# Introduction to Charles Law Sample Problems

Charles law describes the relationship between Temperature and Volume. Gases expand when heated and shrink when cooled. In other words, at constant pressure and amount of the gas, increasing the temperature of the gas results in to proportional increase in the volume of the gas. This can be expressed in the equation form:

Charles law equation V/T = constant

i.e., V1/T1 = V2/T2

Charles law sample problems are being described below.

## Sample Problems Based on Charles Law

Charles Law Sample Problem 1:

A container contains 5 L of nitrogen gas at 25° C. What will be its volume if the temperature increases by 35° C keeping the pressure constant?

Solution:

V1 = 5 L                                              V2 = ?

T1 = (25°C + 273) K = 298 K         T2 = (25°C + 35°C + 273) K = 333 K

V1/T1 = V2/T2

Substituting the values,

5 L / 298 K = V2 / 333 K

V2 = 5 L x 333 K / 298 K

## Sample Problems Based on Charles Law

Charles Law Sample Problem 2:

By what factor the temperature has to be raised to double the volume of a given gas balloon at constant pressure?

Solution:

Let's say the initial Volume is V and the initial temperature is T

V1 = V and T1 = T

So, when volume is doubled,

V2 = 2V and T2 = T + x , where x is the rise in temperature

According to Charles Law, at constant pressure

V1 / T1 = V2 / T2

=> V / T = 2V / (T + x)

By rearranging,

(T + x) / T = 2V / V = 2

=> T + x = 2T

=> x = T

Now, T2 = T + x = T + T = 2T

Therefore, T2 / T1 = 2T / T = 2

Thus, at constant pressure, to double the volume of a gas the temperature has to be raised by two times.

Charles Law Sample Problem 3:

A sample of gas occupies 3 L at 300 K. What volume will it occupy at 200 K?

Solution:

V1 = 3 L                                        V2 = ?

T1 = 300 K                                     T2 = 200 K

Now, according to Charles law

V1/T1 = V2/T2

Substituting the values,

3 L / 300 K = V2 / 200 K

V2 = 3 L x 200 K / 300 K

Charles Law Sample Problem 4:

A sample of oxygen occupies a volume of 1.6 L at 91°C. What will be the temperature when the volume of oxygen is reduced to 1.2 L?

Solution:

V1 = 1.6 L                                        V2 = 1.2 L

T1 = (91°C + 273) K = 364 K             T2 = ?

Now, according to Charles law

V1/T1 = V2/T2

Substituting the values,

1.6 L / 364 K = 1.2 L / T2

T2 = 1.2 L x 364 K / 1.6 L