Page 202 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 12] GASES 191

12.40. Under what conditions of temperature and pressure do the gas laws work best?
Ans. Under low-pressure and high-temperature conditions (far from the possibility of change to the liquid state).

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12.41. If 0.223 g of a gas occupies 2.87 L at 17 C and 700 torr, what is the identity 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

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700 torr = 0.921 atm 17 C = (17 + 273) K = 290 K
760 torr
PV (0.921 atm)(2.87 L)
n = = = 0.111 mol
RT [0.0821 L·atm/(mol·K)](290 K)
We now know the mass of the gas and the number of moles. That is enough to calculate the molar mass:
0.223 g
= 2.01 g/mol
0.111 mol
The gas has a molar mass of 2.01 g/mol. It could only be hydrogen, H 2 , because no other gas has a molar
mass that low.

12.42. Which temperature scale must be used in (a) Boyle’s law problems, (b) ideal gas law problems, (c) combined gas
law problems, and (d) Charles’ law problems?
Ans. (a) No temperature is used in the calculations for Boyle’s law problems since the temperature must be
constant. (b) through (d) Kelvin.

12.43. The total pressure of a mixture of gases is 1.50 atm. The mixture contains 0.10 mol of N 2 and 0.20 mol of O 2 . What
is the partial pressure of O 2 ?
Ans. The oxygen in the mixture is two-thirds of the number of moles, so it exerts two-thirds of the total pressure—
1.00 atm. Or

n N 2 0.10 mol N 2 P N 2 V/(RT ) P N 2
= = = = 0.50
n O 2 0.20 mol O 2 P O 2 V/(RT ) P O 2
P N 2 = 0.50P O 2
= 1.50 atm
P N 2 + P O 2
= 1.50 atm
0.50P O 2 + P O 2 = 1.50P O 2
1.50 atm
= = 1.00 atm
P O 2
1.50
12.44. Calculate the mass of KClO 3 required to decompose to provide 0.728 L of O 2 at 20 C and 1.02 atm.
◦
Ans. The volume, temperature, and pressure given allow us to calculate the number of moles of oxygen:
PV (1.02 atm)(0.728 L)
= = = 0.0309 mol
n O 2
RT [0.0821 L·atm/(mol·K)](293 K)
The number of moles of KClO 3 may be calculated from the number of moles of O 2 by means of the balanced
chemical equation, and that value is then converted to mass.
2 KClO 3 −→ 2 KCl + 3O 2

2 mol KClO 3 122 g KClO 3
0.0309 mol O 2 = 2.51 g KClO 3
3 mol O 2 1 mol KClO 3
12.45. Two samples of gas at equal pressures and temperatures are held in containers of equal volume. What can be stated
about the comparative number of molecules in each gas sample?
Ans. Since the volumes, temperatures, and pressures are the same, the numbers of moles of the two gases are the
same. Therefore, there are equal numbers of molecules of the two gases.