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,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

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:

$\frac{0.995atm\times 500L}{305K}=\frac{0.720atm\times V_2}{261K}$

$V_2=591.3L$

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