Temperature & Pressure?

Question

A spherical weather balloon is designed to expand to a maximum radius of 17 m when in flight at its working altitude where the air pressure is 0.033 atm and the temperature is 175 K. If the balloon is filled at atmospheric pressure at 300 K, what is its radius at lift-off?

Answer 1

Use the combined gas law
P1V1/T1 = P2V2/T2
Where P1 is the starting pressure
V1 is the starting volume
T1 is the starting temperture (in Kelvin)
P2 is the end pressure
V2 is the end volume
T2 is the end Temperature (in kelvin)


This will find the volume of the balloon then use the formula to find the radius.
V2 = 4/3 * pi *r^3

Answer 2

pV = nRt; so at p = .033 V = 4/3 pi A^3, and t = 175 K; where A = 17 m, the radius at Altitude. And at P = 1.000 (sea level) v = 4/3 pr L^3, T = 300K and where L is the radius at Lift off.

Since n and R are the same at both altitudes, we can write pV/t = Pv/T and pA^3/t = PL^3/T; so that L^3 = pA^3T/tP = A^3(p/P)(T/t) will give you L^3 and you can do the math since p = 1.00, P = .033, A = 17, T = 300, and t = 175.

Lesson learned: Start with PV = nRT and use it to plug in values at sea level and then at altitude. Then set up the ratio using the sea level and altitude values. The important thing to realize is that PV = nRT under two different conditions (P and T) will give you a set of two equations to work with.

Answer 3

Assuming that the gas is ideal and the balloon is always perfectly spherical:
PV = nRT
V = 4/3 * pi *r^3
n1 = n2 = n
r1 = 17
P1 = .033
P2 = 1
T1 = 175
T2 = 300

P1V1 / P2V2 = nRT1 / nRT2
(.033 * V1) / (1 * V2) = 175 / 300
V2 = 300*.033*V1 / 175
V2 ~= 0.05657*V1
4/3 * pi *(r2)^3 ~= .05657*4/3*pi*(r1)^3
r2 ~= r1*(.05657)^(1/3)
r2 ~= 17*0.38388
r2 ~= 6.526

Therefore, the radius at lift off is approximately 6.526 metres.