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Thermodynamics: calculate q, w, E,H,S for a 5 step process

by Lisa...
Tags: process, step, thermodynamics
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Lisa...
#1
Nov26-06, 10:44 AM
P: 189
I need to calculate q, w, [tex]\Delta E[/tex], [tex]\Delta H[/tex] and [tex]\Delta S[/tex] for the process of heating a sample of ice weighing 18.02 g (1 mole) from -30.0 °C to 140.0°C at constant pressure of 1 atm.
Given are the temperature independent heat capacities (Cp) for solid, liquid and gaseous water: 37.5 J/K/mol, 75.3 J/K/mol, 36.4 J/K/mol respectively. Also, the enthalpies of fusion and vaporization are 6.01 kJ/mol and 40.7 kJ/mol respectively. Assure ideal gas behavior.


I thought of this process as 5 steps:

I) Solid water of -30°C is heated to 0°C
II) Solid water of 0°C melts to give liquid water at 0°C
III) Liquid water of 0°C is heated to 100°C
IV) Liquid water of 100°C is vaporized to give gaseous water at 100°C
V) Gaseous water of 100°C is heated to 140°C

Then I figured I needed to calculate q, w, [tex]\Delta E[/tex], [tex]\Delta H[/tex] and [tex]\Delta S[/tex] for each step seperately and sum them to give the values of q, w, [tex]\Delta E[/tex], [tex]\Delta H[/tex] and [tex]\Delta S[/tex] for the whole process.

=> So q is calculated for I),III) and V) by q=n Cp[tex]\Delta T[/tex] with the Cp values of respectively solid, liquid and gaseous water. For II) and IV) q= n H with values of H of respectively fusion and vaporization enthalpies.

=> Because the process is carried out at constant pressure, all the q's equal the H's.

=> Entropies are calculated for I), III) and V) by [tex]\Delta S = n C_p ln \frac{T_2}{T_1}[/tex] and for II) and IV) with [tex]\Delta S =\frac{\Delta H}{T}[/tex] with T the melting/boiling point and delta H the fusion/ vaporization enthalpy.

=> Now the point at which I got stuck: calculating w and [tex]\Delta E[/tex]

I know that [tex]\Delta E = w + q [/tex] and [tex]w = -p \Delta V[/tex]. For V) I can calculate w with p= 1 atm and by using the ideal gas law to find delta V and with the delta E formula + known q delta E can be obtained.........

But what about the work that is done in the other four steps? I don't know the changes in volume. Could somebody please please please explain to me how I'd calculate w and delta E for the first 4 steps?

EDIT: Now I figured delta E = 0 for II and IV therefore w=-q, because [tex]\Delta E = n C_v \Delta T [/tex] thus it only depends on the temperature, which remains constant during II and IV, so [tex]\Delta T =0 = \Delta E[/tex] . Is that a correct way of thinking? And now how would I tackle calculation of delta E & w of step I, III and V?
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Lisa...
#2
Nov26-06, 02:07 PM
P: 189
Or perhaps it is necessary to calculate w and E directly for the transition (solid) => (gas) with:

[tex]w = -p \Delta V = - n R \Delta T [/tex] and
[tex]\Delta E = \frac{3}{2} R \Delta T [/tex] (=q +w)
and [tex]q= \Delta H = \frac{5}{2} \Delta T[/tex]
Lisa...
#3
Nov26-06, 03:01 PM
P: 189
On a second thought, I don't think my edit is such a brilliant idea, because then q= n Cp delta T should be 0 too for step II and IV which can't be cause heat is neaded to melt the ice/vaporize the water.

I know I need densities of -30,0°C ice (solid), 0°C ice/water (solid/liquid), 100°C water (liquid). Then I can calculate delta V for all the steps (the volume of gaseous water can be calculated from the ideal gas law) and thus find w and E..... but how would I get that far?

OlderDan
#4
Nov26-06, 04:11 PM
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HW Helper
P: 3,033
Thermodynamics: calculate q, w, E,H,S for a 5 step process

Quote Quote by Lisa... View Post
On a second thought, I don't think my edit is such a brilliant idea, because then q= n Cp delta T should be 0 too for step II and IV which can't be cause heat is neaded to melt the ice/vaporize the water.

I know I need densities of -30,0°C ice (solid), 0°C ice/water (solid/liquid), 100°C water (liquid). Then I can calculate delta V for all the steps (the volume of gaseous water can be calculated from the ideal gas law) and thus find w and E..... but how would I get that far?
Since the whole process takes place at constant pressure, do you need to break the process down into steps to find the total PΔV? The ice maybe expands a bit on warming and then melts to become denser water, which first becomes more dense (to 4°C) and then expands and finally evaporates and expands some more, but the total ΔV from start to finish is the difference between the volume of ice at -30°C and the volume of gas at 140°C. The density of ice is reasonably constant at .92g/mL, so the volume is really small compared to the ~22L of gas at the end. I don't think you need to worry about a small uncertainty in ice density.
Lisa...
#5
Nov27-06, 12:35 AM
P: 189
Thank you so very very much! :) Now I can finish the problem! You're great :D

Oh btw and I guess that delta E can be calculated with this total work and the TOTAL q right?
OlderDan
#6
Nov27-06, 11:40 AM
Sci Advisor
HW Helper
P: 3,033
Quote Quote by Lisa... View Post
Thank you so very very much! :) Now I can finish the problem! You're great :D

Oh btw and I guess that delta E can be calculated with this total work and the TOTAL q right?
I believe that is correct. Conservation of work/energy. The heat either results in some work being done, or it stays in the system


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