Balloon problem, thermodynamics

AI Thread Summary
The discussion revolves around calculating the number of moles of helium gas in a balloon that lifts a load of 110 N, considering the balloon's skin weight and buoyant forces. The net buoyant force is defined as the difference between the buoyant force and the weights of the balloon and its load. The participant clarifies the calculation of volume, arriving at 17.38 m³ by using the densities of air and helium. The densities are provided, with air at 1.293 kg/m³ and helium at 0.1786 kg/m³. The final result confirms the calculations, emphasizing the importance of understanding buoyant forces in thermodynamics.
illidari
Messages
46
Reaction score
0

Homework Statement



A helium balloon is used to lift a load of 110 N. The weight of the balloon's skin is 50 N, and the volume of the balloon when fully inflated is 32 m3. The temperature of the air is 0°C and the atmospheric pressure is 1 atm. The balloon is inflated with sufficient helium gas so that the net buoyant force on the balloon and its load is
30 N. Neglect changes of temperature with altitude.

(a) How many moles of helium gas are contained in the balloon?




Homework Equations



Find V from FB = mg + 30 N
PairVg = 190 N + PHeVg; V = 17.38 m3

The Attempt at a Solution



I am confused on how you get the V=17.38 from using the equation I posted. Could someone explain that part to me.
 
Physics news on Phys.org
First, let me just say that the question is being a bit sloppy in using "net buoyant force" to mean the net upward force (the sum of all contributions in the vertical direction -- which includes the buoyant force due to the air displaced by helium gas).

Moving on...

The helium gas displaces an equal volume of air, thus producing lift (buoyant force). You'll need to look up the density of air and the density of helium gas (you can calculate the latter from the molar mass of helium and the molar volume of a gas at 0 degrees and 1 atmosphere pressure).

The net upward force is then (buoyant force) - weights
 
Didn't realize these were just stuff I had to look up...

Density air = 1.293
density helium = .1786

190/ (1.293-.1786 * 9.81) = 17.38

>.< Well an answer is an answer

Thanks!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top