How do I find the entropy variation?

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Homework Statement


Find the ∆S per mol between liquid water at -5 ºC and ice at -5ºC at 1020hPa
Data:
∆CP,m fusion = 37,3 J K-1 mol-1
∆fusH = 6,01 kJ mol-1
The answer is 21.3 J/K mol

Homework Equations


Usually I solve these problems by steps when they are at P=1 atm but since its at P=1020 hPa i don't really know how to proceed.
I tried using the following equation:
∆S= Q/T + nCpm ln T2/T1
Since I am given the Cpm I interpret that the pression is constant throughout the process

The Attempt at a Solution


I first tried to solve it like I normally would, without taking into account that the press ion a was not 1 atm.
I want to try and find the temperature at which water soldifies at 1020 hPa but I don't know how with the data I am given. I thought that since Qp is the enthalpy maybe I could find it using Q = n∆T specific heat of water
but even thought I know what the specific heat of water is (41814 J/K kg) I'm not given this data so I'm unsure if this is what I have to do.
Handwritten entropy calculation: H2O(l) at -5 C to H2O(s) at -5 C via 0 C, with dS12 = 0.68 J/K, dS23 = -22.01 J/K, and dS34 terms shown.
Handwritten working on lined paper: ΔS34 = 37.3 ln(268/273) = -0.68 J/K, then ΔSr = ΔS12 + ΔS23 + ΔS34, and ΔSr = 0.68 - 0.68 + 22.01 = -22.01 J/K, with a partial line for problem 10 showing cp = 8.62 + 3·10^-3 T J/K·mol.

This is what I first tried doing but after I realized my mistake I did this:
Handwritten derivation of entropy change for liquid water freezing to ice: ΔS liquid-solid = -ΔfusH/T + ΔCp,m ln(T2/T1), then ΔS = -6010/273 + 37.3 ln(268/273) ≈ 22.3 J/K

I didn't obtain the value for entropy I should have with neither of the methods I tried. Where am I wrong?
 
Attachments
  • Handwritten entropy calculation: H2O(l) at -5 C to H2O(s) at -5 C via 0 C, with dS12 = 0.68 J/K, dS23 = -22.01 J/K, and dS34 terms shown.
    25383443_1748247851875564_819929200_o.jpg
    36.7 KB · Views: 569
  • Handwritten working on lined paper: ΔS34 = 37.3 ln(268/273) = -0.68 J/K, then ΔSr = ΔS12 + ΔS23 + ΔS34, and ΔSr = 0.68 - 0.68 + 22.01 = -22.01 J/K, with a partial line for problem 10 showing cp = 8.62 + 3·10^-3 T J/K·mol.
    25383256_1748250738541942_717490943_o.jpg
    50.6 KB · Views: 500
  • Handwritten derivation of entropy change for liquid water freezing to ice: ΔS liquid-solid = -ΔfusH/T + ΔCp,m ln(T2/T1), then ΔS = -6010/273 + 37.3 ln(268/273) ≈ 22.3 J/K
    25401064_1748247905208892_575902280_o.jpg
    27.3 KB · Views: 568
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Discussion

You are correct about thinking of this as a multi-step process.

ice at -5C and 1020 hPa to ice at 0 C and 1020 hPa
ice at 0 C and 1020 hPa to ice at 0 C and 1 atm
ice at 0 C and 1 atm to water at 0 C and 1 atm
water at 0 C and 1 atm to water at 0C and 1020 hPa
water at 0C and 1020 hPa to water at 5 C and 1020 hPa
 
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Chestermiller said:
You are correct about thinking of this as a multi-step process.

ice at -5C and 1020 hPa to ice at 0 C and 1020 hPa
ice at 0 C and 1020 hPa to ice at 0 C and 1 atm
ice at 0 C and 1 atm to water at 0 C and 1 atm
water at 0 C and 1 atm to water at 0C and 1020 hPa
water at 0C and 1020 hPa to water at 5 C and 1020 hPa
Handwritten thermodynamics working showing ΔCp fus = 37.3 J/K·mol and ΔH fus = 6.01 kJ/mol, a cycle of phase-change steps for H2O between -5° and 0° at pressures 1020·10² Pa and 101325 Pa, and the entropy sum ΔS_T = ΔS12 + ΔS23 + ΔS34 + ΔS45 + ΔS56.
Handwritten entropy calculation on lined paper: total entropy change as the sum of five stage terms, each worked out with ln ratios of temperatures or pressures, ending with a total of -20.65 J/K.

I tried doing it this way and it really does make more sense but I can't still obtain the 21.3 J/K I should and I really don't know why. I thought maybe it's because of the Cpm I have been given , but it turns out it doesn't really differ from the one I was given, so I thought I might change the sign since one it's fusion and the other one its for solidification. Still, I haven't obtained the result I need and I don't know why. Could anyone please help me?
 
Attachments
  • Handwritten thermodynamics working showing ΔCp fus = 37.3 J/K·mol and ΔH fus = 6.01 kJ/mol, a cycle of phase-change steps for H2O between -5° and 0° at pressures 1020·10² Pa and 101325 Pa, and the entropy sum ΔS_T = ΔS12 + ΔS23 + ΔS34 + ΔS45 + ΔS56.
    fisica1.jpg
    31.9 KB · Views: 597
  • Handwritten entropy calculation on lined paper: total entropy change as the sum of five stage terms, each worked out with ln ratios of temperatures or pressures, ending with a total of -20.65 J/K.
    fisica2.jpg
    33.9 KB · Views: 663
You did this mostly correctly in your first attempt. The first thing to recognize is that 1020 hPa is essentially 1 atm, so there is no need to account for a pressure effect. So the change in entropy in going from 1 mole of ice at -5 C to water at - 5 C is given by:
$$\Delta S=C_i\ln(273/268)+\frac{\Delta H(0)}{273}+C_w\ln(268/273)=(C_W-C_i)\ln(268/273)+22.0$$
We know that ##C_w-C_i=37.3\ J/mol.K##
 
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I don't quite understand how you're using logarithmic properties here, could you explain it to me please?
 
Upupumiau said:
I don't quite understand how you're using logarithmic properties here, could you explain it to me please?
Are you saying that you are not sure how to determine the change in entropy of a solid or liquid when it is heated or cooled from an initial temperature to a final temperature (and why this should involve natural logs of temperatures)?
 
I do, sorry. I had to revisit the basics. Thank you for your help.
 

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