Help! I Need a Hint on Isothermal Balloon Expansion Problem

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The discussion revolves around solving an isothermal balloon expansion problem crucial for exam preparation. The balloon, filled with helium, rises while maintaining pressure equilibrium with the surrounding atmosphere. The first part of the problem requires establishing the pressure inside the balloon at height h, which is equal to the atmospheric pressure derived from the equation p/p0 = exp(-mgh/kT). For the second part, the volume of the helium is determined using the adiabatic expansion principles, leading to the conclusion that V = V0 exp(3mgh/5kT). The key insight is recognizing the relationship between specific heats, Cp/Cv = 3/5, which simplifies the calculations.
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I got stuck on the following problem, which is very important for my exam preparations - would anyone mind giving me a hint?

A balloon of Volume V(0) is tethered to the ground using a light cable. It is filled with helium gas of density rho (He). At ground level the pressure and density in the atmosphere are p(0) and rho(0) respectively. The atmosphere is isothermal with a temperature T(0). The balloon is allowed to rise to a height h by very slowly letting out the cable. At all heights the gas pressure inside the balloon equals the pressure outside.

I have already shown that for the pressure and density the following is true:

p/p_0 = exp(-mgh/kT) = \rho/\rho_0

QUESTION:

The He in the balloon (an excellent insulator) expands adiabatically. What is the pressure in the balloon at height h and find its volume.

Please, please, can anyone give me a hint ?
 
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You are asked to assume that the atmosphere is isothermal, T = T0 everywhere. So, you know the atmospheric pressure in terms of h, at any point. By force balance, the pressure inside the balloon must be the same as the pressure outside. That answers the first part of the question. For the second part (volume), use the fact the He is a monoatomic (ideal) gas, and find the volume dependence from the adiabatic expression.
 
thanks for this! So in the first part I will simply state the peviously derived equation for p:

p/p_0 = exp(-mgh/kT)


In the second part I still do not get the correct result. Would you mind giving me another hint?

My attempt: T is not constant since the expansion is adiabatic
from the ideal gas equation:

pV= RT

inserting this in the equation gives:

V/V_0 = exp (mgh/kT) = exp(+mghR/kpV)

which is definitely wrong. The answer given the solution manual says

V = V_0exp(3mgh/5kT)
 
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Thanks for your help. I just found the answer myself. It is quite straight forward once you know that Cp/Cv = 3/5
 
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