Answer: The volume of balloon at a height of 20 km is [tex]1.32\times 10^5L[/tex]
Explanation:
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1,V_1\text{ and }T_1[/tex] are the initial pressure, volume and temperature of the gas
[tex]P_2,V_2\text{ and }T_2[/tex] are the final pressure, volume and temperature of the gas
We are given:
[tex]P_1=745torr\\V_1=1.41\times 10^4L\\T_1=21^oC=[21+273]K=294K\\P_2=63.1torr\\V_2=?L\\T_2=-40^oC=[-40+273]K=233K[/tex]
Putting values in above equation, we get:
[tex]\frac{745torr\times 1.41\times 10^4L}{294K}=\frac{63.1torr\times V_2}{233K}\\\\V_2=\frac{745\times 1.41\times 233}{294\times 63.1}=1.32\times 10^5L[/tex]
Hence, the volume of balloon at a height of 20 km is [tex]1.32\times 10^5L[/tex]