# Boyle’s Law and Charles’ Law Problems

## Boyle’s Law Problems:

Match the items from the first column to the correct answer in the second column.

 1. Boyle’s Law When temperature is held constant: a. k 2. Boyle’s law is stated PV = b. p1V1 = p2V2 3. Formula for Boyle’s Law c. the pressure and volume of a gas are inversely proportional

1. C 2. A 3. B

To check the solutions to the problems, click on Step 1, Step 2, and then Step 3 for explanations.

At a pressure of 405 kPa, the volume of a gas is 6.00 cm3. Assuming the temperature remains constant, at what pressure will the new volume be 4.00 cm3?

Step 1

Identify each piece of information given in the problem and identify what the problem is asking for.
P1 = 405 kPa
V1 = 6.00 cm3
P2 = ?
V2 = 4.00 cm3

Step 2

Rearrange Boyle’s Law to solve the equation for the variable that the question is asking for.
P1V1 = P2V2
P2 = (P1V1)/ V2

Step 3

Plug in your data from step 1 into the rearranged equation from step 2.
P2 = (P1V1)/ V2
P2 = (405kPa (6.00 cm3)) / 4.00 cm3
P2 = 608 kPa

Notice that because the units cm3 appear in the numerator and denominator, they cancel and you are left with units of kPa. This makes sense because we are looking for a pressure.

A volume of gas at 1.10 atm was measured at 326 cm3. What will be the volume if the pressure is adjusted to 1.90 atm?

Step 1

Identify each piece of information given in the problem and identify what the problem is asking for.
P1 = 1.10 atm
V1 = 326 cm3
P2 = 1.90 atm
V2 = ?

Step 2

Rearrange Boyle’s Law to solve the equation for the variable that the question is asking for.
P1V1 = P2V2
V2 = (P1V1)/ P2

Step 3

Plug in your data from step 1 into the rearranged equation from step 2.
V2 = (P1V1)/ P2
V2 = (110 atm(326 cm3)) / 1.90 atm
V2 = 189 cm3

Notice that because the units atm appear in the numerator and denominator, they cancel and you are left with units of cm3. This makes sense because we are looking for a volume.

If 36.5 m3 of a gas are collected at a pressure of 755 mm Hg, what volume will the gas occupy if the pressure is changed to 632 mm Hg?

Step 1

Identify each piece of information given in the problem and identify what the problem is asking for.
P1 = 755 mm Hg
V1 = 36.5 m3
P2 = 632 mm Hg
V2 = ?

Step 2

Rearrange Boyle’s Law to solve the equation for the variable that the question is asking for.
P1V1 = P2V2
V2 = (P1V1)/ P2

Step 3

Plug in your data from step 1 into the rearranged equation from step 2.
V2 = (P1V1)/ P2
V2 = (755 mg Hg(36.5 m3)) / 632 mm Hg
V2 = 43.6 m3

Notice that because the units mm Hg appear in the numerator and denominator, they cancel and you are left with units of m3. This makes sense because we are looking for a volume.

Things to note:

In Boyle’s Law, it does not matter what unit your pressure is in, as long as the P1 and P2 are in the same unit.
Likewise, it does not matter what unit your volume is in, as long as the V1 and V2 are in the same unit.
If your pressures or your volumes are not in the same unit, your units will not cancel. You will be required to convert the measurements into the same units. It would not matter which unit you chose to convert to, as long as they were both in the same unit.

## Charles’ Law Problems:

Match the items from the first column to the correct answer in the second column.

 1. Charles’ Law When pressure is held constant: a. Kelvin 2. Charles’ law is stated V/T = b. 3. Formula for Charles’ Law c. the volume and temperature of a gas are directly proportional 4.This temperature scale must be used in all gas law problems d. k

1. C 2. D 3. B 4. A

To check the solutions to the problems, click on Step 1, Step 2, and then Step 3 for explanations.

At 189 K, a sample of gas has a volume of 32.0 cm3. What volume does the gas occupy at 242 K?

Step 1

Identify each piece of information given in the problem and identify what the problem is asking for.
T1 = 189 K
V1 = 32.0 cm3
T2 = 242 K
V2 = ?

Step 2

Rearrange Charles’ Law to solve the equation for the variable that the question is asking for.
V1T2 = V2T1
Solve for V2. Divide both sides by T1.
V2 = (V1T2)/ T1

Step 3

Plug in your data from step 1 into the rearranged equation from step 2.
V2 = (V1T2)/ T1
V2 = (32.0 cm3 (242 K) / 189K
V2 = 41.0 cm3

Notice that the temperature was increased. This caused the volume to also increase. this fits with what we know about temperature and volume. They are directly related.

The gas in a balloon occupies 2.25 L at 298 K. At what temperature will the balloon expand to 3.50 L?

Step 1

Identify each piece of information given in the problem and identify what the problem is asking for.
T1 = 298 K
V1 = 2.25 L
T2 = ?
V2 = 3.50 L

Step 2

Rearrange Charles’ Law to solve the equation for the variable that the question is asking for.
V1T2 = V2T1
Solve for T2. Divide both sides by V1.
T2 = (V2T1)/ V1

Step 3

Plug in your data from step 1 into the rearranged equation from step 2.
T2 = (V2T1)/ V1
T2 = (3.50 L (298 K) / 2.25 L
T2 = 464 K

A sample of gas has a volume of 852 mL at 25°C. What Celsius temperature is necessary for the gas to have a volume of 945 mL?

Step 1

Identify each piece of information given in the problem and identify what the problem is asking for.
T1 = 25 °C + 273 = 298 K
V1 = 852 mL
T2 = ?
V2 = 945 mL

*Notice that the temperature given is in degrees Celsius. Kelvin temperatures must be used in all gas laws. This temperature must be converted to Kelvin by adding 273 to it.

Step 2

Rearrange Charles’ Law to solve the equation for the variable that the question is asking for.
V1T2 = V2T1
Solve for T2. Divide both sides by V1.
T2 = (V2T1)/ V1

Step 3

Plug in your data from step 1 into the rearranged equation from step 2.
T2 = (V2T1)/ V1
T2 = (945 mL (298 K) / 852 mL
T2 = 311 K

The problem asks for the temperature in degrees Celsius. We must convert K to °C by subtracting 273.

T2 = 311 K = 273 = 58 °C

Things to note:

In Charles’ Law, it does not matter what unit your volume is in, as long as the V1 and V2 are in the same unit.
It does matter what unit your temperature is in. All temperatures must be in K to work in a gas law problem.
If your volumes are not in the same unit, your units will not cancel. You will be required to convert the measurements into the same units. It would not matter which unit you chose to convert to, as long as they were both in the same unit.
When working these problems, it was easy to know which law to use, as the questions were grouped by the law. In future problems, you will not be told which law to use. You will proceed with step 1 as above, then look at the data that you are given and what information is missing. Use this to determine which law will work.
Rationalize what you know about the gas laws and the relationships that exist between the variables with your answer. Does it make sense? If not, go back and search for your mistake.

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