Question

You fill a balloon with helium gas to a volume of 2.68L at 23°C and 789mmHg. Then you release the balloon and it rises. What will the volume be when the pressure drops to 632mm Hg if the temperature remains the same?

Answer 1

Answer : The final volume of balloon will be, 3.35 L

Explanation :

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

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure = 789 mmHg

$P_2$ = final pressure = 632 mmHg

$V_1$ = initial volume = 2.68 L

$V_2$ = final volume = ?

Now put all the given values in the above equation, we get:

$789mmHg\times 2.68L=632mmHg\times V_2$

$V_2=3.35L$

Therefore, the final volume of balloon will be, 3.35 L