Pressure of Helium inside a balloon floating in air

AI Thread Summary
The discussion revolves around calculating the absolute pressure of helium inside a spherical balloon that is floating in air. The balloon has a radius of 2.98 m and a mass of 3.16 kg, filled with helium at a temperature of 285 K, while the surrounding air density is 1.19 kg/m3. The buoyant force is calculated based on the volume of the balloon, leading to a total mass of 131.912 kg, from which the mass of helium is derived. The number of moles of helium is then calculated, and the ideal gas law is applied to find the pressure, resulting in a value of 6.88 x 10^5 Pascals. However, the answer was marked incorrect, prompting a request for verification of the final calculation.
yaylee
Messages
22
Reaction score
0
Pressure of Helium inside a balloon "floating" in air

Homework Statement


A spherical balloon of radius R = 2.98 m is made from a material of mass M = 3.16 kg and is filled with helium gas at temperature T = 285 K. Assume the thickness of the balloon is negligible compared to the radius of the balloon, and the balloon just floats on air, neither rising nor falling. If the density of the surrounding air is ρ = 1.19 kg/m3, find P, the absolute pressure of the helium inside the balloon.
ASSUME: The balloon material displaces a negligible amount of air, and therefore creates no measurable buoyancy.

Homework Equations


Volume of a sphere = (4/3)π(r^3)
Buoyant force of an object "floating": F = (mass object)(g) = (ρliquid)(Vobject)(g), and the "g's" cancel.
PV = nRT
moles = grams/ (grams/mol)


The Attempt at a Solution


First: find the mass of the TOTAL object. We can then subtract the mass of the balloon's material from the Total mass to find the mass of the Helium itself.
Use, Buoyant Force equation, where (ρliquid)(Vobject) = (mass object) = (1.19)(4/3)π(2.98^3) = 131.912 kg.

Then: (mass object) - (mass of balloon material) = mass of Helium = 131.912 - 3.16 = 128.75 kg = 128,750 grams.
number of moles of Helium inside balloon: 128, 750 g/4 g/mol = 32187.50 moles.

Now: can use pv = nRT, or P = nRT/V = 32187.50(8.314)(285)/((4/3)(∏)(2.98^3) = 6.88 x 10^5 Pascals.

The answer key marked me incorrect, however, I believe I have reasoned the problem correctly. Does anyone have any suggestions?

My unlimited thanks in advance!
 
Physics news on Phys.org


I think you're an order of magnitude out. Check the final calculation.
 


Thanks!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
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