Calculate new boiling point with change post-pressure change

In summary, Liquid X has a boiling point of 127C at a pressure of 10.6*10^5 Pa with an enthalpy of vaporization of 5000 J/mol. When the pressure is raised to 1.08*10^5 Pa, the temperature at which it will boil is approximately 4*10^13 K. However, this is incorrect and the correct temperature can be found by integrating the ideal gas equation and using the correct form of the equation.
  • #1
teroenza
195
5

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


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

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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  • #2
Your Relevant Equation is incorrect. The left side is flipped upside down. Otherwise, you have the right idea.

Chet
 
  • #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]
 
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  • #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").
 
  • #5
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.
 

1. How does changing the pressure affect the boiling point of a substance?

Changing the pressure can affect the boiling point of a substance because it alters the equilibrium between the liquid and gas phases. Increasing pressure can result in a higher boiling point, while decreasing pressure can result in a lower boiling point.

2. How do I calculate the new boiling point with a change in pressure?

To calculate the new boiling point with a change in pressure, you can use the Clausius-Clapeyron equation which relates the boiling point to the pressure and the enthalpy of vaporization. You will need to know the original boiling point, the original pressure, and the new pressure to calculate the new boiling point.

3. Can the new boiling point be lower than the original boiling point?

Yes, the new boiling point can be lower than the original boiling point if the pressure is decreased. This is because a decrease in pressure leads to a decrease in the boiling point due to the change in equilibrium between the liquid and gas phases.

4. How accurate are the calculations for determining the new boiling point?

The accuracy of the calculations for determining the new boiling point depends on the accuracy of the input values, such as the original boiling point, original pressure, and new pressure. It is important to use precise and accurate measurements to ensure the most accurate result.

5. Can the new boiling point be higher than the original boiling point?

Yes, the new boiling point can be higher than the original boiling point if the pressure is increased. This is because an increase in pressure leads to an increase in the boiling point due to the change in equilibrium between the liquid and gas phases.

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