When is ΔH = ΔU in Thermodynamics?

[abhishek4]

In thermodynamics when can we say that ΔH = ΔU?

 

[Chestermiller]

In thermodynamics when can we say that ΔH = ΔU?

Well, for cases where $\Delta(PV)=0$

 

[abhishek4]

Well, for cases where $\Delta(PV)=0$

That is true but when I asked my teacher she also said that in a closed system this is possible. Why is that?

 

[Chestermiller]

That is true but when I asked my teacher she also said that in a closed system this is possible. Why is that?

Do you think that delta PV can't be zero in a closed system? What about a process involving an ideal gas for which there is vo temperature change?

 

[abhishek4]

Do you think that delta PV can't be zero in a closed system? What about a process involving an ideal gas for which there is vo temperature change?

I know that in a closed system there is no change in volume and we can assume that pressure is remaining constant too but then using the ideal gas equation moles come out to be constant too so in such a case moles formed must be equal to the moles reacted which does not happen in all cases. I just wanted to know if someone could give me an example of this so that i could understand it better. An example of an equation would help.

 

[Chestermiller]

I know that in a closed system there is no change in volume and we can assume that pressure is remaining constant too but then using the ideal gas equation moles come out to be constant too so in such a case moles formed must be equal to the moles reacted which does not happen in all cases. I just wanted to know if someone could give me an example of this so that i could understand it better. An example of an equation would help.

You never mentioned that you were exclusively talking about situations involving chemical reaction. Is that what you want to focus on?

 

[abhishek4]

You never mentioned that you were exclusively talking about situations involving chemical reaction. Is that what you want to focus on?

Yes, i am sorry i forgot to mention that in the post.

 

[Chestermiller]

Yes, i am sorry i forgot to mention that in the post.

If it is a reaction involving ideal gases, then it happens if the number of moles of products is equal to the number of moles of reactants (stoichiometrically).

 

[Ronie Bayron]

That is true but when I asked my teacher she also said that in a closed system this is possible. Why is that?

There is no flow work in a close system ΔPV=0, since ΔH=ΔU+ΔPV. The gas can not expand, so what happens is, internal energy of the gas can be measured by simply the difference in its temperature→ C_vΔT

 

[Chestermiller]

There is no flow work in a close system ΔPV=0, since ΔH=ΔU+ΔPV.

Are you really saying that Δ(PV) = 0 for any process in a closed system?

The gas can not expand, so what happens is, internal energy of the gas could be measured by simply the difference in its temperature→ C_vΔT

Are you really saying that, when a chemical reaction occurs in a closed system, the change in internal energy is C_vΔT?

 

[Ronie Bayron]

Are you really saying that Δ(PV) = 0 for any process in a closed system?

Yes, Chester. Why? Is there something wrong with it?

Are you really saying that, when a chemical reaction occurs in a closed system, the change in internal energy is C_vΔT?

Yes Chester, could there by any more energy unaccounted by it?

 

[Chestermiller]

Yes, Chester. Why? Is there something wrong with it?

Yes Chester, could there by any more energy unaccounted by it?

Hi Ronie,

Actually neither of your answers is correct. I was just checking to see if you really meant what you seemed to indicate. Your answers come very close to bordering on misinformation, which is a no-no at Physics Forums. Please, in the future, if your expertise is not sufficient to address a member's questions, do not offer answers. All this does is confuse the member further. Understand that misinformation can lead to infraction points and eventual banning.

Chet

 

[Ronie Bayron]

Hi Ronie,

Actually neither of your answers is correct. I was just checking to see if you really meant what you seemed to indicate. Your answers come very close to bordering on misinformation, which is a no-no at Physics Forums. Please, in the future, if your expertise is not sufficient to address a member's questions, do not offer answers. All this does is confuse the member further. Understand that misinformation can lead to infraction points and eventual banning.

Chet

How and why? I am delighted and loved to see the reasons why its bordering misinformation. Isn't that suppose the goal of the forum, to educate people in the right track. I have no problem getting wronged, when it's proven.
Here is my reference analysis for "bomb calorimeter" from Ohlone College.

 

[Chestermiller]

How and why? I am delighted and loved to see the reasons why its bordering misinformation. Isn't that suppose the goal of the forum, to educate people in the right track. I have no problem getting wronged, when it's proven.
Here is my reference analysis for "bomb calorimeter" from Ohlone College.

My problem is not with your desire to learn in Physics Forums, but with the authoritative manner in which you gave your answer in post #9 (considering it was incorrect), and your lack of qualifying statements for what you were saying.

With regard to your response in the first statement, Δ(PV) is not generally equal to zero even in an isolated system, let alone in a closed system. If the volume is constant in a closed system, then Δ(PV)= VΔP, and if the pressure is constant in a closed system, then Δ(PV)=PΔV. And then there are all those other cases in which both P and V change in a closed system.

Now for the second statement you agreed to: when a chemical reaction occurs in a closed system, the change in internal energy is C_vΔT. Now, we're not talking about the calorimeter here (which contains the reacting system), we are talking about the actual set of chemical species that constitute the reaction mixture. You are aware that, even if the temperature and pressure of a system are unchanged between the initial and final states of the system, its internal energy and enthalpy can still change, correct?

I assume you have heard of changes in phase which occur at constant temperature and pressure? Examples are melting and vaporization. You are aware that both the enthalpy and internal energy of the material change during a phase change at constant temperature and pressure? You have heard of the terms heat of vaporization and heat of melting, correct? So, for these, the internal energy change is not equal to CvΔT.

In the case of chemical reactions, there are energy effects associated with making and breaking of chemical bonds. You are aware that both the enthalpy and the internal energy of the reacting mixture change during a chemical reaction at constant temperature and pressure (or constant temperature and volume)? You have heard of the terms heat of reaction, heat of combustion, and heat of formation, correct? So, for these, the internal energy change is not equal to CvΔT.

Further questions?

 

[Ronie Bayron]

My problem is not with your desire to learn in Physics Forums, but with the authoritative manner in which you gave your answer in post #9 (considering it was incorrect), and your lack of qualifying statements for what you were saying.

I am sorry Chet, we seemed to have a disconnect here.
In post #9, I was responding to OP's query "In thermodynamics when can we say that ΔH = ΔU?". To cut the chase, my answer was direct to the point. CvΔT(sensible heat) refers to an Isochoric process.

With regard to your response in the first statement, Δ(PV) is not generally equal to zero even in an isolated system, let alone in a closed system. If the volume is constant in a closed system, then Δ(PV)= VΔP, and if the pressure is constant in a closed system, then Δ(PV)=PΔV. And then there are all those other cases in which both P and V change in a closed system.

You caught me like a lawyer on that one, for the term "close" system, probably, lost the case. Isothermal & Isobaric expansion also happens in a close system, I agree, but it's far from the point of ΔH = ΔU which is in the first place the, OP's concern.

Now for the second statement you agreed to: when a chemical reaction occurs in a closed system, the change in internal energy is C_vΔT. Now, we're not talking about the calorimeter here (which contains the reacting system), we are talking about the actual set of chemical species that constitute the reaction mixture. You are aware that, even if the temperature and pressure of a system are unchanged between the initial and final states of the system, its internal energy and enthalpy can still change, correct?

I assume you have heard of changes in phase which occur at constant temperature and pressure? Examples are melting and vaporization. You are aware that both the enthalpy and internal energy of the material change during a phase change at constant temperature and pressure? You have heard of the terms heat of vaporization and heat of melting, correct? So, for these, the internal energy change is not equal to CvΔT.

Yes, I am well aware of latent heats of medium(pure substance) which occurs at constant pressure & temperature regions--" internal energy of the material change during a phase change at constant temperature and pressure"

In the case of chemical reactions, there are energy effects associated with making and breaking of chemical bonds. You are aware that both the enthalpy and the internal energy of the reacting mixture change during a chemical reaction at constant temperature and pressure (or constant temperature and volume)? You have heard of the terms heat of reaction, heat of combustion, and heat of formation, correct? So, for these, the internal energy change is not equal to CvΔT.

Further questions?

Nope and yes, one more. Then, how would you measure or determine the values of specific heat in the lab for "heat of reaction, heat of combustion, and heat of formation"? I suppose you have a good understanding on this.

 

[Chestermiller]

I am sorry Chet, we seemed to have a disconnect here.
In post #9, I was responding to OP's query "In thermodynamics when can we say that ΔH = ΔU?". To cut the chase, my answer was direct to the point. CvΔT(sensible heat) refers to an Isochoric process.

Who says that, in an isochoric process, ΔH=ΔU? For an isochoric process ΔH=ΔU+VΔP.

You caught me like a lawyer on that one, for the term "close" system, probably, lost the case. Isothermal & Isobaric expansion also happens in a close system, I agree, but it's far from the point of ΔH = ΔU which is in the first place the, OP's concern.

ΔH is not equal to ΔU unless Δ(PV)=0.

Nope and yes, one more. Then, how would you measure or determine the values of specific heat in the lab for "heat of reaction, heat of combustion, and heat of formation"? I suppose you have a good understanding on this.

Yes I do have a good understanding of it. You would not need to determine the values of specific heat in the lab for heat of reaction, heat of combustion, and heat of formation, because these changes are defined at constant temperature. They represent the amount of heat (per mole) that must be added to the system to hold it at constant temperature in going from pure reactants to pure products.

Ronie: You need to beef up your background in Thermodynamics (particularly chemical thermodynamics) before you can confidently answer thermo questions on Physics Forums. I don't recommend online sources, because they tend to be incomplete. And Physics Forums cannot provide a complete tutorial on Thermodynamics because of our limited structure. Here are some good references, however, that can be very helpful to you:

Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness
Fundamentals of Engineering Thermodynamics by Moran et al

 

[Ronie Bayron]

Who says that, in an isochoric process, ΔH=ΔU? For an isochoric process ΔH=ΔU+VΔP.

So, the analysis of below source is erroneous or probably lack some consideration of VdP? How would that differ with ΔH=ΔU?

ΔH is not equal to ΔU unless Δ(PV)=0.

Then, when is Δ(PV)=0? Is this condition hypothetical?

Yes I do have a good understanding of it. You would not need to determine the values of specific heat in the lab for heat of reaction, heat of combustion, and heat of formation, because these changes are defined at constant temperature. They represent the amount of heat (per mole) that must be added to the system to hold it at constant temperature in going from pure reactants to pure products.

, you are referring to that (T) temperature right? Generally, for medium involving chemical reaction, right? Consequently, for pure substance that does not undergo chemical reaction the change of number of moles Δn is zero. So, the term becomes zero for pure substance, right? Therefore, analysis for pure substance simplifies into ΔH=ΔU. Do you agree on this? But, for medium which undergoes chemical reaction, Δn≠0 therefore, ΔH=ΔE+Δn_gRT, is this correct?

Thus, we can summarize ΔH for isochoric process as follows
For (Pure Substance - does not undergo chemical reaction) ----> ΔH=ΔU
For (Substance that undergoes chemical reaction)------------------> ΔH=ΔE+Δn_gRT

Do you agree on this?

 

[Chestermiller]

So, the analysis of below source is erroneous or probably lack some consideration of VdP? How would that differ with ΔH=ΔU? View attachment 95979

This derivation is correct. But I don't see any ΔH in there. So, what makes you think that this proved ΔH=ΔU? Here's what I get by continuing this analysis: $$ΔH=ΔU+Δ(PV)=nC_vΔT+nRΔT=n(C_v+R)ΔT=nC_pΔT$$
Do you have a problem with this.

Then, when is Δ(PV)=0? Is this condition hypothetical?

No. For an ideal gas, it is when ΔT=0

, you are referring to that (T) temperature right? Generally, for medium involving chemical reaction, right? Consequently, for pure substance that does not undergo chemical reaction the change of number of moles Δn is zero. So, the term becomes zero for pure substance, right? Therefore, analysis for pure substance simplifies into ΔH=ΔU. Do you agree on this?

No way. The equation you wrote applies to an ideal gas mixture at CONSTANT TEMPERATURE. So, for a pure ideal gas (or a non-reacting ideal gas mixture), even though Δn=0, if the temperature changes, ΔH≠ΔU. For that case, as we showed above, $ΔU=nC_vΔT$ and $ΔH=nC_pΔT$

But, for medium which undergoes chemical reaction, Δn≠0 therefore, ΔH=ΔE+Δn_gRT, is this correct?

Yes, but only at constant temperature.

Thus, we can summarize ΔH for isochoric process as follows
For (Pure Substance - does not undergo chemical reaction) ----> ΔH=ΔU
For (Substance that undergoes chemical reaction)------------------> ΔH=ΔE+Δn_gRT

Do you agree on this?

Obviously not.

 

[Ronie Bayron]

This derivation is correct. But I don't see any ΔH in there. So, what makes you think that this proved ΔH=ΔU?

So, what's the difference of ΔH, ΔU & Q in this set-up.? How do you define ΔH then?

Here's what I get by continuing this analysis: $$ΔH=ΔU+Δ(PV)=nC_vΔT+nRΔT=n(C_v+R)ΔT=nC_pΔT$$
Do you have a problem with this.

No, I don't have the problem with that derived equation but, how it's used, I have an issue.

No way. The equation you wrote applies to an ideal gas mixture at CONSTANT TEMPERATURE.

How can a gas mixture has Δn? Mixture does not react, so Δn=0 and of course ΔH= ΔE only .

I just do not know, I can not somehow agree with you. May be because we are from different field.

Enthalpy for me is a equation of a state basing on the First Law of Thermodynamics which is "arbitrarily" set to sum up(internal energy+flow work)at a point, so if there is no flow work done, enthalpy ΔH reduces to internal energy ΔU. That's is what the book said, the way we are taught and we use to run boilers and steam engines. Otherwise if yours is true-that for a closed system at constant volume( ΔH=ΔU+VΔP) , we will have to revised a new steam tables, probably.

 

[Chestermiller]

So, what's the difference of ΔH, ΔU & Q in this set-up.? How do you define ΔH then?

View attachment 96008

You're saying you have a degree in Mechanical Engineering, and you don't know how ΔH is defined, correct? H = U + PV, and ΔH = ΔU + Δ(PV). In this setup, ΔH=ΔU+VΔP=ΔU+nRΔT

No, I don't have the problem with that derived equation but, how it's used, I have an issue.

I see that you have an issue, but I don't know what that issue is.

How can a gas mixture has Δn? Mixture does not react, so Δn=0 and of course ΔH= ΔE only .

Not if the temperature changes.

I just do not know, I can not somehow agree with you. May be because we are from different field.

No. Your problem is much more fundamental than this. You just haven't learned thermodynamics properly.

Enthalpy for me is a equation of a state basing on the First Law of Thermodynamics which is "arbitrarily" set to sum up(internal energy+flow work)at a point, so if there is no flow work done, enthalpy ΔH reduces to internal energy ΔU. That's is what the book said, the way we are taught and we use to run boilers and steam engines.

Enthalpy is defined in all thermo textbooks as H ≡ U + PV. For an open flow system operating at steady state like the type you are describing, the change in enthalpy of the flowing stream is not equal to the sum of the change in internal energy plus the flow work (sometimes called shaft work). For an incompressible fluid passing between entrance and exit of a system like that, the rate of change of enthalpy for the flowing stream is equal to $\dot{m}\Delta h= \dot {m} (\Delta u+v\Delta P)$, where $\dot{m}$ is the mass flow rate, and the lower case letters represent properties per unit mass. Check your textbook and see.

Otherwise if yours is true-that for a closed system at constant volume( ΔH=ΔU+VΔP) , we will have to revised a new steam tables, probably.

Oh yeah? What if I told you that, if you check your steam tables, you will find that the specific internal energy values u differ from the specific enthalpy values h by the product of the specific volume times the pressure vP? That is h = u + pV. What do you think about that? So even your own steam tables back up what I have been saying.

One more thing. I'm not sure you are aware of it, but I am a Chemical Engineer, not a physicist. And, believe it or not, we ChEs deal with boilers and steam engines all the time. In addition, one other thing we deal with that Mechanical Engineers usually don't are chemically reacting systems. This is our "bread and butter." So please don't tell me about how to model the thermodynamics of chemical reactors. I've modeled many of them during my career.

Chet

 

[Ronie Bayron]

You're saying you have a degree in Mechanical Engineering, and you don't know how ΔH is defined, correct? H = U + PV, and ΔH = ΔU + Δ(PV). In this setup, ΔH=ΔU+VΔP=ΔU+nRΔT

No, you said that. I know what I am doing.

So what's is then your answer for this --"So, what's the difference of ΔH, ΔU & Q in this set-up.? How do you define ΔH then?"
You haven't answered this yet. This is the OP's query. Please focus your answer with this question, no ricochet, please.
Or may be rephrasing it will help - Is ΔH, heat? Is ΔU heat? Is Q heat? Are they not equal for the set-up at post #19?

Enthalpy is defined in all thermo textbooks as H ≡ U + PV. For an open flow system operating at steady state like the type you are describing, the change in enthalpy of the flowing stream is not equal to the sum of the change in internal energy plus the flow work (sometimes called shaft work). For an incompressible fluid passing between entrance and exit of a system like that, the rate of change of enthalpy for the flowing stream is equal to $\dot{m}\Delta h= \dot {m} (\Delta u+v\Delta P)$, where $\dot{m}$ is the mass flow rate, and the lower case letters represent properties per unit mass. Check your textbook and see.

Needless to say, I know that. So you now see where VΔP is applicable(incompressible fluids), yes? Try steam on a turbine if it follows the same formula $\dot{m}\Delta h= \dot {m} (\Delta u+v\Delta P)$

Oh yeah? What if I told you that, if you check your steam tables, you will find that the specific internal energy values u differ from the specific enthalpy values h by the product of the specific volume times the pressure vP? That is h = u + pV. What do you think about that? So even you own steam tables back up what I have been saying.

So, how do we connect ΔH=ΔU +VΔP in there for a constant volume analysis?

 

[Chestermiller]

Hi Ronie,

We seem to be moving in several different directions at the same time, which is driving both of us crazy. So, I'd like to take a step back and simplify. I would like to define a simple focus problem that we can work on involving a constant volume system experiencing a temperature change. This will help us definitively answer your question "So, how do we connect ΔH=ΔU +VΔP in there for a constant volume analysis?" If you agree to work with me on this, I will define a constant volume system involving pure superheated steam that is heated from an initial temperature to a final temperature. I will choose conditions in which the steam exhibits nearly ideal gas behavior, so that we can solve the problem using the steam tables, and then compare with ideal gas predictions. I will specify the volume of the tank, the mass of steam in the tank, and the initial and final temperatures. We will then determine the pressure, the specific volume, the change in internal energy, the change in enthalpy, and the amount of heat added. There are other outputs that I will also ask you to determine, but it is not necessary to specify these now.

The reason I am suggesting going to a simple situation first is that, if we can't solve and agree on that, we will never be able to address a more complicated situation. Plus, we will be able to establish some general conclusions from the results of the analysis.

I am hoping you will agree to pursue this with me. Please let me know if you feel up to the challenge.

Chet

 

[Chestermiller]

So Ronie,

When do you want to start working on that focus problem? I was hoping you would be up to the challenge.

Chet

 

[Ronie Bayron]

So Ronie,

When do you want to start working on that focus problem? I was hoping you would be up to the challenge.

Chet

Hi Chet, sweet! I had a reply on this, posted couple of months ago, but it was gone. I wonder why. Anyhow, yes Chet, go ahead.

 

[Chestermiller]

Hi Chet, sweet! I had a reply on this, posted couple of months ago, but it was gone. I wonder why. Anyhow, yes Chet, go ahead.

OK. Fasten your seat belt.

We are going to be using your own steam tables to make sure that everything we do is exactly right.

I have 1kg of superheated water vapor at 1 bar and 100 C in a closed rigid container. This is going to be the initial condition. From your steam tables, what is the volume of the container? What is the specific volume of the water vapor, the specific internal energy, and specific enthalpy?

We can continue after we have agreed on these answers. The next step is going to involve heating the container to a higher temperature, and determining the changes in internal energy and enthalpy.

Chet

 

[Ronie Bayron]

OK. Fasten your seat belt.

I have 1kg of superheated water vapor at 1 bar and 100 C in a closed rigid container. This is going to be the initial condition. From your steam tables, what is the volume of the container? What is the specific volume of the water vapor, the specific internal energy, and specific enthalpy?

Chet

Yes according to spiraxsarco.com container should have volume of 1.69612 m³per kg of steam and all the rest are given on the attached table of values.

 

[Chestermiller]

Yes according to spiraxsarco.com container should have volume of 1.69612 m³per kg of steam and all the rest are given on the attached table of values.

Good. The two steam table references I have both give H = 2676.2 kJ/kg. So we are pretty much in agreement on that. But I asked you to provide the internal energy U from your steam tables under these conditions, and there is no value given in your table. So what is the value from your source? We can't continue until you provide this.

Chet

 

[Ronie Bayron]

Good. The two steam table references I have both give H = 2676.2 kJ/kg. So we are pretty much in agreement on that. But I asked you to provide the internal energy U from your steam tables under these conditions, and there is no value given in your table. So what is the value from your source? We can't continue until you provide this.

Chet

Yes, let us use this, perhaps.
@ 1 bar, 100 deg C

v = 1.6959 m³/kg
u = 2506.2 kJ/kg
h = 2675.8 kJ/kg

 

[Chestermiller]

Yes, let us use this, perhaps.
@ 1 bar, 100 deg C

v = 1.6959 m³/kg
u = 2506.2 kJ/kg
h = 2675.8 kJ/kg

Excellent! That summarizes the initial state of our system. Now, to get to our final state, we are going to gradually add heat to the 1 kg of vapor in our rigid container until its pressure is 1.5 bars. From your steam tables, what is the final temperature, volume, internal energy, and enthalpy of our vapor?

Chet

 

[Ronie Bayron]

Excellent! That summarizes the initial state of our system. Now, to get to our final state, we are going to gradually add heat to the 1 kg of vapor in our rigid container until its pressure is 1.5 bars. From your steam tables, what is the final temperature, volume, internal energy, and enthalpy of our vapor?

Chet

Let us just make it 1.6 bars to avoid numerous interpolation since its readily available on the table. Again, we recall initial state
@ initial state 1 bar, 100⁰C (we'll include entropy)
v = 1.6959 m3/kg
u = 2506.2 kJ/kg
h = 2675.8 kJ/kg
s = 7.3611 kJ/kg-K
@ Final State 1.6 bar, (s = 7.3611 kJ/kg-K) given process does not lost heat, ideally.
Values of T is in between 400⁰C and 500⁰C

 

[Chestermiller]

Let us just make it 1.6 bars to avoid numerous interpolation since its readily available on the table. Again, we recall initial state
@ initial state 1 bar, 100⁰C (we'll include entropy)
v = 1.6959 m3/kg
u = 2506.2 kJ/kg
h = 2675.8 kJ/kg
s = 7.3611 kJ/kg-K

@ Final State 1.6 bar, (s = 7.3611 kJ/kg-K) given process does not lost heat, ideally.
Values of T is in between 400⁰C and 500⁰C
View attachment 98056

I asked for the results at 1.5 bar, and that's what I want. Also, if the mass of water vapor is 1 kg, and the volume of the rigid container hasn't changed, then has the specific volume of the vapor changed between the initial and final states? Also, regarding your statement that "given process does not lost heat," what did you think I meant when I said "we are going to gradually add heat to the 1 kg of vapor in our rigid container?"

Chet

 

[Ronie Bayron]

I asked for the results at 1.5 bar, and that's what I want. Also, if the mass of water vapor is 1 kg, and the volume of the rigid container hasn't changed, then has the specific volume of the vapor changed between the initial and final states? Also, regarding your statement that "given process does not lost heat," what did you think I meant when I said "we are going to gradually add heat to the 1 kg of vapor in our rigid container?"

Chet

O you're so fast. Haven't got enough time to think and edit my post.
Anyhow, isentropic process was mistake to describe what you want.- since you add up heat gradually and heat is not by some sort of chemical potential, inside a perfect adiabatic enclosure in which case can also be true at conditions you describe.

I realized, that this is isochoric process. It can not be found on the table fixated @ final v = 1.6959 m3/kg which remains as is, no matter what. Assuming water vapor behaves an ideal gas above regions of the saturated vapor curve, we can find T using PV =mRT or T = PV/mR; where R- is specific gas constant of water of course. This holds true unless, you rebutted.
Do you have problem with this?

I asked for the results at 1.5 bar, and that's what I want

It's not so essential to prove your point. Let's just cut the chase. Is that okay with you?

 

[Chestermiller]

O you're so fast. Haven't got enough time to think and edit my post.
Anyhow, isentropic process was mistake to describe what you want.- since you add up heat gradually and heat is not by some sort of chemical potential, inside a perfect adiabatic enclosure in which case can also be true at conditions you describe.

I realized, that this is isochoric process. It can not be found on the table fixated @ final v = 1.6959 m3/kg which remains as is, no matter what. Assuming water vapor behaves an ideal gas above regions of the saturated vapor curve, we can find T using PV =mRT or T = PV/mR; where R- is specific gas constant of water of course. This holds true unless, you rebutted.
Do you have problem with this?It's not so essential to prove your point. Let's just cut the chase, okay with you?

Well, we said we were going to do it strictly by using your steam tables (to make sure I'm not trying to pull a fast one). Later, we can compare with the ideal gas law, but, for now I'd like to stick with the steam tables. Besides, if you assume an ideal gas, you still will have to deal with the problem of determining the internal energy and the enthalpy at the final state.

There are steam tables online that have the properties evaluated at 1.5 bar, if you want to avoid that interpolation (and I deliberately chose 1.5 bar because one of the specific volume values listed in the table at 1.5 bar is almost exactly our specific volume, so you wouldn't have to do any interpolating). However, if you can't find a steam table with 1.5 bar in it, I will settle for your using 1.6 bar. But then you will have to interpolate with respect to the specific volume to get the temperature, internal energy and enthalpy. Your choice.

 

[Ronie Bayron]

Well, we said we were going to do it strictly by using your steam tables (to make sure I'm not trying to pull a fast one). Later, we can compare with the ideal gas law, but, for now I'd like to stick with the steam tables. Besides, if you assume an ideal gas, you still will have to deal with the problem of determining the internal energy and the enthalpy at the final state.

There are steam tables online that have the properties evaluated at 1.5 bar, if you want to avoid that interpolation (and I deliberately chose 1.5 bar because one of the specific volume values listed in the table at 1.5 bar is almost exactly our specific volume, so you wouldn't have to do any interpolating). However, if you can't find a steam table with 1.5 bar in it, I will settle for your using 1.6 bar. But then you will have to interpolate with respect to the specific volume to get the temperature, internal energy and enthalpy. Your choice.

Do you have any to recommend? So what's the value of u and h @ 1.5 bars & v = 1.6959 m³/kg

 

[Chestermiller]

Do you have any to recommend? So what's the value of u and h @ 1.5 bars & v = 1.6959 m³/kg

I got one by googling "steam tables," but it was a PDF file, and I don't know how to get the actual URL. There are also steam tables in thermo books I have (that are consistent with the PDF table). In Moran et al, they give v = 1.695 m^3 at 280 C, with u = 2778.6 kJ/kg and h = 3032.8 kJ/kg. If you are comfortable using these values, we can continue.

Chet