Question

A weather balloon is filled with helium that occupies a volume of 500 L at 0.995 atm and 32.0 ℃. After it is released, it rises to a location where the pressure is 0.720 atm and the temperature is -12 ℃. What is the volume of the balloon at the new location?

Answer 1

Answer : The volume of the balloon at the new location is, 591.3 L

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 = 0.995 atm

`P_2` = final pressure of gas = 0.720 atm

`V_1` = initial volume of gas = 500 L

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas =
`32.0^oC=273+32=305K`

`T_2` = final temperature of gas =
`-12^oC=273+(-12)=261K`

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

`(0.995atm* 500L)/(305K)=(0.720atm* V_2)/(261K)`

`V_2=591.3L`

Therefore, the volume of the balloon at the new location is, 591.3 L