3 Example of Charles Law Problems

The gas law of Charles or Law of constant pressure, is another of the laws of gases, enunciated by Gay-Lussac, who unveiled the work of Jacques Charles, published about 20 years earlier.

Charles’s law predicts the behavior of a mass of gas when the pressure remains constant and the temperature and volume vary.

Charles’s law is stated as follows:

At constant pressure, the volume of a gas is directly proportional to the variation of its temperature.

Constant pressure : refers to the pressure that the gas exerts on the walls of the container will not vary throughout the experience.

Volume : is the occupied space that occupies the gas, in general it is considered a container with walls that do not deform, and whose lid works like a plunger.

Temperature: it  is the increase or loss of heat that the gas suffers during the experimentation. If the temperature increases, the volume increases. If the temperature decreases, the volume also decreases.

Algebraically, Charles’s Law is expressed with the following formula:

Expression of Charles's Law


V = gas volume

T = Gas temperature

k = constant of proportionality for that mass of gas.

This means that for a given mass of gas, at constant pressure, the relationship between volume and temperature variations will always have the same ratio of proportionality, represented by the constant k:

Representation of Charles's Law

So once the constant is determined, we can calculate any of the other values ??from the other known data:

Clearance for each variable V and T

3 Examples of Charles’s Law applied to problems:

Example 1 : Calculate the new volume, if in a container there is a mass of gas that occupies a volume of 1.3 liters, at a temperature of 280 K. Calculate the volume when reaching a temperature of 303 K.

1  = 1.3 l. 
1  = 280 K 
2  =? 
2  = 303 K

Clearance for V2

Substituting values:

Calculation of V2 in Example1

The new volume at 303 K is 1.41 liters.

Example 2 . If we have a gas that at 10 degrees Celsius occupies 2.4 liters, calculate the final temperature, if at the end it occupies 2.15 liters.

1  = 2.4 l 
1  = 10 ° C = 283 K 
2  = 2.15 l 
2  =?

Clearance for T2

Substituting values:

Calculation of T2 in Example 2

The new temperature is 253 K, which is equal to -20 ° C.

Example 3 . We have a gas that we know that its initial temperature is 328 K, the final volume is 3.75 l, and its constant ratio is 0.00885.

1  =? 
1  = 328 K 
2  = 3.75 l 
2  =? 
k = 0.00885

Expression of Charles's Law

Substituting values:

Replacement of the Constant Value k

To know the initial Volume:

Solution of V1 for Example 3

The initial volume is 2.90 l.

To know the final temperature:

Clearance for T2 of Charles Law

Final calculation of T2 in Example 3

The final temperature will be 423 K, which is equal to 150 ° C.


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