Troubleshooting a Weather Balloon Expansion Problem

AI Thread Summary
A weather balloon is designed to expand to a maximum radius of 30 m at an altitude with specific pressure and temperature conditions. The initial calculations using the ideal gas law led to an incorrect radius at lift-off. The correct approach involves using the relationship of volume and temperature, showing that the ground volume is 16% of the final volume. Consequently, the initial radius is determined by taking the cube root of the volume ratio. The discussion highlights the importance of understanding gas laws and volume relationships in solving the problem accurately.
PrideofPhilly
Messages
37
Reaction score
0

Homework Statement



A weather balloon is designed to expand to
a maximum radius of 30 m when in flight at
its working altitude where the air pressure is
0.033 atm and the temperature is 95 K.
If the balloon is filled at atmospheric pres-
sure and 462 K, what is its radius at lift-off?

Homework Equations



PV = nRT
Volume = 4/3πr^3

The Attempt at a Solution



First, I used the ideal gas law to find n:

(0.033)(4/3π(30)^3) = n(8.31)(95)
n = 4.727610453 mol

Then, I solved for r:

V = nRT/P
4/3πr^3 = (4.727610453)(8.31)(462)/1 atm
r = 11.30381236 m (WRONG ANSWER)

What am I doing wrong?
 
Physics news on Phys.org
So PV/T is constant, you don't actually need the number of moles.
V1 = (T1/P1) * (P2V2/T2)
= (462/1) * (0.033*V2/95)
V1 = 0.16 V2

So the ground volume is 16% off the final volume, an the radius goes as the cube root of volume, so the initial radius is 0.16^0.33 of 30m
 
where did you get the 0.33 from?
 
PrideofPhilly said:
where did you get the 0.33 from?

He means to the power of one third (the cube root, as he said).
 
nevermind, i got it! thank you!
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top