Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Gas expansion

  1. May 4, 2006 #1
    I got stuck on the following problem, which is very important for my exam preparations - would anyone mind giving me a hint?

    A ballon 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:

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


    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 ?
  2. jcsd
  3. May 4, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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.
  4. May 5, 2006 #3
    thanks for this! So in the first part I will simply state the peviously derived equation for p:

    [tex] p/p_0 = exp(-mgh/kT) [/tex]

    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:

    [tex] pV= RT [/tex]

    inserting this in the equation gives:

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

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

    [tex]V = V_0exp(3mgh/5kT) [/tex]
    Last edited: May 5, 2006
  5. May 6, 2006 #4
    Thanks for your help. I just found the answer myself. It is quite straight forward once you know that Cp/Cv = 3/5
    Last edited: May 6, 2006
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook