Concept explainers
Interpretation:
A mixture of Ar (0.40 mol), O2 (0.50 mol), and CH4 (0.30 mol) exerts a pressure of 740 mm Hg. If the methane and oxygen are ignited and complete combustion occurs, the final pressure of Ar, CO2, H2O, and the remainder of the excess reactant and total pressure of the system should be calculated.
Concept introduction:
An ideal gas which is known as the perfect gas is a gas whose volume V, Pressure P and temperature T are related through the
Here,
-
n = Number of moles of the gas
R = Universal gas constant
T = Temperature
P = Pressure
Ideal gases are described as the molecules which have negligible size but have an average molar kinetic energy which is dependent on the temperature. When temperature is low most of the gases behave like ideal gases and the ideal
Answer to Problem 5.92PAE
Solution:
Partial Pressure of each gas is as follows:
Total pressure = 740 mmHg
Given:
The number of moles of gases at initial stage are as follows:
Explanation of Solution
Thus, total number of moles =
As given total pressure = P = 740 mmHg = P1 The combustion reaction is as follows:
Argon being a noble gas does not react in the reaction.
Thus, as per the above equation limiting reagent is Oxygen
Now,
0.50 mol of O2 forms no of mol of
And,
0.30 mol of
Thus, oxygen is limiting reagent here,
Calculate the number of moles of
0.50 mol of O2 forms no of mol of
Remaining CH4will be:
Thus, number of moles of gas after combustion will be:
Thus, total number of moles =
Total Pressure = P2that needs to be calculated.
As per the ideal gas equation,
Therefore, the total pressure of the system remains the same that is 740 mmHg
Now, partial pressure is calculated as follows:
Suppose for gas A,
Here,
Ar =
CO2=
H2O =
CH4 =
An ideal gas which is known as the perfect gas is a gas whose volume V, Pressure P and temperature T are related through the ideal gas laws. Based on the ideal gas law concept we calculated that
Partial Pressure of each gas is as follows:
Total pressure = 740 mmHg
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Chapter 5 Solutions
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