Mammoth Memory

Charles' Law In Use

We know that in Charles' law

V1T1=V2T2

NOTE: Pressure remains constant

Also, if you put a balloon inside a bucket and pour cold liquid nitrogen over the balloon, you can watch it shrink.

Charles law pouring liquid nitrogen over a balloon will make it shrink.

This is Charles' Law

V1 and T1 refer to the volume and temperature before you pour the liquid nitrogen on the balloon and V2 and T2 refer to the volume and temperature after you pour the liquid nitrogen on the balloon.

Reduce the temperature of the gas and the volume of the gas reduces too.

Charles law reduce the temperature of the gas and the volume of the gas reduces too.

Reducing the temperature of a gas slows down the molecules and they have less energy (kinetic energy) to hit and push the sides of the balloon. The molecules will exert less force and so the external atmospheric pressure will collapse the balloon.

NOTE 1:

The standard unit used to measure volume is m3 (metre cubed)

The standard unit used to measure temperature is K (Kelvin)

 

NOTE 2:

You can use other units for volume such as m3 or ml as long as you use the same units either side of the equation but you must use Kelvin for temperature.

 

NOTE 3:

Never forget to convert temperature to Kelvin

Temperature in Kelvin = temperature in degrees Celsius ( C)+273 

You must always use temperature measured in Kelvin in any gas law equation. 

 

Examples

1. A container holds 50.0 ml of nitrogen at 25C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35C ?

Reveal answer

1. A container holds 50.0 ml of nitrogen at 25C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 35C ?

V1T1=V2T2

Therefore V2=V1×T2T1

V1=50 ml

V2=?

T1=25C+273=298K

T2=25C+35C+273=333K


Therefore V2=50×333298

V2=55.9 ml

 

2. 568 cm3 of chlorine at 25C will occupy what volume at -25C while the pressure remains constant? 

Reveal answer

2. 568 cm3 of chlorine at 25C will occupy what volume at -25C while the pressure remains constant? 

V1T1=V2T2

V1=568 cm3

V2=?

T1=25C+273=298K

T2=-25C+273=248K

Therefore V2=V1×T2T1

V2=568×248298

V2=473 cm3

 

3. A sample of hydrogen has an initial temperature of 50C. When the temperature is lowered to -5.0C, the volume of hydrogen becomes 212 cm3. What was the initial volume of the hydrogen?

Reveal answer

3. A sample of hydrogen has an initial temperature of 50C. When the temperature is lowered to -5.0C, the volume of hydrogen becomes 212 cm3. What was the initial volume of the hydrogen?

V1T1=V2T2

V1=?

V2=212 cm3

T1=50C+273=323K

T2=-5C+273=268K

Therefore V1=V2×T1T2

V1=212×323268

V1=256 cm3