Partial Pressures
The ideal gas law assumes that all gases behave identically and that their behavior is independent of attractive and repulsive forces. If volume and temperature are held constant, the ideal gas equation can be rearranged to show that the pressure of a sample of gas is directly proportional to the number of moles of gas present:
\[P=n \bigg(\dfrac{RT}{V}\bigg) = n \times \rm const. \tag{10.5.1}\]
Nothing in the equation depends on the nature of the gas—only the amount.
With this assumption, let’s suppose we have a mixture of two ideal gases that are present in equal amounts. What is the total pressure of the mixture? Because the pressure depends on only the total number of particles of gas present, the total pressure of the mixture will simply be twice the pressure of either component. More generally, the total pressure exerted by a mixture of gases at a given temperature and volume is the sum of the pressures exerted by each gas alone. Furthermore, if we know the volume, the temperature, and the number of moles of each gas in a mixture, then we can calculate the pressure exerted by each gas individually, which is its partial pressureThe pressure a gas in a mixture would exert if it were the only one present (at the same temperature and volume)., the pressure the gas would exert if it were the only one present (at the same temperature and volume).
To summarize, the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases. This law was first discovered by John Dalton, the father of the atomic theory of matter. It is now known as Dalton’s law of partial pressuresA law that states that the total pressure exerted by a mixture of gases is the sum of the partial pressures of component gases.. We can write it mathematically as
\[P_{tot}= P_1+P_2+P_3+P_4 \; ... = \sum_{i=1}^n{P_i} \tag{10.5.2}\]
where Pt is the total pressure and the other terms are the partial pressures of the individual gases (Figure 10.5.1).
Figure 10.5.1 Dalton’s Law. The total pressure of a mixture of gases is the sum of the partial pressures of the individual gases.
For a mixture of two ideal gases, A and B, we can write an expression for the total pressure:
\[P_{tot}=P_A+P_B=n_A\bigg(\dfrac{RT}{V}\bigg) + n_B\bigg(\dfrac{RT}{V}\bigg)=(n_A+n_B)\bigg(\dfrac{RT}{V}\bigg) \tag{10.5.3}\]
More generally, for a mixture of i components, the total pressure is given by
\[P_{tot}=(P_1+P_2+P_3+ \; \cdots +P_n)\bigg(\dfrac{RT}{V}\bigg)\tag{10.5.4a}\]
\[P_{tot}=\sum_{i=1}^n{n_i}\bigg(\dfrac{RT}{V}\bigg)\tag{10.5.4b}\]
Equation 10.5.4. restates Equation 10.5.3 in a more general form and makes it explicitly clear that, at constant temperature and volume, the pressure exerted by a gas depends on only the total number of moles of gas present, whether the gas is a single chemical species or a mixture of dozens or even hundreds of gaseous species. For Equation 10.27 to be valid, the identity of the particles present cannot have an effect. Thus an ideal gas must be one whose properties are not affected by either the size of the particles or their intermolecular interactions because both will vary from one gas to another. The calculation of total and partial pressures for mixtures of gases is illustrated in Example 11.
Example 10.5.1
For reasons that we will examine in Chapter 15, deep-sea divers must use special gas mixtures in their tanks, rather than compressed air, to avoid serious problems, most notably a condition called “the bends.” At depths of about 350 ft, divers are subject to a pressure of approximately 10 atm. A typical gas cylinder used for such depths contains 51.2 g of O2 and 326.4 g of He and has a volume of 10.0 L. What is the partial pressure of each gas at 20.00°C, and what is the total pressure in the cylinder at this temperature?
Given: masses of components, total volume, and temperature
Asked for: partial pressures and total pressure
Strategy:
A Calculate the number of moles of He and O2 present.
B Use the ideal gas law to calculate the partial pressure of each gas. Then add together the partial pressures to obtain the total pressure of the gaseous mixture.
Solution:
A The number of moles of He is
\[n_{\rm He}=\rm\dfrac{326.4\;g}{4.003\;g/mol}=81.54\;mol \notag \]
The number of moles of O2 is
\[n_{\rm O_2}=\rm \dfrac{51.2\;g}{32.00\;g/mol}=1.60\;mol \notag \]
B We can now use the ideal gas law to calculate the partial pressure of each:
\[P_{\rm He}=\dfrac{n_{\rm He}RT}{V}=\rm\dfrac{81.54\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=196.2\;atm \notag \]
\[P_{\rm O_2}=\dfrac{n_{\rm O_2}RT}{V}=\rm\dfrac{1.60\;mol\times0.08206\;\dfrac{atm\cdot L}{mol\cdot K}\times293.15\;K}{10.0\;L}=3.85\;atm \notag \]
The total pressure is the sum of the two partial pressures:
\[P_{\rm tot}=P_{\rm He}+P_{\rm O_2}=\rm(196.2+3.85)\;atm=200.1\;atm \notag \]
Exercise
A cylinder of compressed natural gas has a volume of 20.0 L and contains 1813 g of methane and 336 g of ethane. Calculate the partial pressure of each gas at 22.0°C and the total pressure in the cylinder.
Answer: P(CH4) = 137 atm; P(C2H6) = 13.4 atm; Pt = 151 atm.