Calculate new boiling point with change post-pressure change

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Liquid X boils at 127°C under a pressure of 10.6*10^5 Pa, and the discussion revolves around calculating its new boiling point when the pressure is raised to 1.08*10^5 Pa. Participants initially struggle with the specific volumes in the equation and the correct application of the ideal gas law. A key point of confusion is the use of the wrong form of the ideal gas equation, which leads to incorrect calculations. The conversation emphasizes that for small pressure changes, a simpler approach may suffice, but larger changes require integration for accurate results. Ultimately, understanding the proper equations and their applications is crucial for solving the problem accurately.
teroenza
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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.
 
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Your Relevant Equation is incorrect. The left side is flipped upside down. Otherwise, you have the right idea.

Chet
 
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.

\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\}
 
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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").
 
teroenza said:
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.
 

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