Concept explainers
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?
(a)
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-
The concept of Ideal gas law will also be used which is given as-
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.
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-
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-
The Relation between the pressure in Pascal and energy in joules is given as-
Or,
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 =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 =
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.
(b)
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-
Ideal gas law will also be used. The mathematical expression of ideal gas law is given as-
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-
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
Put the above values in Equ (1) to calculate the number of moles before compression-
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
Put the above values in (1) to calculate the number of moles after compression-
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-
Molar mass of water = 18
One mole of water = 18g
Therefore,
0.00416 mol of water in grams = 0.0749 g.
The amount of water condensed = 0.0749 g.
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Chapter 9 Solutions
Chemistry: Principles and Reactions
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