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Compresibility of helium gas

  1. Aug 8, 2017 #1
    1. The problem statement, all variables and given/known data
    A cylinder is fitted with a piston, and the cylinder contains helium gas. The sides of the cylinder are adiabatic, impermeable, and rigid, but the bottom of the cylinder is thermally conductive, permeable to helium, and rigid. Through this permeable wall the system is in contact with a reservoir of constant [itex]T[/itex] and [itex]\mu_\text{He}[/itex] (the chemical potential of He). Calculate the compressibility of the system [itex][-(1/v)(dv/dp)][/itex] in terms of the properties of helium [itex](c_p, v, \alpha, \kappa_T,[/itex] etc) and thereby demonstrate that this compressibility diverges. Discuss the physical reason for this divergence.

    2. Relevant equations

    3. The attempt at a solution
    [itex]v[/itex] is volume per mole. Since it is contacted to a heat and particle reservoir I assume they mean [tex]
    -\frac{1}{v} \frac{dv}{dp}_{T, \mu} [/tex]

    I can rewrite this as
    \frac{1}{v} \frac{(d\mu/dp)_{T,v}}{(d\mu/dv)_{T,p}} [/tex]

    We know that [itex] d\mu = - s dT + vdp [/itex] so the numerator is [itex]v[/itex] and the denominator is 0. This seems to be the divergence they mean, but I am not sure. Also, I do not know the physical reason, which I don't want to consider until I know that the first part is correct. Any thoughts?
  2. jcsd
  3. Aug 8, 2017 #2
    Well, I have these thoughts. There is Helium leaving the cylinder through the permeable bottom when it is compressed. The temperature of the reservoir is held constant, so the temperature of the helium in the cylinder stays at T. The chemical potential of the helium inside the cylinder must match that of the reservoir on the other side of the permeable bottom and, since the temperature is constant, the chemical potential will match if the pressure in the cylinder is constant (since the chemical potential of the helium in the cylinder is determined by its T and p).
  4. Aug 8, 2017 #3
    Thank you for the response. I agree with you on your first point but I do not understand why the chemical potential matches the reservoir only if the pressure is constant. I thought the condition for equilibrium would set the chemical potential to be the reservoir's regardless, and should not the chemical potential be a function of T,p, and N (the number of moles of gas)? I think I don't understand what you mean by p being constant.
  5. Aug 8, 2017 #4
    The chemical potential does not depend on N unless it is a mixture. Even then, for an ideal gas, it only depends on T and partial pressure.$$\mu=\mu^0(T)+RT\ln{(p/p_0)}$$
    Last edited: Aug 8, 2017
  6. Aug 8, 2017 #5
    I feel silly now as I wrote [itex]d\mu =-sdT +vdp [/itex]. If the system is equilibrium should not [tex]
    dU + dU_r = 0 = (T-T_r)dS + (\mu-\mu_r)dN [/tex] mean that [itex]T=T_r [/itex] and [itex] \mu_r = \mu [/itex] if the system undergoes a quasi static change? I guess that would 'explain' why I found the derivative to be infinite as the pressure does not change if the volume changes. Is this correct?
  7. Aug 8, 2017 #6
    I don't really follow your equation, but if you are saying that T = Tr and mu = our, that is correct.
  8. Aug 8, 2017 #7
    So in my answer I found that the compressibility is infinite if [itex]\mu, T[/itex] is fixed which has to be the case if [itex]\mu[/itex] is determined by T and p only. A constant T and [itex]\mu][/itex] implies a constant p irrespective of V. I'm just not sure if that is the answer they want and what the physical significance of it is in that case.
  9. Aug 8, 2017 #8
    Well, that's the only way I can see to interpret what the problem statement says. Hope it's right.
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