Question

A hot air balloon contains 40.0 liters of helium gas at 203 kilopascals. What will be the volume of the helium when the balloon rises to an altitude where the pressure is only 35.0 kilopascals?

Answer 1

Answer : The volume of helium when the balloon rises to an altitude are, 232 liters.

Explanation :

According to the Boyle's law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature of the gas and the number of moles of gas.

`P\propto (1)/(V)`

or,

`P_1V_1=P_2V_2`

where,

`P_1` = initial pressure of gas = 203 kPa

`P_2` = final pressure of gas = ?

`V_1` = initial volume of gas = 40 L

`V_2` = final volume of gas = 35 L

Now put all the given values in the above formula, we get the final volume of gas.

`203kPa* 40L=35kPa* V_2`

`V_2=232L`

Therefore, the volume of helium when the balloon rises to an altitude are, 232 liters.

Answer 2

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

V2 = P1 x V1 / P2

V2 = 203 x 40 / 35

V2 =232 L