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 V1 = V2 V2 = V1T2
T1 = 298 K T1 T2 T1
T2 = 373 K
V2 = x
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 V1 = V2 T2 = T1V2
T1 = 273 K T1 T2 V1
T2 = x
V2 = 325 mL
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 = ???? PV = nRT T = PV = 0.991 (604)
n = 0.0851 moles nR 0.0851(0.08206)
v = 604 L
P = 100.4 kPa = 0.991 atm T = 85700 K
R = 0.08206 L∙atm/mol∙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 PV = nRT Ideal Gas Law
P = 66.7 Pa = 6.58 x 10 –4 atm
T = 10. °C = 283 K n = PV = (6.58 x 10–4atm)(0.486 L)
R = 0.08206 L atm/mol K RT (0.08206 L atm/mol K )(283 K)
n = ?
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 V1 = V2 V2 = V1n2 Avogadro’s Law
n1 = 425 mol n1 n2 n1
n2 = 443 mol
V2 = x
V2 = 9.8 L (443 mol) = 10. L
425 mol
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 = 0.0893 mol of gas
22.4 L
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 P1V1 = P2V2 V2 = T2P1V1
T1 = 298 K T1 T2 P2T1
P1 = 15.0 atm
T2 = 273 K
V2 = x L
P2 = 1.00 atm
V2 = 273 K 15.0 atm 5.00 L = 68.7 L
1.00 atm 298 K