Hot air balloon - Ideal gas law

AI Thread Summary
To estimate the required temperature for a hot air balloon to achieve lift-off, the discussion centers on applying the ideal gas law and relevant equations of buoyancy. The initial approach involves calculating the upward force and relating it to the mass of the air and the total mass of the balloon. The key insight is substituting the mass of hot air with the molar mass and using the ideal gas law to find a relationship between pressure, volume, and temperature. After correcting algebraic errors, a final temperature of approximately 450K is determined as necessary for lift-off. The discussion emphasizes careful algebraic manipulation and the importance of correctly applying physical principles.
andyfry
Messages
6
Reaction score
0

Homework Statement


Estimate the required temperature of the air inside a (rigid) balloon in order for it to achieve lift off.
Volume 2000m3
\summass, mtot=920kg

Homework Equations


pV=NkT
p=\frac{Force}{Area}
\rho=\frac{m}{V}

The Attempt at a Solution


We require
Fup>(mair+mtot)g (assume equal to solve)

and Fup=\rhoVg

Not entirely sure how to relate this to the ideal gas law, but have tried
Fup=\frac{NkTA}{V}
(from p=\frac{F}{A})

so T=\frac{Vg}{NkA}(mair+mtot)

But I'm left with an N which I don't know, and also an mair which I also don't know. I assume mair=N(mass of 1 molecule) but subbing this into the equation would still leave one N and a the mass of 1 molecule of air, which i do not know.

Hopefully my attempt is at least going in the right direction...
 
Physics news on Phys.org
andyfry said:

Homework Equations


pV=NkT
p=\frac{Force}{Area}
\rho=\frac{m}{V}

The Attempt at a Solution


We require
Fup>(mair+mtot)g (assume equal to solve)

and Fup=\rhoVg
This looks good up to this point. Can you find an expression to substitute for mair, the mass of hot air? It involves the ideal gas law and the molar mass of air.
 
well mair=nM
where M=molar mass of air
so subbing into pV=nRT
gives mair=\frac{pVM}{RT}
Is this what you mean?
Can't really see how this is helpful though. Still leaves me with unknowns of p, \rho, and M after subbing into intitial equations.
How does using the molar mass of air, M, instead of mair help? They are both unknown.

EDIT:
\rhoinV=\rhooutV+mtot

\frac{PMV}{RT_{in}}=\frac{PMV}{RT_{out}}+mtot

(assuming P in and out are equal)

Rearanging for Tin

Tin=\frac{PMV}{R}(\frac{RT_{out}}{PMV}+\frac{1}{m_{tot}})

Tin=Tout+\frac{PMV}{Rm_{tot}}

assume Tout=283k, P=101kPa, M=28.97, R=8.314

Tin=283+\frac{101000*28.97*10^{-3}*2000}{8.314*920}=1048k
Which doesn't seem to be quite plausible? Would expect a value of maybe half that...
Can someone check my method please??
 
Last edited:
andyfry said:
EDIT:
\rhoinV=\rhooutV+mtot
Actually, it's

ρinV + mtot = ρoutV​

The mass of the balloon (including the hot air inside it) equals the mass of the displaced air, ρoutV.

Can someone check my method please??
You're on the right track, be careful with the algebra.
 
Redbelly98 said:
Actually, it's

ρinV + mtot = ρoutV​

Ah, that might help!

rearranging (properly!) for Tin now gives an answer of 450k, which I believe is the right answer.

Thanks a lot for your help! :)
 
andyfry said:
450k
Yup, looks good to me :smile:
Thanks a lot for your help! :)
You're welcome.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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