Page 195 - Theory and Problems of BEGINNING CHEMISTRY

184 GASES [CHAP. 12

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.

PV = nRT
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)
n = = = 0.0406 mol
RT [0.0821 L·atm/(mol·K)](300 K)
After mixing:
V (1.00 atm)(1.00 L)
P O 2
= = = 0.0406 mol
n O 2
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
◦
◦
25 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)
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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
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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

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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:
V [(736/760)atm](1.00 L)
P O 2
= = = 0.0396 mol
n O 2
RT [0.0821 L·atm/(mol·K)](298 K)