
#1
Jan1013, 07:43 AM

P: 423

Hello
I was wondering, with regards to the Tds equation Tds = de + pdv: 1. All of my textbooks state that integrating this equation, although derived for a reversible process, will give the entropy change regardless of the process or whether or not the process is reversible. However, I don't understand this concept because if the Tds equation was rederived from first law for a general process, I thought that there would be a Tσ (entropy generation term, zero for reversible) in the equation since ds = δQ/T + σ and entropy generation would be path dependent? 2. With regards to the reversible Tds equation, I have trouble seeing how this equation is path independent since, for two fixed states, I always thought there was more than one possible work path or pdv expression in which state 1 can be used to move to state 2 and if this were true it would end up giving different s2s1 values? 3. I have only found proofs for entropy as a state property for reversible process and argued that by extension it must be a state property for any process but is there a proof that directly show that it is a state property for any process? Thanks very much 



#2
Jan1013, 07:56 AM

Sci Advisor
P: 3,366

The state variables p, V, E, S etc. are only defined for equilibrium states. In classical thermodynamics, all states proceed from one initial equilibrium state to a final equilibrium state. The entropy difference can be calculated for any path in equilibrium state space to the final state.




#3
Jan1013, 08:15 AM

P: 5,462





#4
Jan1013, 08:48 AM

P: 423

Tds Equation and EntropyI understand and have accepted the statement but I have trouble understanding how this is reflected, which is why I posed the questions above. For instance, I imagine that there is more than one polytropic (or any) process from two equilibrium states for which work can be applied, but from Tds = de + pdv, the integral of the p/Tdv term (taking T constant) would result in different values while e2e1 doesn't change if the initial and final states are the same. To me this seems to reflect different s2s1 values despite having different processes from the two states. I have the same confusion regarding the derivation of the Tds equation if entropy generation is added as I see that it would change the form of the Tds equation which many books say is applicable for both reversible or irreversible processes. Any help would be appreciated. Thanks very much 



#5
Jan1013, 12:34 PM

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P: 3,366





#6
Jan1013, 01:00 PM

P: 423

So what about a noninsulated piston cylinder system undergoing reversible quasistatic compression such that there is no irreversibility from heat transfer? In this case regardless of the process the temperature would remain the same and the only difference in the Tds equation for different compression processes would be the pdv term? Thanks 



#7
Jan1013, 01:40 PM

Sci Advisor
P: 3,366

If T is constant, p is a unique function of V, namely p=nRT/V.




#8
Jan1013, 01:48 PM

P: 423

So for different polytropic processes to be possible, then the temperature at the end states must not be the same and if it the temperature throughout the process is the same, then the pdv process must satisfy the ideal gas law (assuming an ideal gas is compressed)? So this would mean that for different reversible processes with the same end states, the sum of the integral of de/T and p/Tdv will always end up working out to be the same value? I'm also wondering if you can take a look at my first question in the OP. Thanks very much for your help 



#9
Jan1013, 03:24 PM

Sci Advisor
P: 3,366

The integrals of de/T and p/TdV will not always be the same when T is variable, but their sum will be. Regarding your first question you have to distinguish between a process, which may be irreversible, and a path in the state space of equilibrium states used for calculation of the change of S. Usually when there is entropy production in an irreversible process, the work is also not equal to pdV. So there is no contradiction with TdS not being dQ. Think of stirring a viscous liquid which is a form of nonvolume work which gives rise to entropy production. 



#10
Jan1013, 07:01 PM

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Thanks
PF Gold
P: 4,424

Your initial equation TdS = dU + PdV applies to the differential change in the parameters from one equilibrium state to another. It doesn't matter whether you got from the initial state to the final state by a reversible path or an irreversible path. However, the changes described by the equation can not be assumed to apply along the path; they can only be assumed to apply to the overall change from the initial to the final end point. However, if the path is reversible, they will also apply at all points along the path.
I hope this makes sense. 



#11
Jan1113, 10:04 AM

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Thanks
PF Gold
P: 4,424

If you would like to learn more about how all this plays out with regard to the contribution of entropy generation over an irreversible path to the total entropy change between the initial and final equilibrium states of a closed system, see my postings on the following PF link:
http://www.physicsforums.com/showthread.php?t=660830 



#12
Jan1213, 02:38 PM

P: 423

Thanks very much for your responses
I was wondering, when the Tds equation is integrated I thought that it was integrated along the path in which a process takes place whether reversible or irreversible? Thanks again 



#13
Jan1213, 03:14 PM

Sci Advisor
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Thanks
PF Gold
P: 4,424

You have to recognize that, in a system undergoing an irreversible process, the key variables, like temperature, pressure, and velocity are not uniform within the system, and vary with spatial position and time. But we definitely have the capabilities for solving for these variations by solving the above differential balance equations. One of the tools that is often used in this regard is computational fluid dynamics. We do not need to throw our arms in the air and simply declare that, because the process is irreversible, there is no useful information we can get about what is going on within the system. Chet 



#14
Jan1213, 03:41 PM

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P: 3,366





#15
Jan1213, 03:52 PM

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P: 3,366





#16
Jan1213, 04:11 PM

P: 423

Basically I found that all textbooks say that the equation Tds = de + pdv is valid for between any equilibrium state reversible, irreversible etc. , which I'm confused about since for cases where irreversibilities are present I thought the equation should be Tds = de + pdv + Tdσ. Thanks very much for answering my questions thus far. 



#17
Jan1213, 04:53 PM

Sci Advisor
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Thanks
PF Gold
P: 4,424

Chet 



#18
Jan1213, 06:34 PM

P: 5,462

The material you require to answer all three of your questions was presented by Caratheodory in 1909.
You will find good discussions of this in books by Wilson (1951) Thermodynamics and Statistical Mechanics (Cambridge University Press) Pippard (1964) Classical Themodynamics (Oxford University Press) Carrington (1994) Basic Thermodynamics (Oxford University Press) All three follow and further develop Caratheodory's highly mathematical method and discuss the conditions under which dw and dq may be integrated and prove mathematically that entropy is a state function. All three are highly readable, but different, and contain penetrating insights into classical thermodynamics. I'm sorry but there are no short form solutions to this. Each proof is many pages long. 


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