Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: Shadow on May 26, 2013, 05:22:37 AM
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What knowledge is required for solving the Clausius-Clapeyron equation: dp/dT=ΔH/TΔV and d(lnp)/dT=ΔH/RT2? Derivatives or integrals, or both? What rules with these operations are required to be known?
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Solving for what? In what context? You can integrate to get a different form, but whether it is equivalent of solving, I can't say.
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Here is the problem:
A British artist Roger Hiorns entirely filled a flat with a supersaturated copper sulfate solution. After removal of the solution, blue crystals remained on the walls, floor, and ceiling. Humidity inside this flat has a constant low level. Using the Clausius-Clapeyron equation, calculate the temperature at which the humidity will be 35% (of the saturated vapor pressure of water at the same temperature). Necessary data is attached.
The equation is:
CuSO4*5H2O :rarrow: CuSO4*3H2O+2H2O(g)
Enthalpy is 105.04 kJ/mol. The formula is dp/dT=ΔH/TΔV. What to do now, I don't have a clue.
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I would look into using the integrated form of the equation.
http://en.wikipedia.org/wiki/Clausius-Clapeyron_equation#Ideal_gas_approximation_at_low_temperatures
Not that I tried to solve the problem, it would be just my line of attack.
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That is something worth memorizing.
When can I apply the approximation dp/dT=Δp/ΔT?
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Here I have another problem:
Calculate the melting point of ice under a pressure of 50 bar given that the melting point at 1 bar is 0°C. Assume that the density of ice under these conditions is 0.92 g/mL and the density of liquid water is 1.00 g/mL. The molar enthalpy of fusion of water is 6.01 kJ/mol.
dp/dT=ΔH/TΔV and they say that dp/dT=Δp/ΔT. How can they say this?
Afterwards, Δp/ΔT=ΔH/TΔV. Then:
ΔT=Δp·ΔV·T/ΔH
Now Δp=4 900 000 Pa, ΔH=6010 J/mol and ΔV=18/1-18/0.92=-1.565·10-3 dm3/mol.
When plugging into the equation: ΔT=-348.34 K which is nonsense. What did I do wrong :-\?
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dp/dT=ΔH/TΔV and they say that dp/dT=Δp/ΔT. How can they say this?
Do you know the definition of the derivative?
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:-[ That's for a straight line. Could you look at my calculations? Where did I went wrong?
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Do you know the definition of the derivative?
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The rate at which a function changes when the input variable changes.
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Close, but not exactly. It is a limit of the rate for the variable change going to zero.
dp/dT=Δp/ΔT
is an approximation - the smaller the Δ, the better the approximation. But if the Δ is too large, approximation can lead to nonsensical results.
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I'm confused now. They used that approximation and got a good result, but I can't understand their calculation. It is maybe the best that I post their work. It can be found here (problem 2): http://chemistry.illinoisstate.edu/standard/che360/homework/360ps10solns.pdf
The thing that confuses me is that they used pressure in bars and then somehow they get 0.35K from 0.00348 for ΔT. Could you explain me how they did this :-\?
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Bar and L - one uses meters, the other doesn't. That was your mistake.
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I should have converted dm3 to m3.
Thanks so much :).