Charle's Law

Derivation of Charles Law

As we are aware of the fact that, at constant pressure, the volume of the fixed amount of the dry gas is directly proportional to absolute temperature according to Charle’s law. We can represent the statement in the following manner.

V∝T

Since V and T are varying directly, we can equate them by making use of the constant k.

V/T=constant =k

In this, the value of k depends on the pressure of the gas, the amount of the gas and also the unit of the volume.

V*T=k——-(1)

Let us consider V1 AND T1 to be the initial volume and the temperature, respectively of an ideal gas.

Then we can write equation (1) as

V1/T1=k——-(2)

After it lets change the temperature of the gas to T2. Alternatively, its volume changes to V2 then we can write

V2/T2=k——–(3)

Equating the above equations that is equation 2 and 3, we get

V1/T1=V2/T2

OR

V1T2=V2T1

You are unaware of the fact that, on heating up a fixed amount of gas, that is, by increasing the temperature the volume also increases. Similarly, lowering the temperature, the volume of the gas decreases. And at 0-degree centigrade, the volume of the also increases by 1/273 of its original volume for a unit degree increases in temperature.

Charles Law Solved Problems

1. A gas occupies a volume of 400cm3 at 0-degree Celcius and 780mm of Hg. How many litres of volume will the gas occupy at 80-degree Celcius and 780mm Hg

Solution- According to the question, V1=400cm3

T1=o degree Celcius= 0+273=273K

T2=80 DEGREE CELCIUS=80+273=353K

You need to find the V2.

Here only the temperature is changing, the pressure remains constant

Using Charle’s law, V1/T1=V2/T2

Putting the above values in the Charle’s law we get,

400/273=V2/353

V2=400*353/273

V2=517.21CM3

Since 1 cubic centimeters = 0.001 litres,

Then 517.21cubic centimeters=517.21*10^-3=0.517 litres.

2. A sample of gas has an initial volume of 30.8L AND an initial temperature of -67 degree Celcius. What will be the temperature of the gas if the volume is 21.0L?

According to the question,

V1= 30.8L

T1=-67 degree celcius=206K

T2=?

V2=21.0L

According to Charle’s law

V1/T1=V2/T2

30.8/67=21/T2

T2= 21*67/30.8

=46.9⁰C

If V1 is the 3.60L, T1=255K, T2=102K, then find the value of V2?

According to the question V2=?

As we are aware of the Charle’s law

V1/T1=V2/T2

3.60/255=V2/102

V2= 1.44L

3. A gas occupies 221cm3 at a temperature of 0 C and pressure of 760mm Hg. What will be the volume at 100 C?

It is very clear that the pressure is constant and the mass of the gas doesn’t change, so we can apply Charle’s law here.

The temperature in the question is given in Celcius, so as per the rule of Charle’s law it must be converted to the absolute temperature, that is Kelvin to apply the formula:

V1=221cm3;T1=273(0+273);T2=373K(100+273)

Now, these given values can be put in the formula to get the final volume.

VI /TI=VF /TF

221/273=VF/373

By rearranging the above equation we can get the final volume

VF=(221)*(373)/273K

VF=302cm3