October 16, 2019, 06:06:24 PM
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Topic: Not a problem of the week, but similar level of difficulty  (Read 20901 times)

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Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #15 on: October 02, 2017, 02:59:56 AM »
I would like to try and solve this, but I was curious to know if all the information needed to solve the problem is already present?

Not exactly, but I assume the freezing point of water and fact that it is almost identical with the triple point of water are a common knowledge. Perhaps it should be spelled out in the problem.

Quote
Also, I was wondering what the latter three pieces of information referred to. Are they for ice/water, or the glucose solution?

Ice/water.
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Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #16 on: October 02, 2017, 03:08:21 AM »
For (1), we additionally need to know the molar weight and normal freezing point for water, as well as the universal gas constant (8.315).

These are between things that I assume to be a general knowledge, easily available. Again, you are right they could be spelled out in the problem.

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I finally tried to find the normal freezing point of water using the saturated vapour pressure piece of information. I tried to use the Clausius-Clapeyron relation, but it didn't seem to be valid...

That's actually the path I used to solve the problem originally - using C-C relation twice to find saturated pressures above water and ice, comparing them and solving for T.

I admit I completely forgot about the Kf formula you listed, the question arose from a long discussion on the vapor pressures above solids and liquids and conclusions we can get from them, so I was a bit narrow minded.

Still, chances are Kf formula can be derived this way. I have long lost my notes related to the problem and sadly I have no time to solve it again at the moment. Perhaps later today.
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Offline XeLa.

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Re: Not a problem of the week, but similar level of difficulty
« Reply #17 on: November 18, 2017, 11:15:40 PM »
Okay, I managed to solve this question using the Clausius-Clapeyron equation...

First, consider the equation...

ln(P1/P2)=(ΔHfus/R) × (1/T2 - 1/T1)

ΔHfus is found to be 6010 J/mol using the provided data.

Let P1 = vapour pressure of saturated water at T1, and P2 = vapour pressure of 1 molal glucose solution at T2.

P1 = 611 Pa
T1 = 273 K
P2 = xsolvent × P°solvent = (1 - 0.0177) × 611 = 600.185 Pa (round for final answer)
T2 = ? K

Find T2...

ln(600.185/611) = (6010/8.314) × (1/273 - 1/T2)

T2 = 271.17 K

T2 would be the effective temperature of the glucose solution at its freezing threshold... Given this, we can use the equation:

ΔT = Kf · m

ΔT = 273 K - 271.17 K = 1.83 K
 m = 1 mol/kg

Kf = 1.83/1 = 1.83 kg · K/mol

As such, the cryoscopic constant for water is 1.83 kg · K/mol.


Offline Borek

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Re: Not a problem of the week, but similar level of difficulty
« Reply #18 on: November 19, 2017, 06:34:32 AM »
Yes, that is close to what I had on mind, just my approach took using Claussius-Clapeyron equation for sublimation and evaporation and an additional assumption (quite easy to defend). Had to solve it again to close the case :)

From the information given we know at the melting point pressure of the vapor is 611 Pa. What we also know (not from the question itself, but I assume it to be a common knowledge), is that the melting point (0 °C) is very close to the triple point (0.01 °C apart) - that means we can safely assume 611 Pa to be also a saturated vapor over the solid.

Next things were already said earlier, so I am just listing them here for completeness. At the melting point saturated vapor pressures over the solid and the liquid are identical:

[tex]p_{liquid} = p_{solid}[/tex]

we also know that the

[tex]p_{liquid} = (1-x)p_{liquid}^0[/tex]

and finally we have two Claussius-Clapeyron equations ([itex]t[/itex] stands for the melting point temperature):

[tex]\log\left(\frac{p_{solid}}{611 Pa}\right) = \frac {\Delta H_{sub}}R\left(\frac 1 {273.16} - \frac 1 t\right)[/tex]

and

[tex]\log\left(\frac{p_{liquid}^0}{611 Pa}\right) = \frac {\Delta H_{vap}}R\left(\frac 1 {273.16} - \frac 1 t\right)[/tex]

plugging [itex]p_{liquid}^0[/itex] into the second equation

[tex]\log\left(\frac{\frac{p_{liquid}}{1-x}}{611 Pa}\right) = \frac {\Delta H_{vap}}R\left(\frac 1 {273.16} - \frac 1 t\right)[/tex]

We have two equations, easy to solve for [itex]t[/itex] and [itex]p_{liquid}[/itex]. I was too lazy to do that manually, so I just plugged them into Wolfram Alpha, and got t=271.33.

Again, solution is 1 molal, so ΔT=273.16-171.33=1.83 K×kg/mol.
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