A weather balloon with a volume of 3.40774 L
is released from Earth’s surface at sea level.
What volume will the balloon occupy at an
altitude of 20.0 km, where the air pressure is
10 kPa?
Answer in units of L.
I don't have time to be a troll. I'm a retired university professor just trying to help students help themselves.
If the temperature at sea level is the same as it is at an altitude of 20.0 km, (it isn't of course) then use P1V2 = P2V2
If the problem gives T at sea level and at 20.0 km, then use
(P1V1/T1) = (P2V2/T2).
thanks but where do i put which number for the formula?
If you're using P1V1 = P2V2, then you can pick and choose but just be consistent. For example, I would use
P1 at sea level an that is 101.325 kPa
V1 = 3.40774 L from the problem.
P2 = 10 kPa at 20.0 km
V2 = ? .Solve for that.
ok thank you. every other website i ask is just full of trolls
To solve this problem, we can use Boyle's Law which states that the product of pressure and volume is a constant as long as the temperature remains constant. The equation for Boyle's Law is:
P1 * V1 = P2 * V2
Where:
P1 = initial pressure (sea level)
V1 = initial volume (given as 3.40774 L)
P2 = final pressure (10 kPa)
V2 = final volume (to be calculated)
In this case, the initial pressure (sea level) is not given, but we can assume it to be the standard atmospheric pressure of 101.3 kPa.
Substituting these values into the equation, we get:
(101.3 kPa) * (3.40774 L) = (10 kPa) * V2
Now we can solve for V2 (final volume):
V2 = (101.3 kPa * 3.40774 L) / 10 kPa
Simplifying the equation, we have:
V2 = 34.485 L
Therefore, the volume of the balloon at an altitude of 20.0 km, where the air pressure is 10 kPa, is approximately 34.485 L.