Question

Please help!
1. The sketch above shows the launching of a weather balloon at sea level (0 km). Make a sketch of the same weather balloon (in the green box below) when it is high in the atmosphere (15 km). In words, (in the red box below) explain why the size of the balloon changed
2. Using Boyle’s Law, calculate the new volume of the weather balloon at a higher altitude. Assume the volume of the balloon at ground level is 10.0 m3 where the atmospheric pressure is 1.00 atm. What will be its new volume at an altitude of 15 km where the atmospheric pressure is 0.350 atm? Fill in the equation with the correct values and solve for the new volume of the weather balloon.

Answer 1

Answer:

p1-1.00atm

V1-10.0m^3

=
P2-0.35

V2-?

the answer is 28.57m^3

Answer 2

Using the Boyle's law, the volume of the gas at 15 Km is $28.6 m^3$.

What is Boyle's law?

Boyle's law states that, the volume of given mass of gas is inversely proportional to its pressure at constant temperature. Since pressure decreases with height, the volume will increase at higher altitude.

Initial volume = 10.0 $m^3$

Initial pressure = 1.00 atm

Final pressure =  0.350 atm

Final volume = ?

Using;

P1V1 = P2V2

V2 =$P1V1/P2$

V2 = $10.0 m^3 * 1.00 atm/0.350 atm$

V2 = $28.6 m^3$

Learn more about Boyle's law: https://brainly.com/question/1437490