Question

A weather balloon is filled with helium that occupies a volume of 3.93 104 L at 0.995 atm and 32.0°C. After it is released, it rises to a location where the pressure is 0.720 atm and the temperature is -11.7°C. What is the volume of the balloon at that new location?

Answer 1

Answer: Thus the volume of the balloon at that new location is
`4.65* 10^4L`

Step-by-step 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 =
`3.93* 10^4L`

`V_2` = final volume of gas = ?

`T_1` = initial temperature of gas =
`32.0^oC=273+32.0=305.0K`

`T_2` = final temperature of gas =
`-11.7^oC=273-11.7=261.3K`

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

`(0.995* 3.93* 10^4)/(305.0)=(0.720* V_2)/(261.3)`

`V_2=4.65* 10^4L`

Thus the volume of the balloon at that new location is
`4.65* 10^4L`