Entropy Calculation: Q/T & Ideal Gas Equation

[tony_engin]

Hi all!
I'm a bit puzzled about this question.
Is the entropy change of the ideal gas equals zero?
And for the second part, is it equal to "Q/T" where Q=200kJ and T=40+273?
I'm not sure about how to handle it...please help.

 

[Nylex]

The entropy change of the ideal gas isn't zero. You know that the change in internal energy is 0 for the gas (because U is a function of T only, so no change in T means no change in U) and from the first law therefore, you can get what the heat input to the gas is, in terms of the work done on/by the gas (and you see you have a numerical value, so you don't need to actually calculate anything) and hence calculate the change in entropy.

 

[OlderDan]

The entropy change of the ideal gas isn't zero. You know that the change in internal energy is 0 for the gas (because U is a function of T only, so no change in T means no change in U) and from the first law therefore, you can get what the heat input to the gas is, in terms of the work done on/by the gas (and you see you have a numerical value, so you don't need to actually calculate anything) and hence calculate the change in entropy.

Why is the entropy change of the gas not zero? There is no change of P, V or T. However you formulate dS (for constant n) it involves differentials of these three variables, so the "entropy content" of any state of an ideal gas can be determined. In this problem, isn't the gas just a conduit for the entropy flow into the surroundings with no entropy change of its own?

Of course the entropy leaving the gas is not the same as the entropy entering the surroundings because the temperatures are different. So "flow" should not be taken to mean the entropy is conserved.

 

[Nylex]

Oh ffs. I hate Physics .

 

[Clausius2]

Why is the entropy change of the gas not zero? There is no change of P, V or T. However you formulate dS (for constant n) it involves differentials of these three variables, so the "entropy content" of any state of an ideal gas can be determined. In this problem, isn't the gas just a conduit for the entropy flow into the surroundings with no entropy change of its own?

Of course the entropy leaving the gas is not the same as the entropy entering the surroundings because the temperatures are different. So "flow" should not be taken to mean the entropy is conserved.

I do think there is a change of entropy of gas. This problem is super-simplified, and maybe the professor looks for a simple solution like this:

\Delta S_{gas}=\frac{Q}{T_{gas}}=\frac{200}{273+40}

\Delta S_{surroundings}=\frac{200}{273+25}

The work done over the gas is converted into heat which outflows the reservoir assuming an steady state (T_{gas}=constant).

I do see your reasonement claiming that P,V, and T remains constant and so \Delta S_{gas}=0. Well, this is not correct. Although the solution of this problem is a bit rudimentary (according with the rudimentary data provided), a more serious problem has to model an increasing of gas entropy. Just at the bottom of the logic of this problem is the effect caused both by the paddle wheel and the heat transfer. Both mechanisms involve entropy generation. When you are stirring the gas sure there are pressure and temperature gradients inside the gas, also assuming a semi-periodic steady state. In addition to that, the proper difference of temperatures between the gas and surroundings will generate entropy.

A complete formulation for this problem must be done via Fluid. Mech equations, in particular the entropy equation of a gas:

\frac{\partial}{\partial t}(\rho s)+\nabla\cdot(\rho\overline{v}s)=-\frac{\nabla\cdot \overline{q}}{T}+\frac{\phi}{T}

where:
\overline{v} velocity field caused by the stirring movement
\overline{q} heat flux generated by temperature gradients
\phi Rayleigh internal viscous dissipation function.

This three mechanisms will enhance a gas entropy generation. The last numbers I have made (and surely what the professor waits) are only a rudimentary model of this one.

 

[tony_engin]

I do think there is a change of entropy of gas. This problem is super-simplified, and maybe the professor looks for a simple solution like this:

\Delta S_{gas}=\frac{Q}{T_{gas}}=\frac{200}{273+40}

\Delta S_{surroundings}=\frac{-200}{273+25}

The work done over the gas is converted into heat which outflows the reservoir assuming an steady state (T_{gas}=constant).

I do see your reasonement claiming that P,V, and T remains constant and so \Delta S_{gas}=0. Well, this is not correct. Although the solution of this problem is a bit rudimentary (according with the rudimentary data provided), a more serious problem has to model an increasing of gas entropy. Just at the bottom of the logic of this problem is the effect caused both by the paddle wheel and the heat transfer. Both mechanisms involve entropy generation. When you are stirring the gas sure there are pressure and temperature gradients inside the gas, also assuming a semi-periodic steady state. In addition to that, the proper difference of temperatures between the gas and surroundings will generate entropy.

A complete formulation for this problem must be done via Fluid. Mech equations, in particular the entropy equation of a gas:

\frac{\partial}{\partial t}(\rho s)+\nabla\cdot(\rho\overline{v}s)=-\frac{\nabla\cdot \overline{q}}{T}+\frac{\phi}{T}

where:
\overline{v} velocity field caused by the stirring movement
\overline{q} heat flux generated by temperature gradients
\phi Rayleigh internal viscous dissipation function.

This three mechanisms will enhance a gas entropy generation. The last numbers I have made (and surely what the professor waits) are only a rudimentary model of this one.

Um..as you have mentioned, there are two production of entropy geneartaoin..one from stirring the the gas and the other is the heat transfer.
The former one probably would produce entropy and the latter one seems to pass entropy to the surrounding. So will the two effects just cancel out so that the entropy change equals zero? This also seem to be justified by the two constant state properties (temperature and specific volume of the gas)

 

[Clausius2]

Um..as you have mentioned, there are two production of entropy geneartaoin..one from stirring the the gas and the other is the heat transfer.
The former one probably would produce entropy and the latter one seems to pass entropy to the surrounding. So will the two effects just cancel out so that the entropy change equals zero? This also seem to be justified by the two constant state properties (temperature and specific volume of the gas)

Not at all. Sum the two quantities and you'll see the the entropy of the universe must increase (I have made a mistake: remove the sign minus beyond surroundings entropy. The surroundings entropy variation must be positive, sorry).

The whole process is transforming mechanical work into heat. This is an irreversible process because of dissipation and finite temperature difference.

 

[tony_engin]

um...heat is transferred from the gas to the surroudnings, so the entropy change by heat transfer of the gas should be -200/(273+40), and that of the surrounding is 200/273+25.
And I don't mean that these two equals zero. I mean the entropy change of the gas alone equals zero since there is entropy leaving (heat transfer to the surroundings) and entropy generated by stirring of peddle. Of course the overall change of entropy (surroundings including the gas) should be postive.

 

[SpaceTiger]

I'm going to have to go with OlderDan and tony_engin on this one. Entropy is a state function, so if the gas is left in exactly the same state (T, V) as before the stirring, it ought to have the same entropy. You seem to know what you're doing, tony_engin, I'm not sure why you think you need our help.

 

[Clausius2]

I'm going to have to go with OlderDan and tony_engin on this one. Entropy is a state function, so if the gas is left in exactly the same state (T, V) as before the stirring, it ought to have the same entropy. You seem to know what you're doing, tony_engin, I'm not sure why you think you need our help.

I have given my opinion. If you super simplify the problem then you will be right. Anyway, I am (I am going to be in few time) an engineer, so calculating something which is far away from reality does not make sense to me. If this problem were real, surely it will be a change of entropy due to the equation I posted (reference: J. Spurk "Fluid Mechanics", Ed. Springer Verlag). If you want to stay with simplified stuffs, then go on.

I don't know, I have tried to do my best answering Tony.

Bye.

 

[SpaceTiger]

I have given my opinion. If you super simplify the problem then you will be right. Anyway, I am (I am going to be in few time) an engineer, so calculating something which is far away from reality does not make sense to me. If this problem were real, surely it will be a change of entropy due to the equation I posted (reference: J. Spurk "Fluid Mechanics", Ed. Springer Verlag). If you want to stay with simplified stuffs, then go on.

So I guess you think there's no value in learning Newtonian mechanics by neglecting frictional losses or air drag? Honestly, just suck it up and admit you were wrong. Don't pretend like you're above the problem and scoff at students trying to learn the fundamentals of thermodynamics.

 

[Clausius2]

So I guess you think there's no value in learning Newtonian mechanics by neglecting frictional losses or air drag? Honestly, just suck it up and admit you were wrong. Don't pretend like you're above the problem and scoff at students trying to learn the fundamentals of thermodynamics.

What's wrong with you, boy?. Do not be so superb and wait till Tony feedback us with the correct answer. Maybe YOU are who is wrong. It seems incredible how one guy can loose the respect for someone other. You are so lucky english is not my native language, if not you will hear some words you won't surely like.

I have tried to justify why >>I<<< think there is an entropy change in the gas. According with you have mentioned, imagine there is a block of some mass sliding down an inclinated plane and I want you to answer me which is the temperature of the block just at the bottom. Maybe you will argue T=constant because you have neglected friction, air drag and all secondary issues. Well, I have read the problem and I have decided to count these secondary issues, if not it is very unreal in my opinion.

If you know something of science (I hope you know) you will know that sometimes one is prone to model with simple equations another difficult effects which could be accomplished with another equations not so easy to solve. This is just what I have tried here. It was not my intention to overwhelm someone.

Just keep it cool, baby, and be afraid about science and not about personal attacks.

 

[tony_engin]

Clausius2, SpaceTiger, Nylex, OlderDan,
Really thank you for all your help!
I think I get a clearer picture on this problem through the nice disscussion here.
I will try to look at the problem later when I learn more about thermodynamics.
Again, thank all of you!

 

[Clausius2]

You are welcome, Tony.

 

[SpaceTiger]

What's wrong with you, boy?.

You know I'm older than you, right? Anyway, I'm not going to exchange insults with you, I think this kind of behavior is inappropriate. You know as well as I do that it doesn't matter whether the situation is real or not. The problem was set up to teach a fundamental concept in thermodynamics, not to be a practical experiment.

 

[Clausius2]

You know I'm older than you, right? .

I do know, but you don't seem so.

Let's wait Tony come back with the right result. Then I will accept my comment (although it is true) was not appropriate for this thread.

Anyway I don't know why you have felt so upset with me. I don't understand.

 

[SpaceTiger]

Anyway I don't know why you have felt so upset with me. I don't understand.

Forget it, if the OP is not upset about it, then I'll leave it be.