Page 193 - Theory and Problems of BEGINNING CHEMISTRY

182 GASES [CHAP. 12

EXAMPLE 12.12. What is the volume of 1.00 mol of gas at STP?
nRT (1.00 mol)(0.0821 L·atm/mol·K)(273 K)
Ans. V = = = 22.4L
P 1.00 atm
The volume of 1.00 mol of gas at STP is called the molar volume of a gas. This value should be memorized.

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EXAMPLE 12.13. How many moles of SO 2 are present in a 765-mL sample at 37 C and 775 torr?
Ans. Since R is defined in terms of liters and atmospheres, the pressure and volume are first converted to those units.
Temperature as usual is given in kelvins. First, change the pressure to atmospheres and the volume to liters:
1 atm 1L

775 torr = 1.02 atm 765 mL = 0.765 L
760 torr 1000 mL
PV (1.02 atm)(0.765 L)
n = = = 0.0307 mol
RT [0.0821 L·atm/(mol·K)](310 K)
Alternately, we can do the entire calculation, including the conversions, with one equation:
PV (775 torr)(1.00 atm/760 torr)(765 mL)(1L/1000 mL)
n = = = 0.0307 mol
RT [0.0821 L·atm/(mol·K)](310 K)

EXAMPLE 12.14. At what temperature will 0.0750 mol of CO 2 occupy 2.75 L at 1.11 atm?
PV (1.11 atm)(2.75 L)
Ans. T = = = 496 K
nR [0.0821 L·atm/(mol·K)](0.0750 mol)

EXAMPLE 12.15. What use was made of the information about the chemical identity of the gas in Examples 12.13 and
12.14?
Ans. None. The ideal gas law works no matter what gas is being used.

EXAMPLE 12.16. What volume will 7.00 g of Cl 2 occupy at STP?
Ans. The value of n is not given explicitly in the problem, but the mass is given, from which we can calculate the number
of moles:

1 mol Cl 2
7.00gCl = 0.0986 mol Cl 2
2
71.0gCl 2
nRT (0.0986 mol)[0.0821 L·atm/(mol·K)](273 K)
V = = = 2.21 L
P 1.00 atm
The identity of the gas is important here to determine the number of moles.
EXAMPLE 12.17. (a) How many moles of O atoms were present in Example 12.11? (b) How many moles of He atoms
are present in 465 mL at 25 C and 1.00 atm?
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2 mol O

Ans. (a) 0.0223 mol O 2 = 0.0446 mol O atoms
1 mol O 2
Note that n in the ideal gas equation in the example refers to moles of O 2 molecules.
PV (1.00 atm)(0.465 L)
(b) n = = = 0.0190 mol He
RT [0.0821 L·atm/(mol·K)](298 K)
The He molecules are individual He atoms, and thus there are 0.0190 mol He atoms present.

As soon as you recognize P, V, and T data in a problem, you can calculate n. (In a complicated problem, if you
are given P, V, and T, but you do not see how to do the whole problem, first calculate n and then see what you can do
with it.)