DALTON’S LAW OF PARTIAL PRESSURES

12.8. DALTON’S LAW OF PARTIAL PRESSURES

When two or more gases are mixed, they each occupy the entire volume of the container. They each have the same temperature as the other(s). However, each gas exerts its own pressure, independent of the other gases. Moreover, according to Dalton’s law of partial pressures, their pressures must add up to the total pressure of the gas mixture.

EXAMPLE 12.20. (a) If I try to put a 1.00-L sample of O 2 at 300 K and 1.00 atm plus a 1.00-L sample of N 2 at 300 K and 1.00 atm into a rigid 1.00-L container at 300 K, will they fit? (b) If so, what will be their total volume and total pressure?

Ans. (a) The gases will fit; gases expand or contract to fill their containers. (b) The total volume is the volume of the container—1.00 L. The temperature is 300 K, given in the problem. The total pressure is the sum of the two partial pressures. Partial pressure is the pressure of each gas (as if the other were not present). The oxygen pressure is

1.00 atm. The oxygen has been moved from a 1.00-L container at 300 K to another 1.00-L container at 300 K, and so its pressure does not change. The nitrogen pressure is 1.00 atm for the same reason. The total pressure is 1.00 atm + 1.00 atm = 2.00 atm.

EXAMPLE 12.21.

A 1.00-L sample of O 2 at 300 K and 1.00 atm plus a 0.500-L sample of N 2 at 300 K and 1.00 atm are put into a rigid 1.00-L container at 300 K. What will be their total volume, temperature, and total pressure?

Ans. The total volume is the volume of the container—1.00 L. The temperature is 300 K, given in the problem. The total pressure is the sum of the two partial pressures. The oxygen pressure is 1.00 atm. (See Example 12.20.) The nitrogen pressure is 0.500 atm, since it was moved from 0.500 L at 1.00 atm to 1.00 L at the same temperature (Boyle’s law). The total pressure is

1.00 atm + 0.500 atm = 1.50 atm

EXAMPLE 12.22. If the N 2 of the last example were added to the O 2 in the container originally containing the O 2 , how would the problem be affected?

Ans. It would not change; the final volume would still be 1.00 L. EXAMPLE 12.23. If the O 2 of Example 12.21 were added to the N 2 in the container originally containing the N 2 , how

would the problem change? Ans.

The pressure would be doubled, because the final volume would be 0.500 L. The ideal gas law applies to each individual gas in a gas mixture as well as to the gas mixture as a whole.

Thus, in a mixture of nitrogen and oxygen, one can apply the ideal gas law to the oxygen, to the nitrogen, and to the mixture as a whole.

P V = n RT

Variables V and T , as well as R, refer to each gas and to the total mixture. If we want the number of moles of O 2 , we use the pressure of O 2 . If we want the number of moles of N 2 , we use the pressure of N 2 . If we want the total number of moles, we use the total pressure.

EXAMPLE 12.24. Calculate the number of moles of O 2 in Example 12.21, both before and after mixing. Ans.

Before mixing:

PV

1.00 atm)(1.00 L)

[0.0821 L·atm/(mol·K)](300 K)

After mixing:

1.00 atm)(1.00 L)

= 0.0406 mol

RT

[0.0821 L·atm/(mol·K)](300 K)

Water Vapor

At 25 ◦

C, water is ordinarily a liquid. However, even at 25 ◦

C, water evaporates. In a closed container at

C, water evaporates enough to get a 24-torr water vapor pressure in its container. The pressure of the gaseous water is called its vapor pressure at that temperature. At different temperatures, it evaporates to different extents to give different vapor pressures. As long as there is liquid water present, however, the vapor pressure above pure water depends on the temperature alone. Only the nature of the liquid and the temperature affect the vapor pressure; the volume of the container does not affect the final pressure.

The water vapor mixes with any other gas(es) present, and the mixture is governed by Dalton’s law of partial pressures, just as any other gas mixture is.

EXAMPLE 12.25. O 2 is collected in a bottle over water at 25 ◦ C at 1.00-atm barometric pressure. (a) What gas(es) is (are) in the bottle? (b) What is (are) the pressure(s)?

Ans. (a) Both O 2 and water vapor are in the bottle. (b) The total pressure is the barometric pressure, 760 torr. The water vapor pressure is 24 torr, given in the first paragraph of this subsection for the gas phase above liquid water at 25 ◦ C.

The pressure of the O 2 is therefore 760 torr − 24 torr = 736 torr

EXAMPLE 12.26. How many moles of oxygen are contained in a 1.00-L vessel over water at 25 ◦ C and a barometric pressure of 1.00 atm?

Ans. The barometric pressure is the total pressure of the gas mixture. The pressure of O 2 is 760 torr − 24 torr = 736 torr. Since we want to know about moles of O 2 , we need to use the pressure of O 2 in the ideal gas law:

2 V = [(736/760)atm](1.00 L)