A gas mixture placed in a container consists of 5 grams of O2 and 8 g of He and has a total pressure of 12 atm. Calculate the partial pressure of 02 in the mixture.
Answers
Answer:
In day-to-day life, we measure gas pressure when we use a barometer to check the atmospheric pressure outside or a tire gauge to measure the pressure in a bike tube. When we do this, we are measuring a macroscopic physical property of a large number of gas molecules that are invisible to the naked eye. On the molecular level, the pressure we are measuring comes from the force of individual gas molecules colliding with other objects, such as the walls of their container.
Let's take a closer look at pressure from a molecular perspective and learn how Dalton's Law helps us calculate total and partial pressures for mixtures of gases.
Ideal gases and partial pressure
In this article, we will be assuming the gases in our mixtures can be approximated as ideal gases. This assumption is generally reasonable as long as the temperature of the gas is not super low (close to 0\,\text K0K0, start text, K, end text), and the pressure is around 1\,\text {atm}1atm1, start text, a, t, m, end text. Can we be more specific about when a gas behaves ideally?
This means we are making some assumptions about our gas molecules:
We assume that the gas molecules take up no volume.
We assume that the molecules have no intermolecular attractions, which means they act independently of other gas molecules.
Based on these assumptions, we can calculate the contribution of different gases in a mixture to the total pressure. We refer to the pressure exerted by a specific gas in a mixture as its partial pressure. The partial pressure of a gas can be calculated using the ideal gas law, which we will cover in the next section, as well as using Dalton's law of partial pressures.
Example 1: Calculating the partial pressure of a gas
Let's say we have a mixture of hydrogen gas, \text H_2(g)H
2
(g)start text, H, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis, and oxygen gas, \text O_2(g)O
2
(g)start text, O, end text, start subscript, 2, end subscript, left parenthesis, g, right parenthesis. The mixture contains 6.7\,\text{mol}6.7mol6, point, 7, start text, m, o, l, end text hydrogen gas and 3.3\,\text{mol}3.3mol3, point, 3, start text, m, o, l, end text oxygen gas. The mixture is in a 300\,\text L300L300, start text, L, end text container at 273\,\text K273K273, start text, K, end text, and the total pressure of the gas mixture is 0.75\,\text{atm}0.75atm0, point, 75, start text, a, t, m, end text.
The contribution of hydrogen gas to the total pressure is its partial pressure. Since the gas molecules in an ideal gas behave independently of other gases in the mixture, the partial pressure of hydrogen is the same pressure as if there were no other gases in the container. Therefore, if we want to know the partial pressure of hydrogen gas in the mixture, \text P_{\text H_2}P
H
2
start text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, we can completely ignore the oxygen gas and use the ideal gas law:
\text P_{\text H_2}\text V = \text {n}_{\text H_2}\text{RT}P
H
2
V=n
H
2
RTstart text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, start text, V, end text, equals, start text, n, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, start text, R, T, end text
Rearranging the ideal gas equation to solve for \text P_{\text H_2}P
H
2
start text, P, end text, start subscript, start text, H, end text, start subscript, 2, end subscript, end subscript, we get:
\begin{aligned}\text P_{\text H_2} &= \dfrac{\text{n}_{\text H_2}\text{RT}}{\text V}\\ \\ &=\dfrac{(6.7\,\text {mol})(0.08206\,\dfrac{\text {atm} \cdot \text L} {\text {mol} \cdot \text K})(273\,\text K)}{300\,\text L}=0.50\,\text {atm}\end{aligned}
P
H
2
=
V
n
H
2
RT
=
300L
(6.7mol)(0.08206
mol⋅K
atm⋅L
)(273K)
=0.50atm
Thus, the ideal gas law tells us that the partial pressure of hydrogen in the mixture is 0.50\,\text {atm}0.50atm0, point, 50, start text, a, t, m, end text. We can also calculate the partial pressure of hydrogen in this problem using Dalton's law of partial pressures, which will be discussed in the next section.