Question

A balloon is partly inflated with 4.70 liters of helium at sea level where the atmospheric pressure is 1.00×103 mbar. The balloon ascends to an altitude of 3.00 × 103 meters, where the pressure is 847 mbar. What is the volume of the helium in the balloon at the higher altitude? Assume that the temperature of the gas in the balloon does not change in the ascent.

Answer 1

Answer : The final volume of the helium in the balloon at the higher altitude is 5.55 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 (1)/(V)`

or,

`P_1V_1=P_2V_2`

where,

`P_1` = initial pressure of gas =
`1.00* 10^3mbar`

`P_2` = final pressure of gas =
`847mbar`

`V_1` = initial volume of gas = 4.70 L

`V_2` = final volume of gas = ?

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

`(1.00* 10^3mbar)* 4.70L=847mbar* V_2`

`V_2=5.55L`

Therefore, the final volume of the helium in the balloon at the higher altitude is 5.55 L