Numerical Problems based on Charles’ Law with solution

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,

V1/T1 = V2/T2 (mass and pressure constant) …… (1)

Alternate presentation of Charles’ law gives us this formula:

∴ Volume at t°C, Vt = Initial volume at 0°C + Increase in volume = V0 + V0 x (1/273) x t

Vt = V0 ( 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, V1/T1 = V2/T2

T2 = (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.

V1/T1 = V2/T2

V1 = 800 mL V2 = ?

T1 = 273 + 27 = 300 K

T2 = 273 + 47 = 320 K

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

V2 = (800x 320) /300 = 853.3 mL

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

Numerical Problems based on Charles’ Law with solution
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