Answers to the Gases Review Sheet
1. 3.21 mm Hg = ___.321_ cm Hg = ___0.00422__ atm
2. 6.9 kPa = ___6900__ Pa = _52_ mm Hg
3. 436 mm Hg = _0.574_ atm = __58.1_ kPa
4. 2.9 kPa = _22_ torr = _22_ mm Hg
5. A sample of air has a volume of 2.25 cm3 when its temperature is 298 K.
If the temperature is increased to 373 K without changing the pressure,
what is the new volume of the sample in air ?
V1 = 2.25 cm3
T1 = 298 K
T2 = 373 K
V2 = x
V1 =
T1
V2
T2
V2 = V1T2
T1
V2 = 2.25 cm3 (373 K) = 2.82 cm3
298 K
6. As the temperature of a sample of nitrogen increases from 273 K, the
volume of the sample changes from 275 mL to 325 mL. Assuming the
pressure is constant, what is the final temperature of the gas ?
V1 = 275 mL
T1 = 273 K
T2 = x
V2 = 325 mL
V1 =
T1
V2
T2
T2 = T1V2
V1
T2 = 325 mL (273 K) = 323 K
275 mL
7. The pressure of a 5.71 L sample of neon gas is 23.4 kPa. Calculate the
new pressure when the volume becomes 3.40 L.
V1 = 5.71 L
P1V1 = P2V2
P2 = P1V1
P1 = 23.4 kPa
V2
P2 = x
V2 = 3.40 L
P2 = 5.71 L(23.4 kPa) = 39.3 kPa = 0.388 atm
3.40 L
8. At what temperature is a gas if 0.0851 moles of it is found in a 604 L
vessel at 100.4 kPa ?
T = ????
n = 0.0851 moles
v = 604 L
P = 100.4 kPa = 0.991 atm
R = 0.08206 L∙atm/mol∙K
PV = nRT
T=
PV = 0.991 (604)
nR
0.0851(0.08206)
T = 85700 K
Ideal Gas Law
9. What volume will be occupied by 17.3 g of oxygen gas at 203 kPa and
11C ?
Moles can be obtained from grams
n = 17.3 g O2 1 mol O2
= 0.541 mol O2
32.00 g O2
P = 203 kPa = 2.00 atm
T = 11 C = 284 K
PV = nRT
so V = nRT
R = 0.08206 L atm/mol K
P
V=?
V = (0.541 mol)(0.08206 L atm/mol K)(284 K)
2.00 atm
V = 6.30 L
10. How many moles of gas will occupy a 486 cm3 flask at 10. C and 66.7
Pa?
V = 486 cm3 = 486 mL = 0.486 L
P = 66.7 Pa = 6.58 x 10 –4 atm
T = 10. C = 283 K
R = 0.08206 L atm/mol K
n=?
PV = nRT
Ideal Gas Law
n = PV = (6.58 x 10–4atm)(0.486 L)
RT
(0.08206 L atm/mol K )(283 K)
n = 1.38 x 10 –5 moles
11. If 425 moles of oxygen are collected in a flask of 9.8 L, what volume will
the gas occupy is you have 443 moles ?
V1 = 9.8 L
n1 = 425 mol
n2 = 443 mol
V2 = x
V1 =
n1
V2
n2
V2 = 9.8 L (443 mol) = 10. L
425 mol
V2 = V1n2
n1
Avogadro’s Law
12. A 2.0 L flask contains a mixture of nitrogen gas and oxygen gas at STP.
Calculate the moles of gas present. Can you determine the amount of
each gas present? (STP is 1 atm and 273 K)
V = 2.0 L for both gases
@ STP so you can assume 1 mol = 22.4 L
2.0 L 1 mol
22.4 L
= 0.0893 mol of gas
BUT you don’t know how much of each gas, not enough
information provided.
13. Consider a sample of hydrogen gas collected over water at 25 C where
the pressure of water is 24 torr. The volume occupied by the gaseous
mixture is 0.500 L, and the total pressure is 0.950 atm. Calculate the
partial pressure of hydrogen gas and the number of moles of hydrogen
gas present.
V = 0.500 L for both gases
T = 25 C = 298 K for both gases
n H2O = ?
n H2 = ?
P total = 0.950 atm
P H2O = 24 torr = 0.032 atm
P H2 = ?
There are two gases in this problem, each with its own pressure, but their
combined pressures would equal the total. Ptotal = P1 + P2 + . . .
So 0.950 atm = 0.032 atm + PH2 therefore P
H2
= 0.918 atm
Now there is enough information to find the nH2 using PV = nRT.
nH2 = PV
= (0.918 atm)(0.500L) = 0.0188 mol H2
RT
(0.08206)(298)
14. Calculate the volume of hydrogen produced at 1.50 atm and 19 C by the
reaction of 26.5 g of zinc with excess hydrochloric acid according to the
equation
Zn + 2 HCl  ZnCl2 + H2
P = 1.50 atm
T = 19 C = 292 K
n = use the mass to find moles
26.5 g Zn
1 mol Zn
1 mol H2
= 0.405 mol H2 = n
65.39 g Zn 1 mol Zn
R = 0.08206 L atm/mol K
V=?
PV = nRT so V = nRT = (0.405 mol H2)(0.08206 L atm/mol K )(292 K)
P
1.50 atm
V = 6.47 L
15. A sample of NH3 occupies 5.00 L at 25 C and 15.0 atm. What volume will
this sample occupy at STP?
There are set conditions called Standard Temperature and Pressure (STP). At
STP the temperature is 273 K and the pressure is 1.000 atm. In this problem T
and P are the second set of conditions T2 and P2.
V1 = 5.00 L
T1 = 298 K
P1 = 15.0 atm
T2 = 273 K
V2 = x L
P2 = 1.00 atm
P1V1 = P2V2
T1
T2
V2 = 273 K 15.0 atm 5.00 L
1.00 atm 298 K
V2 = T2P1V1
P2T1
= 68.7 L