In this post, we will solve numerical problems using **Charles****’ Law. **The formulas to be used are as follows:

If V1 is the volume of a certain mass of a gas at temperature T1 and V2 is the volume of the same mass of the same gas at temperature T2 at constant pressure, then according to **Charles’ law**,

**V _{1}/T_{1} = V_{2}/T_{2} (mass and pressure constant) …… (1)**

Alternate presentation of Charles’ law gives us this formula:

∴ Volume at *t*°C, V_{t }= Initial volume at 0°C + Increase in volume = V_{0} + V_{0} x (1/273) x t

**V _{t }= V_{0} ( 1 + t/273) ………………. (2)**

**[ Find detailed discussion on Charles’ law here ]**

## Solving Numerical Problems using Charles’ Law

**Example 1**

*A sample of gas occupies 1.50 L at 25°C. If the* *temperature is raised to 60°C, what is the new* volume of the gas if the *pressure remains constant?*

**Solution: **V1 = 1.50 L

V2 = ?

T1 = 273 + 25 = 298 K

T2 = 60 + 273 = 333 K

Since pressure remains constant, therefore, by applying Charles’ law

V1/T1 = V2/T2

or V2 = (V1/ T1) x T2

= [(1.50 L) /(298 K)] x (333 K)

= **1.68 L**

### **Example **2

A sample of helium has a volume of 520 mL at 100°C. *Calculate the temperature at which the volume* *will become 260 mL. Assume that pressure is* *constant.*

**Solution:**

V1 = 520 mL

V2 = 260 mL

T1 = 100 + 273 = 373 K

T2 = ?

Since pressure remains constant, therefore, by applying Charles’ law :

V1/T1 = V2/T2

or T2= (T1/ V1) xV2 = (373/520) (260)= 186.5 K

or t = 186.5 – 273 = **–86.5°C**

**Example 3**

*At what centigrade temperature will a given* *volume of a gas at 0°C become double its volume,* *pressure remaining constant?*

**Solution: **Let the volume of the gas at 0°C be V.

V1 = V

T1 = 273 + 0 = 273 K

V2 = 2V

T2 = ?

Since pressure remains constant, therefore, by applying

Charles’ law,** V****1/T1 = V2/T2**

**T****2 =** (T1/ V1) xV2 = (273 x 2V) / V = 546 K

Changing the temperature to centigrade scale,

Temperature = 546 – 273 = **273°C.**

**Example 4**

*On a ship sailing in a pacific ocean where* the temperature is 23.4°C, a balloon is filled with 2L air. What will be the volume of the balloon when the ship reaches the Indian ocean, where the temperature is 26.1°C*?*

* Solution: *According to Charles’ law

V1/T1 = V2/T2

V1 = 2L

V2 = ?

T1 = 273 + 23.4 = 296.4 K

T2 = 273 + 26.1 = 299.1

V2 = (V1/T1). T2 = (V1 x T2 )/ T1 = 2L x 299.1K / 296.4K = 2.018 L

**Example 5**

*What is the increase in volume when the* *temperature of 800 mL of air increases from 27°C* *to 47°C under constant pressure of 1 bar ?*

* Solution: *Since the amount of gas and the pressure remains constant, Charles’ law is applicable.

*i.e.*

**V****1/T1 = V2/T2**

* V*1 = 800 mL V2 = ?

T1 = 273 + 27 = 300 K

T2 = 273 + 47 = 320 K

V2 = (V1/T1). T2 = (V1 x T2 )/ T1

**V****2 = (800x 320) /300 = 853.3 mL**

∴ Increase in volume of air = 853.3 – 800 = **53.3 mL**