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Topic: dG and dS after mixing different quantities of water  (Read 7155 times)

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Offline cseil

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dG and dS after mixing different quantities of water
« on: August 20, 2015, 11:56:56 AM »
Hi,
this is the problem I've got from my book. I have a problem (as usual) with the ΔG.

100g of water at 10°C are mixed with 200g of water at 40°C.
Calculate dS and dG.

Cp=1cal/K*g
S°=16.75 cal/K

Firstly I've calculated the final temperature (303K).
The ΔS is easy to calculate and the result is ok.

The problem is with ΔG.

ΔG=ΔH-Δ(TS)

The ΔH is CpdT.
Now I have to calculate the quantity T2S2-T1S1.

I have calculated this quantity for the 100g of water at 10°C and the other 200g of water at 40°C.

S2=S°+ 100*ln(303/298)
S1=S° +100*ln(283/298)

and so on for the other one.

So I calculate
ΔG1= ΔH-(T2S2-T1S1)

I sum ΔG1+ΔG2 and obtain the ΔGtot. Is it correct?

Offline orthoformate

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Re: dG and dS after mixing different quantities of water
« Reply #1 on: August 20, 2015, 03:38:31 PM »
S2=S°+ 100*ln(303/298)
S1=S° +100*ln(283/298)

Where did this come from? How does it relate to ΔS=((dqrev)/T)

The integral of ((dqrev)/T)dT ?

Offline cseil

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Re: dG and dS after mixing different quantities of water
« Reply #2 on: August 21, 2015, 04:37:45 AM »
q=CpdT
TdS = CpdT
dS = Cp/T dT

Offline mjc123

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Re: dG and dS after mixing different quantities of water
« Reply #3 on: August 21, 2015, 05:26:15 AM »
I think you're right. At first I thought "easy, ΔH = 0 so ΔG = -TΔS". But then what value of T would you use? So I think your approach is correct, but be careful! What does "S° = 16.75 cal/K" refer to? Were you going to use the same value of S° for the 100g cold water and 200g warm water?
Your final answer should be independent of both S° (in appropriate units) and T° - have you checked that?

Offline cseil

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Re: dG and dS after mixing different quantities of water
« Reply #4 on: August 21, 2015, 06:26:56 AM »
I think you're right. At first I thought "easy, ΔH = 0 so ΔG = -TΔS". But then what value of T would you use? So I think your approach is correct, but be careful! What does "S° = 16.75 cal/K" refer to? Were you going to use the same value of S° for the 100g cold water and 200g warm water?
Your final answer should be independent of both S° (in appropriate units) and T° - have you checked that?

I thought the same  ;D
How should I use S° for different quantities of water?

Offline mjc123

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Re: dG and dS after mixing different quantities of water
« Reply #5 on: August 21, 2015, 07:23:05 AM »
Multiply the specific entropy by the mass. Have you got a value for S° in cal/g*K? "cal/K" implies a given mass of water - or was it just a typo? Entropy is an extensive property!

Offline cseil

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Re: dG and dS after mixing different quantities of water
« Reply #6 on: August 21, 2015, 07:48:48 AM »
Multiply the specific entropy by the mass. Have you got a value for S° in cal/g*K? "cal/K" implies a given mass of water - or was it just a typo? Entropy is an extensive property!

It was not a typo. It's written cal/K on the book, so I just copied it.
I didn't think about it, actually, and I used the same value for both of the calculations.

I searched online and found that 16.75 is cal/mol*K, so I have to multiply for 5.55 and 11.11.
I did it and it works: I get 101 cal and the book 99.5 cal, so I think it's ok  ;D

Thank you! I'm glad it was just a little mistake and not a big one  ;D
Now I'm getting in trouble with the Clausius-Clapeyron equation, so be ready for another topic  ;D ;D ;D

Offline orthoformate

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Re: dG and dS after mixing different quantities of water
« Reply #7 on: August 21, 2015, 08:27:34 AM »
S2=S°+ 100*ln(303/298)
S1=S° +100*ln(283/298)

Ok, well just to help me understand, can you tell me where the 100 comes from? and why you chose the ratios you did?

I would have thought the answer would be:

S1= 100gCp*ln(303/283)
S2= 200gCp*ln(303/313)

since S=(Cp/T)dT

Offline cseil

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Re: dG and dS after mixing different quantities of water
« Reply #8 on: August 21, 2015, 08:49:25 AM »
S2=S°+ 100*ln(303/298)
S1=S° +100*ln(283/298)

Ok, well just to help me understand, can you tell me where the 100 comes from? and why you chose the ratios you did?

I would have thought the answer would be:

S1= 100gCp*ln(303/283)
S2= 200gCp*ln(303/313)

since S=(Cp/T)dT

The Cp is 1 cal/g*K so if you have 100g Cp is 100 and if you have 200g is 200.


The calculation of S1 and S2 is done for both the quantities of water.
S changes with T as: S°(298K) + Cpln(T2/T1)

but you know S° at 298K so you have to calculate the entropy change starting from 298K and not the initial temperature of the water. You are calculating S at 303K (and it is S2) and S at 283K (S1).

Offline mjc123

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Re: dG and dS after mixing different quantities of water
« Reply #9 on: August 21, 2015, 12:59:28 PM »
Quote
I did it and it works: I get 101 cal and the book 99.5 cal, so I think it's ok  ;D
Do you see a problem here?
I get a similar answer (100.15 cal). But if I do it the "naive" way I mentioned above, assume a temp of 303K and ΔS = 0.3345 cal/K, I get ΔG = -TΔS = -101.35 cal. Very similar magnitude, but opposite sign. Now for a spontaneous process ΔG should be negative. I've checked my working and I don't think there's a simple sign mistake; it appears to be genuinely the case that, although the increase in entropy of the cold water is greater than the decrease in entropy of the hot water, so ΔS is positive, the increase in TS of the cold water is less than the decrease in TS of the hot water, so Δ(TS) is negative and ΔG is positive. Spot the fallacy, anyone? Is it to do with reversibility? Is it wrong to say dS = dq/T?

Offline cseil

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Re: dG and dS after mixing different quantities of water
« Reply #10 on: August 23, 2015, 04:10:40 PM »
Well, can't be reversible, right?

I know that every heat transfer from a hot object to a cold object is irreversible. An infinitesimal change cannot reverse it or it means that we transferred heat from a cold object to a hot one.

Is that what you mean?

Offline mjc123

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Re: dG and dS after mixing different quantities of water
« Reply #11 on: August 27, 2015, 08:43:56 AM »
Any suggestions, anybody? This is really bugging me!
Does the criterion ΔG ≤ 0 only apply to isothermal processes?

Offline Corribus

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Re: dG and dS after mixing different quantities of water
« Reply #12 on: August 27, 2015, 10:00:35 AM »
I don't think I really understand the question. If the process isn't isothermal, Gibbs energy isn't typically used. T = constant is usually one of the basic assumptions when working with G, no?
Honestly, I'm not sure what is the best way to apply Gibbs energy when there is a large temperature change.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline mjc123

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Re: dG and dS after mixing different quantities of water
« Reply #13 on: August 27, 2015, 12:38:20 PM »
That's probably the answer to my question then. But the original question did ask for ΔG.

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