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A mixture containing 4.33 g of CO2 and 3.11 g of CH4 has a total pressure of 1.09 atm
Chapter 11, Problem 120P(choose chapter or problem)
A mixture containing 4.33 g of \(\mathrm{CO}_{2}\) and 3.11 g of \(\mathrm{CH}_{4}\) has a total pressure of 1.09 atm. What is the partial pressure of \(\mathrm{CO}_{2}\) in the mixture?
Questions & Answers
QUESTION:
A mixture containing 4.33 g of \(\mathrm{CO}_{2}\) and 3.11 g of \(\mathrm{CH}_{4}\) has a total pressure of 1.09 atm. What is the partial pressure of \(\mathrm{CO}_{2}\) in the mixture?
ANSWER:Step 1 of 2
Dalton's law of partial pressures: It states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases.
Partial pressure can be calculated by using ideal gas equation if we assume that each gas acts independently.
Therefore, From the ideal gas equation,
PV = nRT
\(\Rightarrow P_{n}=n_{n} R T / V \dots \dots (1)\)
Therefore, when a mixture contain multi component, then the partial pressure of each component would be.
\(\begin{aligned}
P_{a} & =n_{a} R T / V \ldots \ldots \ldots(2) \\
P_{b} & =n_{b} R T / V \ldots \ldots .
\end{aligned}\)
Thus, total pressure
\(P=P_{a}+P_{b}+\ldots \ldots\)
\(\begin{array}{l}
=R T / V\left(n_{a}+n_{b}+\ldots . .\right) \\
=n_{n} R T / V \ldots \ldots \ldots(3)
\end{array}\)
Since,
Divide the equation (2) by equation (3) we get
\(\frac{P a}{P}=\frac{n a(R T / V)}{n n R T / V}=\mathrm{n}_{\mathrm{a}} / \mathrm{n}_{\mathrm{n}} \dots (4)\)