHg}}\right)\text{×}\left(\frac{\text{1000Pa}}{\text{1kPa}}\right)$

The Relation between the pressure in Pascal and energy in joules is given as-

  $\begin{array}{l}1J=1N.m\\ 1Pa=1N.m^{-2}\end{array}$

  $1Pa=1N.m.m^{-3}$

Or,

  $1Pa=1J{m}^{-3}$

According to given data-

The oxygen is saturated by water vapor and present in the sealed flask. Therefore relative humidity is 100%. So,

RH =1.0

Therefore,

  $RH=\frac{P}{P^{*}}$

RH =1.0

P=P*

The partial pressure of water = saturation pressure at the given temperature

Temperature = 25°C

Therefore vapor pressure of water (table in appendix 1) is = 23.8 mmHg

The pressure in Pascal =

  $\text{P(Pa)=23}\text{.8(mmHg)×}\left(\frac{\text{101}\text{.3kPa}}{\text{760mmHg}}\right)\text{×}\left(\frac{\text{1000Pa}}{\text{1kPa}}\right)$

  $P(Pa)=3.17\times {10}^{3}Pa$

Temperature for the compressed gas mixture = 25°C

As given that oxygen was saturated by the water vapor. The partial pressure cannot be greater than the saturation vapor pressure therefore during compression some of the water will be condensed. So oxygen will be remained saturated by vapor. Therefore,

Partial pressure = 23.8 mm Hg

And, the vapor pressure of water in gas mixture = 23.8 mm Hg

The temperature and partial pressure of water vapor will remain the same before and after compression only volume will change.

Interpretation Introduction

Interpretation:

The amount of O₂ saturated water vapor in a sealed flask with movable piston is 5.25 L at temperature 25°C. The gas is compressed to a volume of 2.00L by depressing the piston at the constant temperature. The amount of water condenses (in grams) during the compressing of the gas mixture is to be determined.

Concept Introduction :

The vapor pressure of water is calculated by the following formula-

  $R{H}_1=\frac{P}{P^{*}}$

Ideal gas law will also be used. The mathematical expression of ideal gas law is given as-

  $PV=nRT$

Answer to Problem 61QAP

The amount of water condenses (in grams) during the compressing of the gas mixture is 0.0749 g.

Explanation of Solution

Ideal gas law is used to the relation between volume, pressure, and temperature. The ideal gas law equation is given as-

  $PV=nRT$.......(1)

Where,

P = pressure

V =volume

n = number of moles

R = gas constant

T = temperature

Given values-

Before compression-

T = 25°C= 25°C+273=298K

P = 3.17×10-³Pa

R = 8.3145J. mol-1.K-1

Volume of gas−

V₁ = 5.25L

  $V_1=5.25L\times \frac{m^3}{1000L}=0.00525m^3$

Put the above values in Equ (1) to calculate the number of moles before compression-

  $3.17\times {10}^{-3}Pa\times 0.00525m^3=n_1\times 8.3145J.mo{l}^{-1}.K^{-1}\times 298K$

  $n_1=0.00672mol$

After compression-

T = 25°C= 25°C+273=298K

P = 3.17×10-³Pa

R = 8.3145J. mol-1. K-1

Volume of gas -

V₂ = 2.00L

  $V_2=2.00L\times \frac{m^3}{1000L}=0.00200m^3$

Put the above values in (1) to calculate the number of moles after compression-

  $\text{3}{\text{.17×10}}^{\text{-3}}\text{Pa×0}{\text{.00200m}}^{\text{3}}{\text{=n}}_{\text{2}}\text{×8}\text{.3145J}{\text{. mol}}^{\text{-1}}{\text{.K}}^{\text{-1}}\text{×298K}$

  $n_1=0.00256mol$

After compression some of the water vapor gets condensed and the number of moles gets reduced. Therefore, the number of water condensed is given as-

  $\text{number of moles reduced}=\left(n_1-n_2\right)$

  $\left(n_1-n_2\right)=(0.00672-0.0025\text{6)mol}$

  $\left(n_1-n_2\right)=0.00\text{416mol}$

Molar mass of water = 18

One mole of water = 18g

  $\left(n_1-n_2\right)=0.00416mol$ in grams =

  $\text{0}\text{.00416 mol in grams =18g×0}\text{.00416mol}$

Therefore,

0.00416 mol of water in grams = 0.0749 g.

The amount of water condensed = 0.0749 g.