A weather balloon is designed to expand to a maximum radius of 20 m at its working altitude, where the air pressure is .030 atm and the temperature is 200 K. If the balloon is filled at atmospheric pressure and 300 K, what is its radius at liftoff?
All I have is PV=nRT and I really don't know where to go with that...
2007-11-12 14:44:48 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Use the combine Law,
P1V1/T1 =P2V2/T2
Given, P1 = 0.030 Atm
V1 = (4/3)pi(20)^3 = 33510 m^3
T1 = 200 K
P2 = 1 Atm
T2 = 300 K
V2 = (4/3)pi(R)^3
V2 =( P1V1)(T2)/(T1(P2))
(4/3 pi) R^3 = 0.03(33510)(300) / (200)(1)
R = 7.11 m ----this is the radius when it lift-off from the ground.
2007-11-12 14:59:19 · answer #1 · answered by dongskie mcmelenccx 3 · 2⤊ 0⤋
To be honest, I don't believe that's correct
2016-07-30 07:04:38 · answer #2 · answered by Anonymous · 0⤊ 0⤋
First solve for initial/final volume ratio V1/V2. nR is constant, so
V1/V2 = (T1/P1) / (T2/P2)
Initial/final radius ratio r1/r2 = (V1/V2)^(1/3)
r1 = 20*r1/r2.
2007-11-12 15:03:41 · answer #3 · answered by kirchwey 7 · 0⤊ 0⤋
Thank you! valuable information and this gives me better knowledge on the topic
2016-08-26 06:37:00 · answer #4 · answered by Anonymous · 0⤊ 0⤋