Practice with the Gas Laws with Answers
STP = 0 oC and 1.0 atm
All temperatures must be in KELVIN
Converting one unit of pressure to another
1. Change 758. mm Hg to Torr
758 mm Hg
x
760 Torr
760 mm Hg
= 758 Torr
2. Change 771.3 mm Hg to atm
771.3 mm Hg x
1atm
= 1.015 atm
760 mm Hg
3. Change 95.6 kPa to mm Hg
95.6 kPa
x
760 mm Hg
101.325 kPa
=
717 mm Hg
4. Is the pressure of the gas sample greater
or less than the pressure of the
atmosphere? Explain.
Less, the mercury line is higher on that side
What is the pressure of the gas in this
manometer (in mm Hg)?
760 mm Hg
-655 mm Hg
105 mm Hg
Convert that pressure to kPa.
105 mm Hg
x
101.325 kPa
760 mm Hg
=
14.0 kPa
Calculating Density of a Gas at STP
1. What is the density of dinitrogen tetroxide at STP?
N2O4
92.01 g = 4.108 g/L
22.4 L
2. What is the density of sulfur trioxide at STP?
SO3
80.06 g = 3.574 g/L
22.4 L
Atmospheric pressure 760 mm Hg
Dalton’s Law of Partial Pressures
PTotal = PA + PB + PC + …
1. What is the total pressure of a gas mixture containing H2 at 0.25 atm, N2 at 0.55 atm,
and NO at 0.30 atm?
PTotal = ?
PH2 = 0.25 atm
PN2 = 0.55 atm
PNO = 0.30 atm
PTotal = PH2 + PN2 + PNO
PTotal = 0.25 atm + 0.55 atm + 0.30 atm
PTotal = 1.10 atm
2. The total pressure of a gaseous mixture of He, Ne and Ar is 1.10 atm. What is the
partial pressure of He if the partial pressures of Ne and Ar are 0.55 atm and 0.30
atm respectively?
PTotal = 1.10 atm
PHe = ? atm
PNe = 0.55 atm
PAr = 0.30 atm
PTotal = PHe + PNe + PAr
1.10 atm = PHe + 0.55 atm + 0.30 atm
PHe = 0.25 atm
3. Oxygen gas from the decomposition of potassium chlorate was collected by water
displacement. The barometric pressure and the temperature during the experiment
were 731.0 mm Hg and 20.0 oC, respectively. Assuming that the water level inside
and outside the gas collection vessel was the same height, what was the partial
pressure of the oxygen collected? (Because the water level is equal inside and
outside the gas collection vessel, Patmosphere = PTotal)
PTotal = 731.0 mm Hg
PH2O (at 20 oC) = 17.5 mm Hg
PO2 = ?
PTotal = PH20 + PO2
731.0 mm Hg = 17.5 mm Hg + PO2
PO2 = 713.5 mm Hg
5. Carbon dioxide gas from the decomposition of copper (II) carbonate was
collected by water displacement. The barometric pressure and the
temperature during the experiment were 98.40 kPa and 20.0 oC, respectively.
Assuming that the water level inside and outside the gas collection vessel
was the same height, what was the partial pressure of the carbon dioxide
collected?
98.40 kPa x
760 mm Hg = 738.1 mm Hg
101.325 kPa
PTotal = 738.1 mm Hg
PH2O (at 20 oC) = 17.5 mm Hg
PCO2 = ?
PTotal = PH20 + PO2
738.1 mm Hg = 17.5 mm Hg + PCO2
PCO2 = 720.6 mm Hg
Boyle’s Law
P1V1 = P2V2
1. A 7.00 L sample of gas at 800. Torr is changed at constant temperature until its
pressure is 1,000. Torr. What is its new volume?
V1 = 7.00 L
P1 = 800. torr
V2 = ? L
P2 = 1000. torr
P1V1 = P2 V2
(800.)(7.00) = (1000.)(V2)
V2 = 5.60 L
2. To what pressure must a sample of gas be subjected at constant temperature in
order to compress it from 500. mL to 300. mL if its original pressure is 1.71 atm?
V1 = 500. mL
P1 = 1.71 atm
V2 = 300. mL
P2 = ? atm
P1V1 = P2 V2
(500.)(1.71) = (P2)(300.)
P2 = 2.85 atm
3. A flask contains 155 cm3 of hydrogen collected at a pressure of 22.5 kPa. Under
what pressure would the gas have a volume of 90.0 cm3 at the same temperature?
(Recall that 1cm3 = 1mL)
V1 = 155 cm3
P1 = 22.5 kPa
V2 = 90.0cm3
P2 = ? kPa
P1V1 = P2V2
(22.5)(155) = (P2)(90.0)
P2 = 38.8 kPa
4. If the pressure exerted on a 300.0 mL sample of hydrogen gas at constant
temperature is increased from 0.500 atm to 0.750 atm, what will be the final volume
of the sample?
V1 = 300.0 mL
P1 = .500 atm
V2 = ? cm3
P2 = .750 atm
P1V1 = P2V2
(.500)(300.0) = (.750)(V2)
V2 = 200. mL
5. A helium balloon has a volume of 5.0 L at a pressure of 101.3 kPa. The balloon is
released and reaches an altitude of 6.5 km at a pressure of 50.7 kPa. If the gas
temperature remains the same, what is the new volume of the balloon? Assume
that pressures are the same inside and outside the balloon.
V1 = 5.0L
P1 = 101.3 k Pa
V2 = ? mL
P2 = 50.7 kPa
P1V1 = P2V2
(101.3)(5.0) = (50.7)(V2)
V2 = 10. L
6. A sample of oxygen gas has a volume of 150. mL at a pressure of 0.947 atm. What
will the volume of the gas be at a pressure of 1.00 atm if the temperature remains
constant?
V1 = 150. mL
P1 = .947 atm
V2 = ?
P2 = 1.00 atm
P1V1 = P2V2
(.947)(150.) = (1.00)(V2)
V2 = 142 mL
Charles’ Law
V1 = V2
T1
T2
1. A 4.50 L sample of gas is warmed at constant pressure from 300. K to 350. K. What
will its final volume be?
V1 = 4.50 L
T1 = 300. K
V2 = ? L
T2 = 350. K
V1 = V2
T1
T2
4.50 L =
V2
300. K
350. K
V2 = 5.25 L
2. A 4.50 L sample of gas is warmed at constant pressure from 27 oC to 77 oC. What
will its final volume be?
V1 = 4.50 L
T1 = 300 K
V2 = ? L
T2 = 350 K
V1 = V2
T1
T2
4.50 L =
V2
300. K
350. K
V2 = 5.25 L
(remember, temperature must be in KELVINS!)
3. Gas in a balloon occupies 2.5 L at 300.0 K. The balloon is dipped into liquid
nitrogen that is at a temperature of 80.0 K. What volume will the gas in the balloon
occupy at this temperature?
V1 = 2.5 L
T1 = 300.0K
V2 = ? L
T2 = 80.0K
V1 = V2
T1
T2
2.5 L =
300.0 K
V2
80. K
V2 = .67 L
4. A helium-filled balloon has a volume of 2.75 L at 20. oC. The volume of the balloon
decreases to 2.46 L after it is placed outside on a cold day. What is the outside
temperature in oC?
V1 = 2.75 L
T1 = 20 oC 293 K
V2 = 2.46 L
T2 = ? oC
V1 = V2
T1
T2
2.75 L = 2.46 L
293 K
T2
T2 = 262 K
262 K
-273
-11 oC
5. A gas at 65 oC occupies 4.22 L. At what Celsius temperature will the volume be
3.87 L, assuming the same pressure?
V1 = 4.22 L
T1 = 65 oC 338 K
V2 = 3.87 L
T2 = ? oC
V1 = V2
T1
T2
4.22 L = 3.87 L
338 K
T2
T2 = 310 K
310 K
-273
37 oC
6. Gay-Lussac’s Law
P1
T1
= P2
T2
1. The gas in a sealed can is at a pressure of 3.00 atm at 25 oC. A warning on the can
tells the user not to store the can in a place where the temperature will exceed 52 oC.
What would the gas pressure (in atm) in the can be at 52 oC?
P1 = 3.00 atm
T1 = 25 oC 298 K
P2 = ? atm
T2 = 52 oC 325 K
P1 = P2
T1 T2
3.00 atm = P2
298 K
325 K
P2 = 3.27 atm
2. Before a trip from New York to Boston, the pressure in an automobile tire is 1.8 atm
at 20. oC. At the end of the trip, the pressure gauge reads 1.9 atm. What is the new
Celsius temperature of the air inside the tire? (Assume tires with constant volume.)
P1 = 1.8 atm
T1 = 20 oC 293 K
P2 = 1.9 atm
T 2 = ? oC
P1 = P2
T1 T2
1.8 atm = 1.9 atm
293 K
T2
T2 = 309 K
309 K
-273
36 oC
3. At 120. oC, the pressure of a sample of nitrogen is 1.07 atm. What will the pressure
be at 205 oC, assuming constant volume?
P1 = 1.07 atm
T1 = 120 oC 393 K
P2 = ? atm
T2 = 205 oC 478 K
P1 = P2
T1 T2
1.07 atm = P2
393 K
478 K
P2 = 1.30 atm
4. A sample of helium gas has a pressure of 1.20 atm at 22 oC. At what Celsius
temperature will the helium reach a pressure of 2.00 atm?
P1 = 1.20 atm
T1 = 22 oC 295 K
P2 = 2.00 atm
T 2 = ? oC
P1 = P2
T1 T2
1.20 atm = 2.00 atm T2 = 492 K
295 K
T2
492 K
-273
219 oC
5. The gas in a vessel exerts 1.4 atm at 22 oC. What pressure would be exerted if you
doubled the temperature? (Remember what to do with temperatures!)
P1 = 1.4 atm
T1 = 22 oC 295 K
P2 = ? atm
T2 = 590 K
P1 = P2
T1 T2
1.4 atm = P2
295 K
590 K
P2 = 2.8 atm
Combined Gas Law
P1V1
T1
= P2V2
T2
1. A helium-filled balloon has a volume of 50.0 L at 25 oC and 1.08 atm. What volume
will it have (in L) at 0.855 atm and 10. oC?
P1 = 1.08 atm
V1 = 50.0 L
T1 = 25oC 298 K
P2 = .855 atm
V2 = ? L
T2 = 10.oC 283 K
P1V1
T1
= P2V2
T2
(1.08 atm)(50.0 L) = (.855 atm)(V2)
298 K
283 K
V2 = (1.08)(50.0)(283) : (298)(.855) = 60.0 L
2. The volume of a gas is 27.0 mL at 22.0 oC and 0.974 atm. What will the volume be
(in mL) at 15.0 oC and 0.993 atm?
P1 = .974 atm
V1 = 27.0 mL
T1 = 22oC 295 K
P2 = .993 atm
V2 = ? L
T2 = 15.oC 288 K
P1V1
T1
= P2V2
T2
(.974 atm)(27.0 mL) = (.993 atm)(V2)
295 K
288 K
V2 = (.974)(27.0)(288) : (295)(.993) = 25.9 mL
3. A 700. mL gas sample at STP is compressed to a volume of 200. mL and the
temperature is increased to 30.0 oC. What is the new pressure of the gas in Pa?
(Remember, STP = 0 oC and 1.0 atm)
P1 = 1.0 atm
V1 = 700. mL
T1 = 0oC 273 K
P2 = ? kPa
V2 = 200. mL
T2 = 30.0oC 303 K
P1V1
T1
= P2V2
T2
(1.0 atm)(700. mL) = (P2)(200. mL)
273 K
303 K
V2 = (1.0)(700.)(303) : (273)(200.) = 3.88 atm x 101325 Pa = 390, 000 Pa
1 atm
4. At 22 oC a sample of gas occupies 2.35 L of space and exerts 750 mm Hg of
pressure. What will the new volume be (in L) at STP?
P1 = 750 mm Hg
V1 = 2.35 L
T1 = 22oC 295 K
P2 = 1 atm 760 mm Hg
P1V1 = P2V2
V2 = ? L
T1
T2
T2 = 0oC 273 K
(750 mm Hg)(2.35 L) = (760 mm Hg)(V2)
295 K
273 K
V2 = (750)(2.35)(273) : (295)(760) = 2.1 L
A

B
Graham’s Law of Effusion
MolarMass B
MolarMass A
1. At a certain temperature, molecules of chlorine gas travel at 0.380 km/s. What is the
speed of sulfur dioxide gas molecules under the same conditions?
 A = .380 km/s
Gas A = SO2
Gas B = Cl2
 B = ? km/s
MMA = 64.06
MMB = 70.906
 SO 2

 Cl 2
MolarMass Cl 2
 SO2
MolarMass SO 2
.380

70.906
64.06
 SO 2 = .400 km/s
2. Given that atoms of neon gas effuse at 800. m/s at a given temperature, calculate
the rate of effusion for molecules of butane gas, C4H10, at the same temperature.
Gas A = Ne
 A = 800. km/s
Gas B = C4H10
 B = ? km/s
MMA = 20.174
MMB = 58.123
 Ne
 C 4 H 10

MolarMass C 4 H 10
800.
MolarMass Ne
 C 4 H 10

58.123
20.174
 C 4H 10 = 471 km/s
3. The average speed of helium atoms is 1.20 x 103 m/s at a certain temperature.
What is the average speed of HCl molecules at the same temperature?
Gas A = He
Gas B = HCl
3
=1.20x10
km/s
A
 B = ? km/s
MMA = 4.0026
MMB = 36.461
 He

 HCl
MolarMass HCl
1.20 x10 3
 HCl
MolarMass He

36.461
4.0026
 HCl = 3.98 X102 km/s
4. Compare the rate of effusion of molecules of water vapor with that of molecules of
carbon dioxide gas at the same pressure and temperature. (Which one travels
faster, and how many times faster does it travel?)
Gas A = H2O
Gas B = CO2
MMA = 18.02
MMB = 44.01
 H 2O

 CO 2
MolarMass CO 2
MolarMass H 2O

44.01
18.02
= ~1.6
H2O vapor travels
1.6 times faster than
CO2 gas
You have just determined that the lighter gas, H2O, travels 1.6 times faster than the heavier
gas, CO2. If you get an answer of less than 1 (which means the gas travels slower) when you
solve for how much faster one gas travels than the other, switch your Gas A and Gas B
5. Hydrogen sulfide has a very strong rotten-egg odor. Methyl salicylate, C8H10O3, has
a wintergreen odor. Benzaldehyde, C7H6O, has an almond odor. If vapors for these
three substances were released at the same time from across the room, which
would you smell first? Why?
You should smell the H2S first. It is lighter (smaller molar mass), so it would
travel faster to your nose!
PV = nRT
Ideal Gas Law
State the characteristics of an ideal gas…





Gas molecules are very small (negligible size), and are spread very far apart
Gas molecules are in constant random motion
Gas molecules have an average KE directly proportional to their Kelvin temp
Gas molecules have elastic collisions
There are no forces of attraction or repulsion between gas molecules
Under what conditions do real gases act most like ideal gases? Explain your logic.
Conditions of high temperature (because at low temp, gases turn into
liquids!) and high pressure (because at high pressure, gas molecules
can be compressed into liquids)
1. Suppose you have a 500.0 mL container that holds 15.0 g of oxygen gas
at a temperature of 32 oC. What is the pressure inside the container in
mm Hg?
P=?
PV = nRT
V = 500.0 mL .500 L
(P)(.500 L) = (.469 mol)(62.4 L.atm/mol.K)(305 K)
n = 15.0 g x 1 mol O2 = .469 mol O2
P = 17,900 mm Hg
32.00 g O2
R = 62.4 L.atm
The label for R determines
mol.K
your other labels!
o
T = 32 C 305 K
2. How many grams of oxygen gas must be in a 980 mL container to exert a
pressure of 9.70 kPa at a temperature of 25 oC?
P = 9.70 kPa
V = 980 mL .980 L
n=
R = 8.314 L.kPa
mol.K
o
T = 25 C 298 K
PV = nRT
(9.70 kPa)(.980 L) = (n)(8.314 L.kPa/mol.K)(298 K)
n = .00384 mol O2 x 32.00 g O2 = .123 g O2
1 mol O2
3. What is the volume in milliliters of .125 mol of carbon monoxide gas at 20.0o
C and 101.33 kPa of pressure?
P = 101.33 kPa
V = ? mL
n = .125 mol O2
R = 8.314 L.kPa
mol.K
o
T = 20 C 293 K
PV = nRT
(101.33 kPa)(V) = (.125 mol)(8.314 L.kPa/mol.K)(293 K)
V = 3.01 L = 3010 mL
3. What is the percent yield if 93.5 mL of gas is collected over 27 oC water at 751
mm Hg atmospheric pressure if a 1.50 g sample of Aluminum is placed in 75.0
mL of 1.25 M HCl?
2Al + 6HCl  2AlCl3 + 3H2
1.50 g Al x 1 mol Al x 3 mol H2 = .0834 mol H2
26.98 g Al 2 mol Al
75.0 mL HCl x
1 L HCl
x 1.25 mol HCl x 3 mol H2 = .0469 mol H2
1000 mL HCl
1 L HCl
6 mol HCl
PH2 = 751 mm Hg – 26.7 mm Hg
VH2 = .0935 L
nH2 = ? mol H2
R = 62.4 L.mm Hg/mol.K
TH2 = 27oC 300. K
Actual yield
Theoretical yield
TY
= 724.3 mm Hg
PV=nRT
(724.3 mm Hg)(.0935 L) = (n)(62.4 L.mm Hg/mol.K)(300. K)
n = .00362 mol H2
x 100 = % yield
.00362 mol H2 x 100 = 7.71% yield
.0469 mol H2
5. What is the percent yield if 1.74 L of hydrogen gas are collected over 22 oC
water at 745 mm Hg if 4.00 g of sodium reacts with 40.0 g water?
2Na + 2HOH  2NaOH + H2
4.00 g Na x 1 mol Na x 1 mol H2 = .0870 mol H2 TY
22.99 g Na 2 mol Na
40.0 g H2O x 1 mol H2O x 1 mol H2 = 1.11 mol H2
18.02 g H2O 2 mol H2O
PH2 = 745 mm Hg – 19.8 mm Hg
VH2 = 1.74 L H2
nH2 = ? mol H2
R = 62.4 L.mm Hg/mol.K
TH2 = 22oC 295 K
Actual yield
Theoretical yield
= 725.2 mm Hg
PV=nRT
(725.2 mm Hg)(1.74 L) = (n)(62.4 L.mm Hg/mol.K)(295 K)
n = .069 mol H 2
x 100 = % yield
.069 mol H2 x 100 = 79.3% yield
.0870 mol H2