MathConnections
CH 3 CHCH 2 CH3CH2CH2CH2
5.4 Partial Pressure
of the sample. And since the problem involves a gas, we can attempt to use the ideal gas law to fi nd the number of moles. We are given values for V, P, and T, so we will be able to solve for n. Some of the units given are unusual, so we will need to be careful to handle them properly.
Solution We can begin by converting some of the given information into units more commonly used for gases.
T = (72 − 32) × 5/9 = 22°C = 295 K V = 575 cm3= 575 mL = 0.575 L
For P, we will use the value in torr and choose the corresponding value for R. We could also convert from torr to atm, of course.
n = — PV
RT = (752 torr)(0.575 L)
—— (62.37 L torr mol−1 K−1)(295 K) = 0.0235 mol The gas in our sample is CO2, with a molar mass of 44.0 g/mol, so
0.0235 mol × 44.0 g
— mol = 1.03 g CO2
Analyze Your Answer The temperature and pressure are near typical room con- ditions, and the volume is a little more than a half liter. So this should correspond to a small balloon fi lled with CO2. One gram is a pretty small mass but seems reasonable for the gas in a small balloon.
Check Your Understanding Find the volume in ft3 of 1.0 mole of an ideal gas at −25°C and 710 torr.
present. If the nitrogen were the only gas present, nO2 would be zero, and we would be left with
P = PN2= nN2RT
— V
If only oxygen were present, we could write a similar result for PO2. For a mixture, the observed pressure is the sum of the hypothetical partial pressures of the individual gases.
This concept, first established by John Dalton, is sometimes referred to as Dalton’s law of partial pressures: The pressure of a mixture of gases is the sum of the partial pressures of the component gases. In simple terms, the partial pressure is what the pressure would be if only that component were present. As long as the component gases behave ideally and do not actually react with one another, this relationship will hold for any gas mixture.
For an arbitrary gas mixture, we can write the partial pressure of each component as Pi= niRT
— V (5.5)
where the index, i, indicates which gas we are talking about (for a mixture of four gases, i would vary from 1 to 4 yielding, P1, P2, P3, and P4) and ni is the number of moles of component i. The total pressure of the mixture, P, is the sum of these partial pressures:
P =
∑
i Pi =
∑
i ni RT —
V (5.6)
The total number of moles is the sum of the numbers of moles of the individual components:
P = ntotal — RT
V (5.7)
These expressions suggest another way to understand partial pressures. We might choose to express the concentration of each component gas in terms of its mole fraction, defi ned as
Xi= ni
— ntotal (5.8)
For a pure gas, ni= ntot and the mole fraction is one. Any mixture will have ni < ntot, so the value of the mole fraction has a range from zero to one. If we take the ratio of a partial pressure to total pressure, using Equations 5.5, 5.7, and 5.8,
Pi
— P = ni (RT/V )
— ntotal (RT/V ) = ni
— ntotal = Xi (5.9)
So the partial pressure is given by the mole fraction times the total pressure.
Pi= Xi P (5.10)
This set of equations provides the numerical means for us to use partial pressures, as shown in Example Problems 5.4 and 5.5.
E X A M P L E P RO B L E M 5 . 4
Not all pollution is due to human activity. Natural sources, including volcanoes, also contribute to air pollution. A scientist tries to generate a mixture of gases similar to those found in a volcano by introducing 15.0 g of water vapor, 3.5 g of SO2 and 1.0 g of CO2 into a 40.0-L vessel held at 120.0°C. Calculate the partial pressure of each gas and the total pressure.
John Dalton is also known for his model proposing the existence and nature of atoms.
John Dalton is also known for his model proposing the existence and nature of atoms.
The upper case Greek letter sigma (Σ) is used to indicate the sum of a number of similar terms.
The upper case Greek letter sigma (Σ) is used to indicate the sum of a number of similar terms.
Strategy As usual, we will begin by assuming that all the gases behave ideally. Then we can treat each gas separately to determine its partial pressure. Finally, sum these partial pressures to calculate the total pressure.
Solution For all of the gases, T = 120°C = 393 K, and V = 40.0 L.
H2O: 15.0 g H2O × 1 mol H2O
— 18.0 g H2O = 0.833 mol PH2O= (0.833 mol)(0.08206 L atm mol−1K−1)(393 K)
——— 40.0 L = 0.672 atm
SO2: 3.5 g SO2× 1 mol SO2
— 64.1 g SO2 = 0.055 mol
P SO 2 = (0.055 mol)(0.08206 L atm mol−1K−1)(393 K)
——— 40.0 L = 0.044 atm
CO2: 1.0 g CO2× 1 mol CO2
— 44.0 g CO2 = 0.023 mol
P CO 2 = (0.023 mol)(0.08206 L atm mol−1K−1)(393 K)
——— 40.0 L = 0.018 atm
Total pressure is the sum of these:
Ptot= P H2O + P SO2 + P CO2 = 0.672 + 0.044 + 0.018 = 0.734 atm
Note that you could also have used the total number of moles to fi nd Ptot and then used mole fractions to fi nd the partial pressures.
Analyze Your Answer It is possible to gain a feeling for quantities such as pres- sure and number of moles if you work a lot of these problems. In this case, we have almost a mole of gas in a volume of 40 L, and the temperature is somewhat high. Our result of 0.7 atm is not unreasonable.
Check Your Understanding A mixture containing 13.5 g of oxygen and 60.4 g of N2 exerts a pressure of 2.13 atm at 25°C. What is the volume of the container, and what are the partial pressures of each gas?
E X A M P L E P RO B L E M 5 . 5
An experiment is designed to determine the effect of sulfur dioxide, one of the EPA criteria pollutants, on plants. Among the variations used is a mixture that has the mole fractions given in the following table.
Gas N2 O2 H2O SO2
Mole fraction 0.751 0.149 0.080 0.020
If the desired total pressure is 750. torr, what should the partial pressures be? If the gas is to be in a 15.0-L vessel held at 30°C, how many moles of each substance are needed?
Strategy We know the mole fractions and the desired total pressure. So we can calculate partial pressures using the relationship defi ned in Equation 5.10. From the total pressure and the volume, we can calculate the total number of moles and thereby the number of moles of each gas.
Solution
Pi= XiPtotal
P N2= (0.751)(750. torr) = 563 torr P O2= (0.149)(750. torr) = 112 torr P H2O= (0.080)(750. torr) = 60 torr P SO2= (0.020)(750. torr) = 15 torr
The desired total pressure of 750 torr can also be expressed as 0.987 atm. So ntotal= — PV
RT = (0.987 atm)(15.0 L)
—— (0.08206 L atm mol−1 K−1)(303 K) = 0.595 mol ni= Xintotal
nN2= (0.751)(0.595) = 0.447 mol nO2= (0.149)(0.595) = 8.87 × 10−2 mol nH2O= (0.080)(0.595) = 4.8 × 10−2 mol
nSO2= (0.020)(0.595) = 1.2 × 10−2 mol
Check Your Understanding A mixture of SO2(g) and SO3(g) is to be prepared with a total pressure of 1.4 atm. If the mole fractions of the gases are 0.70 and 0.30, respectively, what are the partial pressures? If the mixture is to occupy 2.50 L at 27°C, what mass of each gas is needed?