Deriving the vapor pressure latent heat dependence

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Homework Help Overview

The discussion revolves around deriving the relationship between vapor pressure and latent heat dependence using the Clausius-Clapeyron equation. Participants are exploring the mathematical formulation of latent heat in relation to temperature, specifically focusing on the latent heat of vaporization at different temperatures.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to derive the latent heat of vaporization in terms of absolute temperature and constants associated with specific temperatures (0 °C and 100 °C). There are discussions about the linear relationship and the correct formulation of the equation.

Discussion Status

The conversation includes various attempts to express the latent heat equation and questions about the correctness of these expressions. Some participants are checking their understanding of the relationships and the implications of their derived equations. There is ongoing exploration of the integration process and the variables involved.

Contextual Notes

Participants are working under the constraints of deriving equations based on the Clausius-Clapeyron equation and are discussing specific values for latent heat at defined temperatures. There is a focus on integrating to find vapor pressure at a specific temperature, with some uncertainty about the limits of integration.

il27

Homework Statement


There is a picture attached showing the entire problem.
Equation 2.78 is the Clausius Clapeyron equation.

Homework Equations


Clausius Clayperon equation.
L = L'T (since there is a linear dependence on temperature)

The Attempt at a Solution


$$ \frac{de_s}{dT} = \frac{L}{T\alpha} $$
$$ pV = RT $$ $$ V= RT/p $$
$$ \frac{dp_s}{dT} = \frac{L}{(RT^2)/p} $$

i am not sure how to continue with this though.
 

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il27 said:

Homework Statement


There is a picture attached showing the entire problem.
Equation 2.78 is the Clausius Clapeyron equation.

Homework Equations


Clausius Clayperon equation.
L = L'T (since there is a linear dependence on temperature)

The Attempt at a Solution


$$ \frac{de_s}{dT} = \frac{L}{T\alpha} $$
$$ pV = RT $$ $$ V= RT/p $$
$$ \frac{dp_s}{dT} = \frac{L}{(RT^2)/p} $$

i am not sure how to continue with this though.
Let ##L_0## be the latent heat of vaporization at 0 C and ##L_{100}## be the latent heat of vaporization at 100 C. In terms of the absolute temperature T, what is the latent heat of vaporization at absolute temperature T?
 
Chestermiller said:
Let ##L_0## be the latent heat of vaporization at 0 C and ##L_{100}## be the latent heat of vaporization at 100 C. In terms of the absolute temperature T, what is the latent heat of vaporization at absolute temperature T?

In terms of absolute temperature, the latent heat of vaporaization would be:
$$ L = 19T + 334 $$ 334 being the latent heat of water at 0 deg C. is that right?
 
il27 said:
In terms of absolute temperature, the latent heat of vaporaization would be:
$$ L = 19T + 334 $$ 334 being the latent heat of water at 0 deg C. is that right?
I want it algebraically in terms of ##L_0## and ##L_{100}##. And I asked for it in terms of the absolute temperature T, not the centigrade temperature.
 
Chestermiller said:
I want it algebraically in terms of ##L_0## and ##L_{100}##. And I asked for it in terms of the absolute temperature T, not the centigrade temperature.

Would it be: $$ L = L_{100}*T + L_0 $$ ?
 
il27 said:
Would it be: $$ L = L_{100}*T + L_0 $$ ?
Do you not know how to determine the equation for the straight line L vs T that passes through the points (273,##L_0##) and (373, ##L_{100}##)?
 
Chestermiller said:
Do you not know how to determine the equation for the straight line L vs T that passes through the points (273,##L_0##) and (373, ##L_{100}##)?

oh sorry!
that was dumb. yes of course. the equation i get for L is $$ L = (\frac{L_{100}-L_0}{100})T $$ no y-intercept.
what would be the next step?
 
il27 said:
oh sorry!
that was dumb. yes of course. the equation i get for L is $$ L = (\frac{L_{100}-L_0}{100})T $$ no y-intercept.

Incorrect. Try again.
 
Chestermiller said:
Incorrect. Try again.

sorry! i think i forgot the y-intercept.
$$ L =
il27 said:
oh sorry!
that was dumb. yes of course. the equation i get for L is $$ L = (\frac{L_{100}-L_0}{100})T $$ no y-intercept.
what would be the next step?
Chestermiller said:
Incorrect. Try again.

sorry! the slope is fine right? i think i forgot the y-intercept:
$$ L = (\frac{L_{100}-L_0}{100})T + 3.73L_0 - 2.73L_{100} $$
 
  • #10
il27 said:
sorry! the slope is fine right? i think i forgot the y-intercept:
$$L = (\frac{L_{100}-L_0}{100})T + 3.73L_0 - 2.73L_{100} $$
OK. So now you have $$\frac{d\ln{p_s}}{dT}=\frac{L}{RT^2}$$
Now all you need to do is substitute for L, and integrate from T1 to T2.
 
  • #11
Chestermiller said:
OK. So now you have $$\frac{d\ln{p_s}}{dT}=\frac{L}{RT^2}$$
Now all you need to do is substitute for L, and integrate from T1 to T2.
great, thank you! I think when I did the $$ RT^2 $$, that was wrong. i will just integrate:

$$ \frac{dp_s}{dT} = \frac{de_s}{dT} = \frac{L}{T\alpha_v} $$ for the pressure. i will plug in for L. would that be okay?
 
  • #12
il27 said:
great, thank you! I think when I did the $$ RT^2 $$, that was wrong. i will just integrate:

$$ \frac{dp_s}{dT} = \frac{de_s}{dT} = \frac{L}{T\alpha_v} $$ for the pressure. i will plug in for L. would that be okay?
The version with ##RT^2## was correct, and the equation I wrote was correct.
 
  • #13
Chestermiller said:
The version with ##RT^2## was correct, and the equation I wrote was correct.

oh okay, i will use that one then! thank you.
just wondering, what is the $$ \alpha_v $$ that is a constant right?
 
  • #14
il27 said:
oh okay, i will use that one then! thank you.
just wondering, what is the $$ \alpha_v $$ that is a constant right?
No. It’s RT/ps
 
  • #15
Chestermiller said:
No. It’s RT/ps
oh okay. it's 1/density? which is RT/ps?
 
  • #16
il27 said:
oh okay. it's 1/density? which is RT/ps?
Yes, it’s the molar volume of the vapor.
 
  • #17
Chestermiller said:
Yes, it’s the molar volume of the vapor.
great thank you! and m
Chestermiller said:
Yes, it’s the molar volume of the vapor.

and i just wanted to check but my euqation for L is correct?:
$$ L = (\frac{L_{100}-L_0}{100})T + 3.73L_0 - 2.73L_{100} $$
 
  • #18
also, would I be integrating from p0 to p, p0 being 101325 pa, and T1 to T2?
because if i want to find the vaporation pressure using the new equation with the temperature of 20 deg. C, what would be my T1 and T2?
 
  • #19
like if i wanted to:
Use the expression derived in part (a) to recalculate the saturation vapor pressure at 20 °C.
 

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