A weather ballon is designed to expand to a maximum radius of 20 m at its working altitude, where the air pressure is 0.030 atm and the temperature is 200 K. If the ballon is filled at atmospheric pressure and 300 K, what is its radius at liftoff?
Oh ok sweetness! Thanks. Figured out the answer. That makes sence. The only thing I had seen wrong with the equation you put is that you would need to find the volume, which is
(4/3)ðr^3 not jut r^3. But still helped out so thanks =)
Well, let's break it down, shall we? At its working altitude, the weather balloon has a radius of 20 m. We know that the air pressure at this altitude is 0.030 atm and the temperature is 200 K.
Now, if we consider the ideal gas law (PV = nRT), we can assume that the number of moles (n) remains constant throughout the process. At liftoff, the balloon is filled at atmospheric pressure (let's say 1 atm) and the temperature is 300 K.
So, if we rearrange the equation, we get:
V1/T1 = V2/T2
Where:
V1 = initial volume (unknown)
T1 = initial temperature (300 K)
V2 = final volume (20 m^3)
T2 = final temperature (200 K)
Now, solving for V1:
V1 = (V2 * T1) / T2
= (20 * 300) / 200
= 30 m^3
So, at liftoff, the radius of the balloon would be the cube root of the initial volume (V1 = 30 m^3) divided by (4/3π):
r = [3V1 / (4π)]^(1/3)
= [3(30) / (4π)]^(1/3)
= (90 / 4π)^(1/3)
≈ 1.49 m
Therefore, the radius of the balloon at liftoff would be approximately 1.49 meters. Gosh, it's always quite a blow to see the balloon shrink!
To find the radius of the weather balloon at liftoff, we need to use the ideal gas law. The ideal gas law states that the product of the pressure, volume, and temperature of a gas is a constant.
First, we need to find the pressure at liftoff. The pressure at liftoff is equal to the atmospheric pressure, which is typically around 1 atm. Let's convert this to the SI unit of pressure, which is Pascal (Pa). 1 atm is approximately equal to 101325 Pa.
Next, we can plug the given values into the ideal gas law equation:
P1 * V1 / T1 = P2 * V2 / T2
Where:
P1 = Pressure at liftoff in Pascal (Pa)
V1 = Volume at liftoff (unknown)
T1 = Temperature at liftoff in Kelvin (K)
P2 = Pressure at working altitude in Pascal (Pa) = 0.030 atm * 101325 Pa/atm
V2 = Maximum volume at working altitude (known) = (4/3) * π * (20 m)^3
T2 = Temperature at working altitude in Kelvin (K) = 200 K
Plug in the known values into the equation and solve for V1:
(1 atm * 101325 Pa/atm) * V1 / 300 K = (0.030 atm * 101325 Pa/atm) * (4/3) * π * (20 m)^3 / 200 K
Simplifying the equation:
V1 / 300 K = (0.030 atm * 101325 Pa/atm) * (4/3) * π * (20 m)^3 / (1 atm * 101325 Pa/atm)
V1 / 300 K = (0.030) * (4/3) * π * (20 m)^3 / (1 * 101325)
V1 / 300 K = (4 * π * (20 m)^3) / (3 * 101325)
V1 / 300 K ≈ 0.0077
Finally, solve for V1 by multiplying both sides by 300 K:
V1 ≈ 0.0077 * 300 K
V1 ≈ 2.31 m^3
So, the volume of the weather balloon at liftoff is approximately 2.31 cubic meters. To find the radius, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3
Plugging in the known volume at liftoff:
2.31 m^3 = (4/3) * π * r^3
Simplifying the equation:
r^3 = (2.31 m^3 * 3) / (4 * π)
r^3 ≈ 0.5486 m^3 / π
r ≈ (0.5486 m^3 / π)^(1/3)
r ≈ 0.4607 m
Therefore, the radius of the weather balloon at liftoff is approximately 0.4607 meters.
They didn't ask for the volume. Anyway, I'm glad you got the right answer.
PV/T is constant. That means PR^3/T is also constant. T must be in Kelvin. If subscript "1" denotes sea level and "2" denotes the working altitude,
(0.03 atm)*(20m)^3/200K = (1.0 atm)*(R1^3)*/300K
Solve for R1