Page 194 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 12] GASES 183

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EXAMPLE 12.18. If 4.58 g of a gas occupies 3.33 L at 27 C and 808 torr, what is the molar mass of the gas?
Ans. If you do not see at first how to solve this problem to completion, at least you can recognize that P, V, and T data are
given. First calculate the number of moles of gas present:

1 atm
808 torr = 1.06 atm
760 torr
27 C + 273 = 300 K
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PV (1.06 atm)(3.33 L)
n = = = 0.143 mol
RT [0.0821 L·atm/(mol·K)](300 K)
We now know the mass of the gas and the number of moles. That is enough to calculate the molar mass:
4.58 g
= 32.0 g/mol
0.143 mol
EXAMPLE 12.19. What volume is occupied by the oxygen liberated by heating 0.250 g of KClO 3 until it completely
decomposes to KCl and oxygen? The gas is collected at STP.
Ans. Although the temperature and pressure of the gas are given, the number of moles of gas is not. Can we get it
somewhere? The chemical reaction of KClO 3 yields the oxygen, and the rules of stoichiometry (Chap. 10) may be
used to calculate the number of moles of gas. Note that the number of moles of KClO 3 is not used in the ideal gas
law equation.
heat
2 KClO 3 −→ 2 KCl + 3O 2

1 mol KClO 3 3 mol O 2
0.250 g KClO 3 = 0.00307 mol O 2
122 g KClO 2 mol KClO 3
3
Now that we know the number of moles, the pressure, and the temperature of O 2 , we can calculate its volume:
nRT (0.00307 mol)[0.0821 L·atm/(mol·K)](273 K)
V = = = 0.0688 L
P 1.00 atm

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