Question

A weather balloon is filled to a volume of 2 liters with helium gas at a pressure of 101 kPa, at a temperature of 22oC. The balloon is released and floats to a height where the pressure is 90 kPa and the air temperature is now 16oC. What is the new volume of the weather balloon?

Answer 1

Answer: The new volume of weather balloon will be 2.2 Liters

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of gas = 101 kPa

`P_2` = final pressure of gas = 90 kpa

`V_1` = initial volume of gas = 2 L

`V_2` = final volume of gas = ? L

`T_1` = initial temperature of gas =
`22^oC=273+22=295K`

`T_2` = final temperature of gas =
`16^oC=273+16=289K`

Now put all the given values in the above equation, we get the final pressure of gas.

`(101* 2L)/(295K)=(90* V_2)/(289K)`

`V_2=2.2L`

Therefore, the volume will be 2.2 Liters.