Hot air balloon - Ideal gas law

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andyfry
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Homework Statement


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

Homework Equations


pV=NkT
p=[tex]\frac{Force}{Area}[/tex]
[tex]\rho[/tex]=[tex]\frac{m}{V}[/tex]

The Attempt at a Solution


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

and Fup=[tex]\rho[/tex]Vg

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

so T=[tex]\frac{Vg}{NkA}[/tex](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...
 
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andyfry said:

Homework Equations


pV=NkT
p=[tex]\frac{Force}{Area}[/tex]
[tex]\rho[/tex]=[tex]\frac{m}{V}[/tex]

The Attempt at a Solution


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

and Fup=[tex]\rho[/tex]Vg
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=[tex]\frac{pVM}{RT}[/tex]
Is this what you mean?
Can't really see how this is helpful though. Still leaves me with unknowns of p, [tex]\rho[/tex], 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:
[tex]\rho[/tex]inV=[tex]\rho[/tex]outV+mtot

[tex]\frac{PMV}{RT_{in}}[/tex]=[tex]\frac{PMV}{RT_{out}}[/tex]+mtot

(assuming P in and out are equal)

Rearanging for Tin

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

Tin=Tout+[tex]\frac{PMV}{Rm_{tot}}[/tex]

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

Tin=283+[tex]\frac{101000*28.97*10^{-3}*2000}{8.314*920}[/tex]=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:
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! :)