Thermodynamics -- Entropy of an Isometric Process

In summary: T1s2g-s2f=(h2g-h2f)/T2In summary, the conversation discusses finding the entropy change in a process where a well-insulated rigid container containing a saturated liquid-vapor mixture of water is heated until all the liquid is vaporized. The process is divided into three steps: first, using thermodynamic steam tables to find the specific entropy at the given pressure and quality; second, assuming the specific entropy of the water after heating and pressurizing is equal to the specific entropy of the vapor at the final state; and finally, finding the total
  • #1
AnotherParadox
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In order to better explain my question let me give a precise situation and then state my question

Say I have a well insulated rigid container containing some mass m of a saturated liquid-vapor mixture of water at some pressure P1. Initially it's at some quality x1. An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporized.

To find the entropy change of this process I'm told to

1) Look up the values for water in a thermodynamic steam table for H2O and use the given (observed) property values (for the given property at the given pressure P1) of specific entropy from s1=(1-x1)s1,f + x1*s1,g

2) Assume the specific entropy of the water after it's done heating and further pressurizing to its fully vaporized state to be s1,g=s2

3) Multiply the mass by the change in specific entropy m(s2 - s1) thus obtaining total entropy change of the process.

Now that I have established where this curiosity originated from let me ask my question

During step 2 I assumed (without knowing why) that it is simply O.K. to treat s2 as s1,g this seems highly odd to me since I'm heating a rigid, insulated, container increasing the content's temperature as well as its pressure. By the time the water reaches state two it is no longer the same information on the table for sg at some new pressure and temperature P2 and T2 respectively..

Won't using the entropy values for state 2 from state 1 in the table result in inaccurate results since specific entropy should be different for a higher pressure and temperature?
 
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  • #2
They expect you to be assuming that the system pressure is being held constant (not the volume).
 
  • #3
Chestermiller said:
They expect you to be assuming that the system pressure is being held constant (not the volume).
Rigid container problems are typically constant volume whereas moving boundary problems are constant pressure.
 
  • #4
OK. So the constraint on this problem is that the volume of the container is constant. But, in that case, ##s_2## is not equal to ##s_{1g}##. Do you know how to find the final state of the water vapor in the case where the total volume is constant using the steam tables? In terms of the initial specific volumes of the liquid and vapor, what is the volume of the container? What is the final specific volume of the vapor if it now fills the container?
 
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  • #5
Chestermiller said:
OK. So the constraint on this problem is that the volume of the container is constant. But, in that case, ##s_2## is not equal to ##s_{1g}##. Do you know how to find the final state of the water vapor in the case where the total volume is constant using the steam tables? In terms of the initial specific volumes of the liquid and vapor, what is the volume of the container? What is the final specific volume of the vapor if it now fills the container?
Edit:
Oh man was I wrong. (and so were the online solution help sites)

I can determine s2 from the specific volume by looking at the table and finding where the specific volume vg matches v1.

_____I wouldn't be able to find the final state of the saturated liquid-vapor mixture (as far as I know) since there aren't two intensive properties of which I can determine about it.

I can determine the initial volume and final volumes since they are the same and are based off of the vf and vg values given at the first state. Since neither mass nor volume are changing the specific volume of the system must be the same at both states.

I thought about this last night and I suppose the process could be isothermal if we suggest that all of the heat entering the system is used to convert the liquid part of the mixture to vapor, this same suggestion may be extended to the pressure not increasing as this energy is used to change the phase of the (pure)substance and not increase the temperature or the pressure.

If these suggestions are correct for this model, which I suspect they are, then I have no problem seeing the 2nd state at the same temperature and pressure and thus the conclusion sg,1=s2 ...

Would suffice for me but it would be interesting to see if there's another angle to this or if anyone can confirm these suggestions etc.
 
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  • #6
What do the words "until all the liquid in the tank is vaporized" mean to you?
 
  • #7
Chestermiller said:
What do the words "until all the liquid in the tank is vaporized" mean to you?

Since those words are used in the context of "saturated liquid-vapor mixture of water" I would predict the words "until all the liquid in the tank is vaporized" to mean that the saturated liquid-vapor mixture with some given quality between 0 and 1 is now at quality of 1 where it is purely a saturated vapor.
 
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  • #8
Oh man was I wrong. (and so were the online solution help sites)

I can determine s2 from the specific volume by looking at the table and finding where the specific volume vg matches v1.

Pressure is increasing several times and temperature by around 15% (abs scale)
 
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What is an isometric process in thermodynamics?

An isometric process, also known as an isochoric process, is a thermodynamic process in which the volume of a system remains constant while other parameters such as pressure, temperature, and internal energy may change.

How is entropy defined in an isometric process?

Entropy, denoted by S, is a thermodynamic property that measures the randomness or disorder of a system. In an isometric process, the entropy change (ΔS) is equal to the amount of heat transferred (Q) divided by the system's absolute temperature (T), i.e. ΔS = Q/T.

What is the relationship between entropy and energy in an isometric process?

In an isometric process, the change in entropy (ΔS) is directly proportional to the change in energy (ΔU) and inversely proportional to the temperature (T). This relationship is described by the equation ΔS = ΔU/T.

Can entropy be negative in an isometric process?

No, entropy can never be negative in any thermodynamic process. This is because the second law of thermodynamics states that the total entropy of a closed system always increases or remains constant, but it can never decrease.

How does the entropy of an isometric process relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system always increases or remains constant. In an isometric process, the entropy may increase or remain constant, but it can never decrease, thus following the second law of thermodynamics.

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