Calculate new boiling point with change post-pressure change

[teroenza]

Homework Statement

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?

Homework Equations

\frac{dT}{dP}=\frac{l}{T(v_{B}-v_{A})}

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.

 

[Chestermiller]

Your Relevant Equation is incorrect. The left side is flipped upside down. Otherwise, you have the right idea.

Chet

 

[teroenza]

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 Boltzmann's constant, the term with it in the denominator dies.

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

\left\{\left\{\text{Tb}\to \frac{l \text{Pb} \text{Ta}}{k \text{Ta} (\text{Pa}-\text{Pb})+l \text{Pb}}\right\}\right\}

 

[teroenza]

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").

 

[Chestermiller]

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").

For a small change in pressure, it's OK, but, for a larger change in pressure, you have to integrate.