Deriving the vapor pressure latent heat dependence

[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.

 

[Chestermiller]

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?

 

[il27]

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?

 

[Chestermiller]

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.

 

[il27]

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: $$ L = L_{100}*T + L_0 $$ ?

 

[Chestermiller]

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, $L_{100}$)?

 

[il27]

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?

 

[Chestermiller]

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.

 

[il27]

Incorrect. Try again.

sorry! i think i forgot the y-intercept.
$$ L =

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?

Incorrect. Try again.

sorry! the slope is fine right? i think i forgot the y-intercept:
$$ L = (\frac{L_{100}-L_0}{100})T + 3.73L_0 - 2.73L_{100} $$

 

[Chestermiller]

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.

 

[il27]

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?

 

[Chestermiller]

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.

 

[il27]

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?

 

[Chestermiller]

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

 

[il27]

No. It’s RT/ps

oh okay. it's 1/density? which is RT/ps?

 

[Chestermiller]

oh okay. it's 1/density? which is RT/ps?

Yes, it’s the molar volume of the vapor.

 

[il27]

Yes, it’s the molar volume of the vapor.

great thank you! and m

Yes, it’s the molar volume of the vapor.

and i just wanted to check but my euqation for L is correct?:
$$ L = (\frac{L_{100}-L_0}{100})T + 3.73L_0 - 2.73L_{100} $$

 

[il27]

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?

 

[il27]

like if i wanted to:
Use the expression derived in part (a) to recalculate the saturation vapor pressure at 20 °C.