# Calculation questions on Charles's law

### Worked Examples on Charles's Law:

Charles's law states that the volume of a given mass of gas is directly proportional to the absolute temperature of the gas (in kelvin), provided the pressure remains constant.

French physicist, Jacques Charles (1746-1823) formulated this law in 1780.

Please read the details of Charles's law alongside its graph and equations here.

• #### Question 1

A tube is filled with hydrogen gas to a volume of 10dm3 at a temperature of 30oC. If this tube is heated to a temperature of 100oC, what will be the new volume of the tube?

• Solution

Step 1: List the known quantities from the question.

• Initial Volume (V1) =10dm3

• Initial Temperature (T1) = 30oC

• Final Volume (V2) = Unknown

• Final Temperature (T2)= 100oC

Step 2: When solving questions relating to Charles's law, the temperature must be in Kelvin. From our question, the temperature was given in degree celcius (oC). Therefore, we need to covert the temperature from degree celcius to kelvin. To do this, we simply add 273.15 to the celcius temperature.

• T1 = 30 + 273.15 = 303.15K

• T2 = 100 + 273.15 = 373.15K

Step 3: Apply the formula for Charles's law to find the final volume (V2). Below is the formula:

• V1/T1 = V2/T2

Step 4: Make V2 the subject of the formula.

• V1/T1 = V2/T2

• V1 x T2 = V2 x T1

• V1 x T2 / T1 = V2

• V2 = (V1 x T2) ÷ T1

Step 5: Substitute values into the formula to find the final volume (V2).

• V2 = (10 x 373.15) ÷ 303.15

• V2 = 3731.5 ÷ 303.15

• V2 = 12.3dm3

• Notice that the final volume (V2) increases just as the final temperature (T2) increases. This is in accordance with Charles's law since both are always directly proportional.

• #### Question 2

200L of oxygen gas was compressed at a temperature of 37oC to 20L. Calculate is the temperature of oxygen at the final volume.

• Solution

Step 1: List the known quantities from the question.

• Initial Volume (V1) = 200L

• Initial Temperature (T1)= 37oC

• Final Volume (V2) = 20L

• Final Temperature (T2) = Unknown

Step 2: Covert the temperature from degree celcius to kelvin.

• T1 = 37 + 273.15 = 310.15K

Step 3: Apply the formula for Charles's law to find the final temperature (T2).

• V1/T1 = V2/T2

Step 4: Make T2 the subject of the formula.

• V1/T1 = V2/T2

• V1 x T2 = V2 x T1

• T2 = V2 x T1 / V1

• T2 = (V2 x T1) ÷ V1

Step 5: Substitute values into the formula to find the final temperature (T2).

• T2 = (20 x 310.15) ÷ 200

• T2 = 6203 ÷ 200

• T2 = 31.02K

• Notice that the final temperature (T2) decreases just as the final volume (V2) decreases. This is in accordance with Charles law.

• #### Question 3

What volume of gas will be changed to 250mL if its temperature changes from 5K to 10K.

• Solution

Step 1: List the known quantities from the question

• Initial Volume (V1) = ?

• Initial Temperature (T1) = 5K

• Final Volume (V2) = 250mL

• Final Temperature (T2) = 10K

Step 2: Apply the formula for Charles's law to find the initial volume (V1).

• V1/T1 = V2/T2

Step 3: Make V1 the subject of the formula.

• V1/T1 = V2/T2

• V1 x T2 = V2 x T1

• V1 = V2 x T1 / T2

• V1 = (V2 x T1) ÷ T2

Step 4: Substitute values into the formula to find the initial volume (V1).

• V1 = (250 x 5) ÷ 10

• V1 = 1250 ÷ 10

• V1 = 125mL

Notice that the initial volume (V1) decreases just as the initial temperature (T1) decreases. This is in accordance with Charles's law.

• #### Question 4

A noble gas occupies 3000mL of tube at a temperature of 373.15K. What volume will it occupy in litres at 100oC?

• Solution

Step 1: List the known quantities from the question

• Initial Volume (V1) = 3000mL

• Initial Temperature (T1) = 373.15K

• Final Volume (V2) = ?

• Final Temperature (T2) = 100oC

Step 2: Covert the final temperature (T2) from degree celcius to kelvin.

• T1 = 100 + 273.15 = 373.15K

Step 3: Since our answer (final volume) is required in litres, we should convert our initial volume from millilitres to litres by dividing it by 1000.

V1 = 3000 ÷ 1000 = 3L

Step 4: Apply the formula for Charles law to find the final volume (V2).

• V1/T1 = V2/T2

Step 5: Make V2 the subject of the formula.

• V1/T1 = V2/T2

• V1 x T2 = V2 x T1

• V1 x T2 / T1 = V2

• V2 = (V1 x T2) ÷ T1

Step 6: Substitute values into the formula to find the final volume (V2).

• V2 = (3 x 373.15) ÷ 373.15

• V2 = 1119.45 ÷ 373.15

• V2 = 3L

• Notice that the final volume (V2) equals initial volume (V1). This is so because there isn't any change in value of both temperatures. Final temperature and initial temperature are the same.

• #### Question 5

At -30oC, a gas has a volume of 500mL. What will be its new temperature if the volume decreases by 450mL?

• Solution

Step 1: List the known quantities in the question.

• Initial Volume (V1) = 500mL

• Initial Temperature (T1) = -30oC

• Final Volume (V2) = 500mL - 450mL -> 50mL

• Final Temperature (T2) = Unknown

Step 2: Covert initial temperature (T1) to kelvin.

• T1 = -30 + 273.15 = 243.15K

Step 3: Make T2 the subject of the formula. You can do this by cross multiplying.

• V1/T1 = V2/T2

• V1 x T2 = V2 x T1

• T2 = V2 x T1 / V2

• T2 = (V2 x T1) ÷ V1

Step 4: Substitute values into the formula to find the final temperature (T2).

• T2 = (50 x 243.15) ÷ 500

• T2 = 12157.5 ÷ 500

• T2 = 24.32K

You can read on Graham's law of diffusion with worked examples here.

• Notice that the final temperature (T2) decreases just as the final volume (V2) decreases. This is in accordance with Charles's law. 