Question

Dwayne filled a small balloon with air at 298.5 K. He put the balloon into a bucket of water, and the water level in the bucket increased by 0.54 liter.

If Dwayne puts the balloon into a bucket of ice water at 273.15 K and waits for the air inside the balloon to come to the same temperature, what will the volume of the balloon be? Assume the pressure inside the balloon doesn’t change.

Answer 1

Answer : The volume of the balloon will be, 0.494 liters

Solution :

Charles' Law : It is defined as the volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$

or,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1$ = initial volume of gas = ?

$V_2$ = final volume of gas = 0.54 L

$T_1$ = initial temperature of gas = 273.15 K

$T_2$ = final temperature of gas = 298.5 K

Now put all the given values in the above equation, we get the initial volume of balloon.

$\frac{V_1}{273.15K}=\frac{0.54L}{298.5K}$

$V_1=0.494L$

Therefore, the volume of the balloon will be, 0.494 liters