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Pressure and boiling point

  1. Aug 18, 2014 #1
    I'm given an expression for the pressure of air with height z, as

    [itex]P(z)=P(0)e^{(-mgz/k_bT)}[/itex]

    Where m is the molecular weight of air. It's know that the pressure of water varies with height above sea-level due to the variation of the pressure of the air. The excercise is about to show that the rate of change of the boiling point, Tb, with height, z, is given by the following expression,

    [itex]\frac{dT_b}{dz}=-\frac{P(z)mg}{K_bT}\frac{dT_b}{dP}=-\frac{P(z)mgT_b}{K_bT}\frac{T_b(\rho_L-\rho_G)}{\rho_L\rho_GL_{LG}}[/itex]

    Where [itex]\rho_L[/itex] is the mass density of water, [itex]\rho_G[/itex] is the mass vapour density and [itex]L_{LG}[/itex] is the latent heat per unit mass.

    Relevant equation,

    [itex]L_{LG}=T_b\Delta S[/itex] (1)


    [itex]\frac{dP}{dT}=\frac{L_{LG}}{T_b\Delta V}[/itex] (2)

    My attempt:

    I differentiate the given expression for the pressure with respect to the temperature, according to equation (2) to get the following,

    [itex]\frac{dP}{dT}=\frac{d}{dT}P(z)=\frac{mgz}{k_bT^2}P(z)=\frac{L_{LG}}{T_b\Delta V}[/itex]

    Solving for the boiling point, Tb, gives,


    [itex]T_b=\frac{L_{LG}k_bT^2}{\Delta V mgzP(z)}[/itex]

    To find the relation which is asked in the excercise, I differentiate Tb with respect to z,

    [itex]\frac{dT_b}{dz}=\frac{L_{LG}k_bT^2}{\Delta V mg}(-\frac{1}{z^2P(z)}+\frac{mg}{k_bTzP(z)})=\frac{L_{LG}T(mgz-k_BT)}{\Delta V mgz^2P(z)}[/itex]

    Which is basically as far as I'm coming with this expression. I would appreciate help/hints about how to get further with this expression.
     
  2. jcsd
  3. Aug 21, 2014 #2
    I'm sorry you are not finding help at the moment. Is there any additional information you can share with us?
     
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