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
