Charles’ Law Problem

Charles’ law is applicable only at low pressures and high temperatures. Only perfect gases are subject to Charles’ law. At high pressures, the relationship between quantity and temperature isn’t linear. In Charles’ law, the pressure is maintained at a constant level. If the pressure remains constant, the volume of the gas is proportional to the temperature. 

Charles’ Law

According to Charles’ law, the volume of a particular gas is proportional to the temperature when the pressure is kept constant. 

VT = k is the Charles’ law equation.

As a result, V = kT. The law is frequently expressed as V=kT when comparing an equivalent substance under two different sets of circumstances.

V1V2 = T1T2

V1T2= V2T1

As seen in this equation, with the increase in temperature of the gas, its volume also increases. This means Charles’ law may be used to compare volume and temperature. 

At a given temperature (-266.66° C according to Gay-Lussac’s figure), Charles’ law appears to imply that the volume of gas will decrease to zero. Since the gas has no energy at room temperature, the molecules are unable to move. The direct link between temperature and volume is known as Charles’ Law. If the pressure remains constant and the quantity of molecules remains constant while the temperature of the molecules rises, the molecules travel faster, generating greater pressure on the gas container, increasing the volume

Charles’ Law Problems

• At 30°C, a gas sample takes up 2.50 L. If the temperature is raised to 70°C and the pressure remains constant, then what will be the new volume of the gas?

Solution: V1 = 2.50 L

V2 = ?

T1 = 273 + 30 = 303 K

T2 = 273 + 70 = 343 K

We will be applying Charles’ Law as pressure is constant here:

V1/T1 = V2/T2

or V2 = (V1/ T1) x T2

= [(2.50 L) /(303 K)] x (343 K)

= 2.83 L

• At 90°C, a helium sample has a volume of 500 mL. Determine the temperature at which the volume of the liquid will become 240 mL. Assume that the pressure stays the same.

Solution: V1 = 500 mL

V2 = 240 mL

T1 = 90 + 273 = 363 K

T2 = ?

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

V1/T1 = V2/T2

or T2= (T1/ V1) xV2 = (363/500) (240)= 174.2 K

or t = 174.2 – 273 = –98.8°C

• If the pressure constant is given, then find the temperature at which the volume of the gas is tripled at 0 °C?

Solution: Allow V to be the volume of the gas at 0°C.

V1 = V

T1 = 273 + 0 = 273 K

V2 = 3V

T2 = ?

Because the pressure remains constant, applying

Charles’ law, V1/T1 = V2/T2

T2 = (T1/ V1) xV2 = (273 x 3V) / V = 819 K

Changing the temperature to centigrade scale,

Temperature = 819 – 273 = 546°C.

• A balloon is inflated with 4L air on a ship cruising in the Pacific Ocean with a temperature of 23.4°C. Calculate the volume of the balloon at the location of the Indian Ocean? The temperature of the Indian Ocean to be taken is 26.1°C.

Solution: According to Charles’ law,

V1/T1 = V2/T2

V1 = 4L

V2 = ?

T1 = 273 + 23.4 = 296.4 K

T2 = 273 + 26.1 = 299.1

V2 = (V1/T1). T2 = (V1 x T2 )/ T1 = 4L x 299.1K / 296.4K = 4.036 L

• If the temperature of the gas is continuously increasing from 30°C to 50°C, at a constant pressure of 1 bar, then what will be the increase in the volume of gas if its initial volume is 600 ml?

Solution: We will be using Charle’s law here:

V1/T1 = V2/T2

V1 = 600 mL, V2 = ?

T1 = 273 + 30 = 303 K

T2 = 273 + 50 = 323 K

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

V2 = (600x 323) /303 = 639.6 mL

∴ Increase in volume of air = 639.6 – 600 = 39.6 mL

• Volume of a given amount of a gas at 47 oC and constant pressure is 225 cm3. If the temperature is decreased to 27 oC at constant pressure, then what would be the volume of gas?

Solution: According to Charles’ Law,

V1V2=T1T2

Here

V1 = 225 cm3

V2 = ?

T1 =273+ 47 = 320 K

T2 =273+ 27 = 300 K

So,

V2= 225×320300

V2= 240 Cm3

• If 3.00 L of a gas is being compressed to 2.00 L at a given 31.0 °C, what would be the reduction in temperature on the Celsius scale?

Solution:

(3.00 L) / 304.0 K) = (2.00 L) / (x)

cross multiply to get:

x = 202.6 K

• A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0 °C?

Solution:

(900.0 mL) / (300.0 K) = (x) / (405.0 K)

x = 1215 mL

• Find the change in volume of the given gas, where 40.0 ml of gas is being cooled from 23.0 °C to 2.00 °C?

Solution:

(40.0 mL) / (296.0 K) = (x) / (275.00 K)

Cross multiply to get:

296x = 11000

x = 37.1 mL 

Change in Volume = 40 – 37.1

The volume decreases by 2.9 mL.

• A gas occupies 2.00 L at standard temperature. What is the volume at 233.0 °C?

Solution:

In cross-multiplied form, it is this:

V1T2 = V2T1

V2 = (V1) [T2 / T1] <— notice how I grouped the temperatures together

x = (2.00 L) [(506.0 K) / (273.0 K)]

x = 3.7 L

Conclusion

According to Charles’ law, the volume of particular gas is proportional to the temperature when the pressure is kept constant. 

In this article, we have mentioned various Charles’ Law-related questions with solutions to make it easier for students to understand the concept.