## Review Session-Thermo-files- Wednesday-Q11-clarification [ENDORSED]

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

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### Review Session-Thermo-files- Wednesday-Q11-clarification

Hello everyone,

This is Edgar (the TA that the held the thermo review session on wednesday evening)
I am uploading the worksheet that I went over along with the solutions.
I just wanted to make a clarification on question 11. I added the question about drawing the P-T diagram to a problem on the book. I actually intended to put V-T diagram (please see solutions). This was intended to illustrate that when you calculate a change in entropy you can split this in iso-processes because you get to the same final point. Entropy is a state function
In this case, you split the process in two steps, (1) change in V isothermally and (2) change in T isovolumetrically (this is why you use Cv=3/2R for monoatomic gas)
If the problem stated that the second process happens at constant P then (2) the change in T would be isobaric (you use Cp=5/2R for monoatomic gas) and then it would make sense for you to be asked to draw a P-T diagram. Some of you pointed this out.
Note: the solutions go in reverse order Q12 --> Q1
Attachments
Solutions-Thermo-12-1.pdf
Solutions go from Q12--> Q1
Review Session-Thermodynamics-converted.pdf

bonnie_schmitz_1F
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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

In problem 10, how do we know that ∆H = q?

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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

Problem 10 asks about the internal energy change of a reaction. When you deal with reactions and also phase changes (which usually happen in reactions), you may assume const T and const P. The T and P are also stated in the problem.
If that is the case q_p=deltaH
Heat at constant pressure is equal to the change of enthalpy of the reaction

Ethan Baurle 1A
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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

Number 8 asks for the enthalpy change per mol so wouldn't you have to convert 2000g to moles of BaSo4 and then solve per mol?

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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification  [ENDORSED]

Since 1L of 1M Ba(NO3)2 contains one mole of Ba2+ and 1L of 1MNa2SO4 contains 1mole of (SO4)2- ions, 1 mole of solid BaSO4 is formed in this experiment. So the enthalpy of this reaction is the enthalpy per mole of BaSO4 formed.

Emmaraf 1K
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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

I wanted to clarify since it's not entirely clear to me about question 11:
We wouldn't know whether to use the Cv or Cp value unless we were given the V-T or P-T diagram and saw whether volume or pressure was held constant, right?

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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

You don’t have to necessarily be given a diagram. But it has to be stated in words that it is a constant pressure process while changing T.

annabel 2A
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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

For #5b, would the reaction only be reversible if deltaG=0?

annabel 2A
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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

For #8, can someone explain why deltaH is negative? Shouldn't the enthalpy change of the reaction be positive since the temperature of the solution increases? So the solution absorbed heat?

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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

For #11, why is Cv used for the change in temp instead of Cp? How do we distinguish when to use what?

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### Re: Review Session-Thermo-files- Wednesday-Q11-clarification

dStotal = dSpressurechange + dStemperaturechange

For dSpressurechange, we use S=nRln(P1/P2), to see how entropy changes as a result of pressure changing along with something (either V or T). I assumed that V would be changing in this part 1 of the question, because if T changed, then the gas wouldn't be at 423K which was the starting temp for part 2 of the question. In other words, the starting temperature of the gas is 423K, and it should stay at that temperature as P changes (so the pressure change is isothermal).

Is the assumption that dP causes dT, and final temperature of part 1 IS the starting temp for the 2nd calculation? In that case, what's actually happening is dP causes a gas to rise to temperature 423 K, but the gas itself isn't starting at 423 K.

I thought if the first change (pressure) was isothermal, then the second change (temp) was isobaric, then that would make more sense. Since the VT graph part of the question was kind of a last minute addition, is it simply not appropriate for this question?

What part of my understanding is wrong here?