Answer Key to Gas Laws Problem Set
The first four problems use the ideal gas law. Notice that pressure, volume, temperature and moles are
not changing. When using the ideal gas law, PV = nRT, the constant, R = 0.08206 L∙atm/mol∙K,
dictates the units of pressure, volume and temperature. The pressure unit must be in atmospheres, the
volume unit must be in liters, and the temperature unit must be in Kelvin.
1. What is the pressure in atmospheres of 0.0233 mol an ideal gas at 27°C in a 52.9 mL container?
Let’s collect the terms we need and make sure they are in the proper units:
n = 0.0233 mol
(no conversion is necessary here)
T = 27°C + 273 = 300K
V = 52.9 mL x
(temperature MUST be in Kelvin)
1 x 10−3 L
= 0.0529 L (volume MUST be in liters)
1 mL
R = 0.08206 L∙atm/mol∙K
Rearrange PV = nRT and solve for pressure by dividing through by V:
P =
nRT
(0.0233 mol)(0.08206 L∙atm/mol∙K)(300K)
=
= 10.8 atm
V
0.0529 L
2. What is the volume in liters of 2.33 mol of an ideal gas at 27°C at a pressure of 872 torr?
Let’s collect the terms we need and make sure they are in the proper units:
n = 2.33 mol
(no conversion necessary here)
T = 27°C + 273 = 300K
P = 872 torr x
(temperature MUST be in Kelvin, as stated above)
1 atm
= 1.15 atm
760 torr
(pressure MUST be in atmospheres)
R = 0.08206 L∙atm/mol∙K
Rearrange PV = nRT and solve for volume by dividing through be P:
V =
nRT
(2.33 mol)(0.08206 L∙atm/mol∙K)(300K)
=
= 49.9 L
P
1.15 atm
3. What is the temperature in Kelvin of 0.0233 mol of an ideal gas at a pressure of 458 torr and a
volume of 1870 mL?
Let’s collect the terms we need and make sure they are in the proper units:
n = 0.0233 mol
(no conversion necessary here)
3. (continued)
P = 458 torr x
1 atm
= 0.603 atm
760 torr
V = 1870 mL x
(pressure MUST be in atmospheres)
1 x 10−3 L
= 1.870 L (volume MUST be in liters)
1 mL
R = 0.08206 L∙atm/mol∙K
Rearrange PV = nRT and solve for temperature by dividing through by nR:
T =
PV
(0.603 atm)(1.870L)
=
= 590 K
nR
(0.0233 mol)(0.08206 L∙atm/mol∙K)
4. How many moles of ideal gas are present in a 347 mL container with a pressure of 3.88
atmospheres at a temperature of 383 °C?
Let’s collect the terms we need and make sure they are in the proper units:
V = 347 mL x
P = 3.88 atm
1 x 10−3 L
= 0.347 L
1 mL
(volume MUST be in liters)
(no converstion is needed)
T = 383°C + 273 = 656 K
(temperature MUST be in Kelvin)
Rearrange PV = nRT and solve for moles by dividing by RT:
n =
PV
(3.88 atm)(0.347 L)
=
= 0.0250 mol
RT
(0.08206 L∙atm/mol∙K)(656 K)
The next series of problems deal with the various gas laws in which pressure, volume, moles and or
temperature conditions change. One way to determine which gas law to use is to consider a gas law you
can name after yourself, in which only R is constant:
(Insert you name here!) Law:
(I’ll call it Ladon’s Law, after myself for the purpose of this tutorial)
P1 V1
P2 V2
=
n1 T1
n2 T2
As you read the problem, cross out any variable not being used to leave behind the equation you need.
You’ll see how this applies in the problems and solutions that follow.
These equations are ratios, thus pressure and volume units may be any unit you choose as long as they are
consistent. However, temperature still must be converted to Kelvin. The temperature cannot be negative,
or have a value of zero as these will lead to pressures and volumes that are impossible. Pressure and
volume can decrease to a value of zero, but they cannot be negative. If you are wondering about 0K, a
gas does not exist at absolute zero, so that will not be an issue.
5. An ideal gas initially at a pressure of 0.233 atm in a 5.32 L piston is compressed to a volume of
3.88 L. What is the new pressure of the gas in atmospheres and in torr?
Notice that only pressure and volume are changing in this problem. Cross out moles (n) and temperature
(T) in Ladon’s Law to reveal the equation you need to solve the problem:
P1 V1
P2 V2
=
n1 T1
n2 T2
The gas law you need is P1V1 = P2V2 , known as Boyle’s Law. Assign pressures and volumes to the
variables:
P1 = 0.233 atm
P2 = ? atm
V1 = 5.32 L
V2 = 3.88 L
Solve for the unknown variable, P2, by dividing through by V2:
P1 V1
(0.233 atm)(5.32 L)
=
= 0.319 atm
V2
3.88 L
P2 =
Notice how the liter units cancel to leave pressure units in atm.
6. An ideal gas initially at a pressure of 459 torr in a 1438 mL piston has its pressure adjusted to
1.79 atm. What is the new volume of the piston?
Notice that only pressure and volume are changing in this problem. Cross out moles (n) and temperature
(T) in Ladon’s Law to reveal the equation you need to solve the problem:
P1 V1
P2 V2
=
n1 T1
n2 T2
The gas law you need is P1V1 = P2V2 , known as Boyle’s Law. Assign pressures and volumes to the
variables:
P1 = 459 torr
P2 = 1.79 atm
V1 = 1438 mL
V2 = ? mL
Notice the pressure units are not consistent. Let’s convert P2 = 1.79 atm to torr:
P2 = 1.79 atm x
760 torr
= 1360 torr
1 atm
Rearrange the equation to solve for V2 by dividing through by P2:
V2 =
P1 V1
(459 torr)(1438 mL)
=
= 485 mL
P2
1360 torr
Notice that we did not need to have volume in liters and pressure in atmospheres. This is because the
equation is a ratio, and the gas law constant, R, is not involved. The torr units cancel to leave volume
units of milliliters.
7. An ideal gas initially at a temperature of 18.44°C in a 5.00 L container is heated to 127.33°C.
What is the new volume of the container in L?
Notice that only volume and temperature are changing. Cross out moles and pressure in Ladon’s Law to
reveal the equation you need to solve the problem:
P1 V1
P2 V2
=
n1 T1
n2 T2
The gas law required is Charles’ Law:
V1
V2
=
T1
T2
Assign the volume and temperatures to the variables. You must convert the temperatures to Kelvin:
V1 = 5.00 L
V2 = ? L
T1 = 18.44°C + 273.15 = 291.59 K
T2 = 127.33°C + 273.15 = 400.48 K
Rearrange Charles’ Law to solve for V2 by multiplying through by T2:
V2 =
V1 T2
(5.00 L)(400.48 K)
=
= 6.87 L
T1
291.59K
Notice how the Kelvin units cancel out to leave the volume in L. Convert these volume units the
milliliter units asked for in the problem:
V = 6.87 L x
1 mL
= 6.78 x 103 L
1 x 10-3 L
8. An ideal gas initially at a temperature of 345°C and a volume of 478mL is compressed to a
volume of 233 mL. What is the new temperature of the gas in Celsius?
Notice that only volume and temperature are changing. Cross out moles and pressure in Ladon’s Law to
reveal the equation you need to solve the problem:
P1 V1
P2 V2
=
n1 T1
n2 T2
The gas law required is Charles’ Law:
V1
V2
=
T1
T2
Assign the volume and temperatures to the variables. You must convert the temperatures to Kelvin:
V1 = 478 mL
V2 = 233 mL
T1 = 345°C + 273 = 618 K
T2 = ?
8. (continued)
Rearrange Charles’ Law to solve for T2. One easy way is to cross-multiply and divide through by V1:
V1
V2
=
upon cross-multiplication becomes 𝑉1 𝑇2 = 𝑉2 𝑇1
T1
T2
Dividing through by V1 yields:
𝑇2 =
𝑉2 𝑇1
(233 𝑚𝐿)(618 𝐾)
=
= 301 𝐾
𝑉1
478 𝑚𝐿
Notice how the volume units in milliliters cancel to leave temperature units of Kelvin. We now need to
convert the Kelvin units back into Celsius, as asked for in the problem.
T2 = 301 K – 273 = 28°C
9. An ideal gas initially at a temperature of 445K at a pressure of 553 atmospheres is cooled to a
temperature of 37°C in a rigid container. What is the new pressure of the gas in the container in
atmosphere and torr units?
Notice that only temperature and pressure are changing. Cross out moles and volume in Ladon’s Law to
reveal the equation you need to solve the problem:
P1 V1
P2 V2
=
n1 T1
n2 T2
The gas law required is the Guy-Lussac Law:
P1
P2
=
T1
T2
Assign the pressure and temperatures to the variables. You must convert the temperatures to Kelvin:
P1 = 553 atm
P2 = ? atm
T1 = 445 K (already in K)
T2 = 37°C + 273 = 310 K
Rearrange the Guy-Lussac Law to solve for the pressure, P2, by multiplying through by the temperature, T2.
P2 =
P1 T2
(553 atm)(310 K)
=
= 385 atm
T1
445 K
Notice how the Kelvin units cancel to leave pressure units of atmospheres.
The pressure in torr would be:
P = 385 atm x
760 torr
= 2.93 x 105 torr
1 atm
10. An ideal gas initially at a temperature of 388K and a pressure of 3.87 atmospheres in a 288 mL
container is cooled to a temperature of 125K and the pressure is raised to 6.33 atmospheres. What
is the new volume of the container in milliliters?
Notice that pressure, volume and temperature are changing. Cross out the mole term in Ladon’s Law to
reveal the equation you need to solve the problem:
P1 V1
P2 V2
=
n1 T1
n2 T2
The equation required is often referred to as the combined gas law:
P1 V1
P2 V2
=
T1
T2
Assign the pressures, volumes and temperatures to the variables.
P1 = 3.87 atm
V1 = 288 mL
T1 = 388 K
P2 = 6.33 atm
V2 = ? mL
T2 = 125 K
Rearrange and solve for V2 by multiplying through by T2 and dividing through by P2:
V2 =
P1 V1 T2
(3.87 atm)(288 mL)
125 K
x
=
x
= 56.7 mL
T1
P2
388 K
6.33 atm
Notice how the pressure of atmospheres and temperature units of Kelvin cancel to leave volume units of
milliliters.
11. If 15.0 moles of an ideal gas has a pressure of 3.55 atmospheres at constant volume and
temperature, how many moles of gas will have a pressure of 9.45 atmospheres?
Notice that only pressure and moles are changing. Cross out volume and temperature from Ladon’s Law
to reveal the equation you need to solve the problem:
P1 V1
n1 T1
The equation required is:
=
P2 V2
n2 T2
(name, if any, is unknown to me!)
P1
n1
=
P2
n2
Assign the pressures and moles to the variables:
P1 = 3.55 atm
P2 = 9.45 atm
n1 = 15.0 mol
n2 = ?
11. (continued)
Rearrange and solve for n2 by cross-multiplying the equation and then dividing through by P1:
Cross multiplication yields:
P1n2 = P2n1
Dividing through by P1 gives the equation needed to solve the problem:
n2 =
P2 n1
(9.45 atm)(15.0 mol)
=
= 39.9 mol
P1
3.55 atm
Notice how the pressure units of atmospheres cancel out to leave mole units.
12. What is the density of 55.2 grams of argon gas at a pressure of 3.95 atmospheres and a
temperature of 35°C?
The density of a gas is usually expressed in units of grams/liter, (g/L), which is a mass/volume. From the
information in the problem, the mass is known, but the volume must be calculated from the ideal gas law.
Let’s gather the information we need ( moles, pressure and temperature). Since we are using the ideal gas
law, the pressure must be in atmospheres and the temperature must be in Kelvin. The moles of argon will
be found using the atomic weight of argon, which is 39.95 g/mol.
P = 3.95 atm
(already in atmospheres, no conversion needed)
V = ?L
n = 55.2 g x
1 mol Ar
= 1.38 mol Ar
39.95 g Ar
R = 0.08206 L∙atm/mol∙K
T = 35°C + 273 = 308 K
Rearrange PV = nRT and solve for volume by dividing through be P:
V =
nRT
(1.38 mol)(0.08206 L∙atm/mol∙K)(308K)
=
= 8.83 L
P
3.95 atm
Notice the pressure, mol, and temperature units cancel to leave volume units of liters. I’ll show you this
as a complex fraction:
mol∙L∙atm∙K
1
mol∙K
atm
x
= L
atm
1
1
atm
Use the mass and volume of the argon to find the density:
D=
m
55.2 g
=
= 6.25 g/L
V
8.83 L
13. A mixture of gas have the following partial pressures: 233 torr N2, 353 torr of He and 832 torr
of O2. (a) What is the total pressure of the mixture of gases? (b) What are the mole fractions of
each of these gases in the mixture?
(a) What is needed here is Dalton’s Law of Partial Pressures. Dalton’s Law states that the total pressure
is equal to sum of the partial pressures of all the gases in a mixture.
Ptotal = pN2 + pHe + pO2 = 233 torr + 353 torr + 832 torr = 1418 torr
(b) A fraction is a “part” over a “whole”. The mole fraction, X, represents the moles of an individual gas
divided by the total moles of all the gases in the mixture. Since moles and pressure are directly
proportional, we can use the partial pressures and total pressure in place of moles:
X N2 =
pN2
233 torr
=
= 0.164
Ptotal
1418 torr
X He =
pHe
353 torr
=
= 0.249
Ptotal
1418 torr
X O2 =
pO2
832 torr
=
= 0.586
Ptotal
1418 torr
Notice mole fraction does not have any units; it is a fraction.
14. A mixture of gases contains 3.44 mol of He, 1.39 mol of Ne and 7.33 mol of N 2. (a) What are
the mole fractions of each of the gases in the mixture? (b) What are the partial pressures of each of
these gases if the total pressure is 5.87 atmospheres?
(a) Here we will use the moles of each gas to find the mole fraction of each gas. The total moles of gas is
the sum of the individual moles:
ntotal = nN2 + nHe + nNe = 7.33 mol N2 + 3.44 mol He + 1.39 mol Ne = 12.16 mol
The mole fraction, X, represents the moles of an individual gas divided by the total moles of all the gases
in the mixture.
X N2 =
nN2
7.33 mol
=
= 0.603
ntotal
12.16 mol
X He =
nHe
3.44 mol
=
= 0.283
ntotal
12.16 mol
X Ne =
nNe
1.39 mol
=
= 0.114
ntotal
12.16 mol
Notice once again, that mole fractions do not have units.
14. (continued)
(b) The partial pressures of each gas can be calculated by multiplying the total pressure by the mole
fraction of each gas:
pN2 = XN2 Ptotal = 0.603(5.87 atm) = 3.54 atm
pHe = X He Ptotal = 0.283(5.87 atm) = 1.66 atm
pNe = XNe Ptotal = 0.114(5.87 atm) = 0.669 atm
15. An antacid tablet is analyzed for %CaCO 3 by measuring the pressure, volume and
temperature of the CO2 generated by adding acid to the sample according to the reaction:
CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)
A 0.200 gram tablet of antacid produces 34.55 mL of gas at 26.80°C. The gas was collected over
water. Water at 26.8°C has a vapor pressure of 26.43 torr. The atmospheric pressure at the time of
the analysis was 758.33 torr. What is the %CaCO 3 in the antacid sample?
This is fundamentally a stoichiometry problem. Most stoichiometry problems follow the strategy:
Quantity A → mol A → mol B → Quantity B
We will use this strategy to solve this problem. Our first task is to find the moles of carbon dioxide.
Previously you have seen grams being converted to moles using a molar mass or a molarity and volume
being used to find moles. Here the moles of gas is found by using the ideal gas law, PV = nRT.
Collect and convert the pressure, volume an temperature to the proper units. The pressure of the carbon
dioxide alone must be found using Dalton’s Law:
pCO2 = Ptotal - pH2 O = 758.33 torr - 26.43 torr = 731.9 torr
Next, convert the pressure of carbon dioxide into atmospheres:
pCO2 = 731.9 torr x
1 atm
= 0.963 atm
760 torr
Convert the volume of carbon dioxide from milliliters to liters:
VCO2 = 34.55 mL x
1 x 10-3 L
= 0.03455 L
1 mL
Convert the temperature in Celsius to Kelvin:
TCO2 = 26.80°C + 273.15 = 299.95 K
15. (continued)
We are now ready to find the moles of carbon dioxide using the ideal gas law. Rearrange PV = nRT to
solve for moles by dividing through by RT:
n=
PV
(0.963 atm)(0.03455 L)
=
= 0.00135 mol CO2
RT
(0.08206 L∙atm/mol∙K)(299.95 K)
We have just completed the first part of the stoichiometry strategy (Quantity A to mol A.) Let’s complete
the remainder of the strategy. We will convert moles of carbon dioxide to moles of calcium carbonate to
grams of calcium carbonate and finally to percent calcium carbonate:
0.00135 mol CO2 x
1 mol CaCO3 100.1 g CaCO3
100
x
x
= 67.57% CaCO3
1 mol CO2
1 mol CaCO3
0.200 g antacid
Here is the problem summarized according the stoichiometry strategy:
Quantity A
P, V. T of CO2
→
→
mol A
mol CO2
→
→
mol B
→
mol CaCO3 →
Quantity B
g CaCO3 to %CaCO3
In the last step, we divided the grams of CaCO3 by the grams of antacid table and multiplied by 100 to
obtain a percent. A percent is the quantity of a “part” divided by a “whole” multiplied by 100. See how
the units in the above expression relate this!
16. What is the density of sulfur dioxide gas at 974 torr and a temperature of 55.7 °C?
From the ideal gas law we can derive an expression that relates, molar mass, pressure, density and
temperature. Start with PV = nRT.
The moles, n, can be substituted with mass in grams (g) divided by molar mass, (ℳ):
n=
PV =
Multiply through by molar mass:
g
ℳ
g
x RT
ℳ
ℳPV = gRT
Divide by volume, V:
ℳP =
g
x RT
V
Since density, D, is mass divided by volume (g/V), substitute D for g/V:
ℳP = DRT
An easy way to remember this equation is the mnemonic “Mud Pies are made from DiRT”.
16 (continued)
Using this equation, collect your terms and convert them to the appropriate unit:
P = 974 torr x
1 atm
= 1.28 atm
760 torr
ℳ = 64.07 g/mol
D = ? g/L
R = 0.08206 L∙atm/mol∙K
T = 55.7 °C + 273.15 = 328.8 K
Rearrange ℳ𝑃 = 𝐷𝑅𝑇 and solve for density:
D=
ℳP
(64.07 g/mol)(1.28 atm)
=
= 3.04 g/L
RT
(0.08206 L∙atm/mol∙K)(328.8 K)
17. What pressure is needed to have neon gas at 56.9 °C have a density of 0.555 g/L?
We will use ℳP = DRT to solve this problem. Collect terms and convert the temperature to the
appropriate unit:
ℳ = 20.18 g/mol
P = ? atm
D = 0.555 g/L
R = 0.08206 L∙atm/mol∙K
T = 56.9 °C + 273.15 = 330.0 K
Rearrange ℳP = DRT to solve for P by dividing through by the molar mass, ℳ:
P=
DRT
(0.555 g/L)(0.0826 L∙atm/mol∙K)(330.0 K)
=
= 0.745 atm
ℳ
20.18 g/mol
18. An unknown gas has a density of 1.39 g/L at a temperature of 56.8°C and a pressure of 1.35
atmospheres. What is the molar mass of the gas?
We will use ℳP = DRT to solve this problem. Collect terms and convert the temperature to the
appropriate unit:
ℳ = ? g/mol
R = 0.08206 L∙atm/mol∙K
P = 1.35 atm
D = 1.39 g/mol
T = 56.8 °C + 273.15 = 330.0 K
18. (continued)
Rearrange ℳP = DRT to solve for the molar mass, ℳ by dividing through by pressure, P:
ℳ=
DRT
(1.39 g/L)(0.08206 L∙atm/mol∙K)(330.0 K)
=
= 28.9 g/mol
P
1.35 atm
19. An unknown gas with a mass of 2.56 g is in a 5.00 L container with a temperature of 345 K and
a pressure of 0.0445 atmospheres. What is the molar mass of the gas?
First find the density of the gas:
D=
2.56 g
= 0.512 g/L
5.00 L
Rearrange ℳP = DRT to solve for the molar mass, ℳ by dividing through by pressure, P:
ℳ=
DRT
(0.512 g/L)(0.08206 L∙atm/mol∙K)(345 K)
=
= 32.6 g/mol
P
0.0445 atm
20. A gas has a density of 1.39 g/L a56.8°C and a pressure of 1.35 atmospheres. What density will
the gas have if the pressure is changed to 5.33 atmospheres and temperature remains constant?
We can manipulate ℳP = DRT in the same way as PV = nRT. Only density and pressure are changing,
so we can set up two expressions for each set of conditions:
ℳP1 = D1 RT
and
ℳP2 = D2 RT
Rearrange both equations so that the variables that are changing are on one side of the expression and the
variables that are constant are on the other side of the expression. To do this, divide both sides by the
molar mass, ℳ, and divide through both sides of the expression by the density term to get:
𝑃1
𝑅𝑇
=
𝐷1
ℳ
𝑎𝑛𝑑
𝑃2
𝑅𝑇
=
𝐷2
ℳ
Since both pressure/density expressions now equal to the same constant, they are equal to each other:
𝑃1
𝑃2
=
𝐷1
𝐷2
Cross-multiply and solve for D2 by dividing through by P1:
P1 D2 = P2 D1 becomes
D2 =
P2 D1
(5.33 atm)(1.39 g/L)
=
= 5.49 g/L
P1
1.35 atm
© Copyright, 2001, L. Ladon. Permission is granted to use and duplicate these materials
for non-profit educational use, under the following conditions: No changes or
modifications will be made without written permission from the author. Copyright
registration marks and author acknowledgement must be retained intact.