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Calculate new boiling point with change post-pressure change

  1. Oct 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Liquid X boils at 127C, at a pressure of 10.6*10^5 Pa. Its enthalpy of vaporization is 5000 J/mol. At what temperature will it boil if the pressure is raised to 1.08*10^5 Pa?

    2. Relevant equations
    [itex]\frac{dT}{dP}=\frac{l}{T(v_{B}-v_{A})}[/itex]


    3. The attempt at a solution
    It seems so simple, but I don't know what to do about the specific volumes in the denominator. I have tried saying that because the second phase is a gas v_B>>v_A, neglecting v_A, then using the ideal gas equation to substitute for v_B = V_B/N = K*T_B/P_B.

    I get T_B (final temp.) as about 4*10^13 K. Which has got to be way way off.
     
    Last edited: Oct 5, 2013
  2. jcsd
  3. Oct 5, 2013 #2
    Your Relevant Equation is incorrect. The left side is flipped upside down. Otherwise, you have the right idea.

    Chet
     
  4. Oct 6, 2013 #3
    Thank you. I have tried solving dt/dp*(ΔP)=ΔT for Tb, using the approximation above (ideal gas law, v_B >> v_A), but get Tb ≈ Ta. Because of the small size of Boltzman's constant, the term with it in the denominator dies.

    [itex]\frac{dT}{dP} = \frac{k*T_{B}*T_{A}}{P_{B}*l}[/itex], It contains the initial temp. because I wanted the slope at the original temp.

    [itex]\left\{\left\{\text{Tb}\to \frac{l \text{Pb} \text{Ta}}{k \text{Ta} (\text{Pa}-\text{Pb})+l \text{Pb}}\right\}\right\}[/itex]
     
    Last edited: Oct 6, 2013
  5. Oct 6, 2013 #4
    Ok I got it. I looked at the relevant Wikipedia article. My problem was that I was using the wrong form of the ideal gas equation, and substituting Vb/n = R*Tb/Pb instead of using the general R*T/P. If I am substituting for vb, I don't get why I use general values (then integrate), instead of using the constants (with subscript "b").
     
  6. Oct 6, 2013 #5
    For a small change in pressure, it's OK, but, for a larger change in pressure, you have to integrate.
     
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