Thermodynamics: calculate q, w, E,H,S for a 5 step processby Lisa... Tags: process, step, thermodynamics 

#1
Nov2606, 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 (C_{p}) 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 C_{p}[tex]\Delta T[/tex] with the C_{p} 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? 



#2
Nov2606, 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] 



#3
Nov2606, 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? 



#4
Nov2606, 04:11 PM

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P: 3,033

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



#5
Nov2706, 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? 



#6
Nov2706, 11:40 AM

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