Combined Gas Law
Problems #1 - 10

Ten Examples

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The form of the Combined Gas Law most often used is this:

(P1V1) / T1 = (P2V2) / T2

Most commonly V2 is being solved for. The rearrangement looks like this:

V2 = (P1V1T2) / (T1P2)

A reminder: all these problems use Kelvin for the temperature. I will not usually comment on the change from °C to K. I will use 273 but be aware that your teacher (or computer lesson) may insist on using 273.15.

When you use the combined gas law paired with Dalton's Law, remember that a gas collected over water is always considered to be saturated with water vapor. The vapor pressure of water varies with temperature and must be looked up in a reference source.


Problem #1: A gas has a volume of 800.0 mL at minus 23.0 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure?

Solution:

1) The combined gas law is rearranged to isolate V2:

    P1V1T2
V2 = ––––––
    P2T1

2) Values are inserted into the proper places:

    (300.0 torr) (800.0 mL) (500.0 K)
V2 = –––––––––––––––––––––––––––
    (250.0 K) (600.0 torr)

V2 = 800.0 mL


Problem #2: 500.0 liters of a gas in a flexible-walled container are prepared at 700.0 mmHg and 200.0 °C. The gas is placed into a tank under high pressure. When the tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the volume of the gas?

Solution:

1) The combined gas law is rearranged to isolate V2:

V2 = (P1V1T2) / (T1P2)

2) Values are inserted into the proper places:

V2 = [(0.92105) (500) (293)] / [(473) (30)]

V2 = 9.51 L

Note the use of square brackets to communicate the correct order of operations.

Note that the problem provides different pressure units for the starting and ending values. I used 700/760 to convert from mmHg to atm.

Note that I paid scant attention to setting up the problem with correct sig figs in the problem. This happens often in gas law problems. Note also I omitted all the units.


Problem #3: 690.0 mL of oxygen are collected over water at 26.0 °C and a total pressure of 725.0 mm of mercury. What is the volume of dry oxygen at 52.0 °C and 800.0 mm pressure?

Solution:

1) Use Dalton's Law to remove the pressure of the water vapor:

Ptotal = PO2 + PH2O

PO2 = Ptotal - PH2O

PO2 = 725.0 mmHg - 25.2 mmHg = 699.8 mmHg

The 25.2 value came from here. I looked up the value associated with 26.0 °C and converted it from kPa to mmHg following the instructions given.

2) Here are the values in a solution matrix:

P1 = 699.8 mmHgP2 = 800.0 mmHg
V1 = 690.0 mL     V2 = x
T1 = 299.0 KT2 = 325.0 K

A common student error is to use Dalton's Law, but then use the total pressure value in the combined gas law instead of using the correct value.

The correct pressure to use for P1 is the 699.8 value, not the 725 value. The 725 is the pressure of an oxygen/water mixture and we want ONLY the oxygen (which is the 699.8 value).

3) Use the combined gas law:

x = [(699.8) (690.0) (325)] / [(299) (800.0)]

x = 656 mL (to three sig figs)


Problem #4: What is the volume of gas at 2.00 atm and 200.0 K if its original volume was 300.0 L at 0.250 atm and 400.0 K.

Solution:

1) Here are the values in a solution matrix:

P1 = 0.250 atmP2 = 2.00 atm
V1 = 300.0 L     V2 = x
T1 = 400.0 KT2 = 200.0 K

Note how the problem statement is worded so as to give the starting values last.

2) The combined gas law rearranged to isolate V2:

V2 = (P1V1T2) / (T1P2)

x = [(0.25) (300) (200)] / [(400) (2)] <--- note lack of units

x = 18.75 L

To three sig figs, this is 18.8 L


Problem #5: At conditions of 785.0 torr of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 745.0 torr and 30.0 °C?

Solution:

V2 = [(785 mmHg) (45.5 mL) (303 K)] / [(288 K) (745 mmHg)]

V2 = 50.3757 mL

To three sig figs, the answer is 50.4 mL


Problem #6: What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0 °C to 30.0 °C and a pressure change from standard pressure to 360.0 mmHg?

Solution:

We are looking to determine V2 in this problem. Here's the set up:

P1 = 760.0 mmHgP2 = 360.0 mmHg
V1 = 400.0 L     V2 = x
T1 = 295 KT2 = 303 K

V2 = [(760 mmHg) (400 mL) (303 K)] / [(295 K) (360 mmHg)]

V2 = 867 mL (to three sig figs)


Problem #7: 400.0 mL of hydrogen are collected over water at 18.0 °C and a total pressure of 740.0 mm of mercury.

(a) What is the partial pressure of H2?
(b) What is the partial pressure of H2O?
(c) What is the volume of DRY hydrogen at STP?

Solution:

1) We will use Dalton's Law to determine the partial pressure of the dry hydrogen gas. We look up the vapor pressure of water in a reference source.

Ptotal = PH2 + PH2O

PH2 = Ptotal - PH2O

740.0 - 15.5 = 724.5 mmHg

I used a different reference source than previously used for the vapor pressure of water. There are many available on the Internet.

2) The partial pressure of the water is its vapor pressure of 15.5 mmHg.

3) Combined gas law rearranged to show V2 isolated:

    P1V1T2
V2 = ––––––
    P2T1

    (724.5 mmHg) (400.0 mL) (273 K)
V2 = ––––––––––––––––––––––––––––
    (291 K) (760.0 mmHg)

V2 = 358 mL (to three sig figs)


Problem #8: The pressure of a gas is reduced to 75% of its initial value and the volume is increased by 40% of its initial value. Find the final temperature, given that the initial temperature was -10 °C. This is a combined gas law problem.

Solution:

Let us assign P1 = 1, therefore P2 = 0.75
Let us assign V1 = 1, therefore V2 = 1.4

I won't bother with units on P or V. Your teacher may want the units added in, so I'll do that below.

T1 = -10 C = 263 K

P1V1/T1 = P2V2/T2

[(1 atm) (1 L)] / 263 K = [(0.75 atm) (1.4 L)] / x

(1 atm) (1 L) (x) = (263 K) (0.75 atm) (1.4 L)

x = 276.15 K = 3.15 °C


Problem #9: The pressure of 8.06 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half. What is the final volume of the gas?

Solution:

1) Assign values as follows:

P1 = 3.00 atmP2 = 1.00 atm
V1 = 8.06 LV2 = x
T1 = 2.00 KT2 = 1.00 K

Note the made up values for P and T.

2) Insert values into the combined gas law equation and solve for x:

P1V1 / T1 = P2V2 / T2

[(3.00 atm) (8.06 L)] / 2.00 K = [(1.00 atm) (x)] / 1.00 K

x = 12.1 L (to three sig figs)


Problem #10: A balloon of air now occupies 10.0 L at 25.0 °C and 1.00 atm. What temperature was it initially, if it occupied 9.40 L and was in a freezer with a pressure of 0.939 atm?

Solution:

1) Assign values as follows:

P1 = 0.939 atmP2 = 1.00 atm
V1 = 9.40 LV2 = 10.0 L
T1 = xT2 = 298 K

Note how the problem gives you the ending conditions regarding PVT and asks you for a starting condition. Also, note that temperature is asked for. Compare this to the usual case of asking for the final volume.

2) Let's rearrange the combined gas law equation for T1:

P1V1   P2V2
––––– = –––––
T1   T2

T1P2V2 = P1V1T2

    P1V1T2
T1 = ––––––
    P2V2

3) Put values in and solve:

    (0.939 atm) (9.40 L) (298 K)
T1 = ––––––––––––––––––––––––
    (1.00 atm) (10.0 L)

T1 = 263 K

The form of the final temperature was not specificed, but it is usually given in Celsius, so:

T1 = -10. °C


Problem #11: A gas occupies an 8.00 mL flexible-walled container. The pressure is doubled, the absolute temperature is quadrupled, and 15% of the gas leaks out. What is the new volume?

Solution:

1) This problem involves the combined gas law that has all four variables in it:

P1V1   P2V2
––––– = –––––
n1T1   n2T2

2) The changes are all expressed in a relative way, so I will assume 1.00 atm and 1.00 K:

P1 = 1.00 atmP2 = 2.00 atm
V1 = 8.00 mLV2 = x
n1 = 1.00 moln2 = 0.85 mol
T1 = 1.00 KT2 = 4.00 K

By the way, having a gas at 1.00 K is a very silly thing. The point, of course, is that the absolute temperature quadruples. We could use 100 K and 400 K and get the same answer. Or use 200 K and 800 K. The key point is that the temperature quadruples. And, note that it is the absolute temperature (in K) that quadruples, not the temperature in degrees Celsius.

Notice how I interpreted the 15% to mean 15% of the moles of gas escaped. If I decided that 15% of the mass escaped, the problem answer would be the same. I will leave it to you to figure out, if you so desire.

3) I won't bother to isolate V2 this time:

(1.00 atm) (8.00 mL)   (2.00 atm) (x)
––––––––––––––––– = –––––––––––––––
(1.00 mol) (1.00 K)   (0.85 mol) (4.00 K)

4) Cross-multiply:

(1.00 atm) (8.00 mL) (0.85 mol) (4.00 K) = (2.00 atm) (x) (1.00 mol) (1.00 K)

4) Divide:

    (1.00 atm) (8.00 mL) (0.85 mol) (4.00 K)
x = –––––––––––––––––––––––––––––––––
    (2.00 atm) (1.00 mol) (1.00 K)

x = 13.6 mL


Bonus Problem: A sample of neon has a volume of 0.730 dm3 at a temperature of 21.0 °C. and pressure of 102.5 kPa. If the density of neon is 0.900 g/dm3 at 0 °C and 101.3 kPa, what is the mass of the sample?

Solution:

1) Convert gas conditions to STP:

Here's the cross-multiplied form of the combined gas law:

P1V1T2 = P2V2T1

(102.5 kPa) (0.730 dm3) (273 K) = (101.3 kPa) (V2) (294 K)

V2 = 0.685887 dm3

2) Determine mass:

0.685887 dm3 times 0.900 g/dm3 = 0.617 g

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