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 $4.65\times 10^4L$

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 = $3.93\times 10^4L$

$V_2$ = 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:

$\frac{0.995\times 3.93\times 10^4}{305.0}=\frac{0.720\times V_2}{261.3}$

$V_2=4.65\times 10^4L$

Thus the volume of the balloon at that new location is $4.65\times 10^4L$