Thermodynamics Energy Balance: Solving for Work Input in an Isothermal Process

[mastermechanic]

Question
A mass of 1.5 kg of air at 120 kPa and 24 C is contained in a piston-cylinder device. The air is compressed to a final pressure of 600 kPa. During the process heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input during the process.

My Solution

I've solved the problem by the isothermal PV equation the answer is -206 kJ but have some problems with the substituting the result in the first law.

We know that $$ E_(In) - E_(Out) = \Delta U$$ and also $$ Q(In) - Q_(Out) - W_b = \Delta U$$

Ok, there is no heat input in the question so Q(in) = 0 and the Q(out) is the energy that is transferred from the air. W (b) is the boundry work that we found -206 kJ which means work input and finally because the process is isothermall Delta U = 0.

Now let's substitute, $$ - Q_(Out) = -206 kJ $$ If the signs are correct what exactly that means.

We have done 206 kJ work on the system and to be able to keep the temperature constant, there must be taken 206 kJ energy from the air? I am confused here can you please explain the result.

Thanks,

 

[Chestermiller]

Where did you learn to do this with the notation Qout - Qin?

 

[mastermechanic]

Where did you learn to do this with the notation Qout - Qin?

Actually it is shown like Q net - W net = Delta U in the book. But I made an arrangement to ease solution. Q(in) is if there is some heat given from the surrounding to the system, Q(out) if there is heat loss, energy loss from the system except for boundry work, Wb is the boundry work itself. Isn't it correct?

 

[Chestermiller]

Well, your arrangement to ease solution actually complicated it and confused you. Working with the way it is done in your book, you have $$Q_{net}=W_{net}=-206\ kJ$$ This tells you that the net heat added to the system is negative. That is, heat is removed from the system to keep the temperature constant. Isn't this what you expected?

 

[mastermechanic]

Well, your arrangement to ease solution actually complicated it and confused you. Working with the way it is done in your book, you have $$Q_{net}=W_{net}=-206\ kJ$$ This tells you that the net heat added to the system is negative. That is, heat is removed from the system to keep the temperature constant. Isn't this what you expected?

Yes this is true, the point I'm confused is aren't we doing the compression work? I mean we ( a person out of the system) push the piston and compress the air. So, the whole work we did was lost, in other words, the work we did physically was transferred as heat to the surrounding. Is it correct?

 

[Chestermiller]

Yes this is true, the point I'm confused is aren't we doing the compression work? I mean we ( a person out of the system) push the piston and compress the air. So, the whole work we did was lost, in other words, the work we did physically was transferred as heat to the surrounding. Is it correct?

Yes. When we pump air into a bicycle tire, the air gets hot. We need to remove the heat (or it naturally gets conducted away) to keep the temperature down close to the initial temperature.

 

[mastermechanic]

Yes. When we pump air into a bicycle tire, the air gets hot. We need to remove the heat (or it naturally gets conducted away) to keep the temperature down close to the initial temperature.

Ok now I understand thanks your for your help :)

 

[JBA]

It appears the problem requests the total work input and not all of the input work is lost, only that due to the heat of compression. Part of the input work input is stored as potential energy in the increased pressure in the vessel.

 

[Chestermiller]

It appears the problem requests the total work input and not all of the input work is lost, only that due to the heat of compression. Part of the input work input is stored as potential energy in the increased pressure in the vessel.

How would you quantify this stored potential energy? What thermodynamic function would you use?

 

[JBA]

If you are asking me.
W = - mRT ln(P1/P2)

 

[mastermechanic]

If you are asking me.
W = - mRT ln(P1/P2)

But I think the all compression work is already stored as potential energy. Actually we can't say "Potential energy", it is internal energy. The is system is stationary, kinetic and potential energies are zero. So the compression work is stored as internal energy during the process and also this internal energy is transferred to the air by the same rate in the same time. That's what I think.

 

[JBA]

Yes this is true, the point I'm confused is aren't we doing the compression work? I mean we ( a person out of the system) push the piston and compress the air. So, the whole work we did was lost, in other words, the work we did physically was transferred as heat to the surrounding. Is it correct?

The total concern I had that I was addressing was your apparent misconception that all of your input work was lost. There is still stored energy in the cylinder that can be utilized to do work even without a temperature change i.e with isothermal compression. It can be expressed as internal energy in the stored gas; but, in the end, your total work of compression is a sum of the stored energy plus the energy lost due to the transfer of the heat of compression out of the container.
Work = Energy = Force x Distance (lbf-in) or (Newton-meter).

 

[Chestermiller]

If you are asking me.
W = - mRT ln(P1/P2)

But that is just the reversible isothermal work that is done. That is equal to the amount of heat that is removed. So there is no increase in the internal energy of the gas. We know that there is no change in internal energy because, for an ideal gas, the internal energy is a function only of temperature, and the temperature is unchanged.

 

[JBA]

My misstatement, I accept that, but I am addressing the problem statement:

 

[JBA]

The above is incomplete due to a problem with my quote input as follows:

Calculate the work input during the process.

The problem statement does not limit itself to only the the irreversible component of work input.

 

[Chestermiller]

The above is incomplete due to a problem with my quote input as follows:
The problem statement does not limit itself to only the the irreversible component of work input.

The tacit implication is that the work is being done reversibly. At the OP's level of development, this is the only reasonable implication. The statement says that "the air is being compressed to a final pressure of 600 kPa," not that it is being compressed at a constant external pressure of 600 kPa, which would then certainly be suggestive of an irreversible process. In any event, either way, the change in internal energy is zero. The only difference would be that the work and heat would be different (but equal).

 

[JBA]

As to the academic level, in reviewing the original statement he states he has already solved the problem using the PV equation for an isothermal process so he may have already used the equation I posted in getting his -206 Kj value; and, all of the remaining is his own academic curiosity and unrelated to the actual problem.
My only focus remains upon his "all input energy is lost" statement. Everything else in the problem discussion is irrelevant to that specific issue.

 

[Chestermiller]

As to the academic level, in reviewing the original statement he states he has already solved the problem using the PV equation for an isothermal process so he may have already used the equation I posted in getting his -206 Kj value; and, all of the remaining is his own academic curiosity and unrelated to the actual problem.
My only focus remains upon his "all input energy is lost" statement. Everything else in the problem discussion is irrelevant to that specific issue.

I thought I understood you to say that the internal energy increased.

 

[mastermechanic]

My only focus remains upon his "all input energy is lost" statement.

The word "lost" was actually wrong but I corrected it like this.

So, the whole work we did was lost, in other words, the work we did physically was transferred as heat to the surrounding...

 

[JBA]

OK, but that that statement is also incorrect, by using the isothermal PV equation you calculated the energy retained in the compressed gas not the heat of compression energy that was lost. An adiabatic PV calculation would have given you a higher work value and subtracting the isothermal PV work from that adiabatic PV work would give you the work lost.

Chestermiller, I really don't know how else to respond on this issue. In your view, is this a realistic representation of the situation?

 

[mastermechanic]

OK, but that that statement is also incorrect, by using the isothermal PV equation you calculated the energy retained in the compressed gas not the heat of compression energy that was lost.

But you're not clear, you have to specify the name of energy. What do you mean by the retained energy in the compressed gas? Is it internal energy? According to $$ Q_ Net - W_Net = \Delta U $$
The process is isotermal so Delta U must be zero. And the only thing we should calculate is Q net which is equal to W net and we can find its value by isothermal PV equation.

 

[Chestermiller]

OK, but that that statement is also incorrect, by using the isothermal PV equation you calculated the energy retained in the compressed gas not the heat of compression energy that was lost. An adiabatic PV calculation would have given you a higher work value and subtracting the isothermal PV work from that adiabatic PV work would give you the work lost.

Chestermiller, I really don't know how else to respond on this issue. In your view, is this a realistic representation of the situation?

As one of the resident experts in thermodynamics at Physics Forums, I am confused and my head is spinning over what you are trying to say. If you are trying to say that, once the gas has been compressed, you can (1) recover a fraction of the work that was required to compress it by allowing it to expand adiabatically and reversibly, I agree, but its final temperature will be lower than the original temperature or (2) recover all the work that was required to compress it by causing it to expand isothermally and reversibly, I agree too, and, in this case its final temperature will be the same as the original temperature. Either way you are able to recover a significant fraction of the work that was required to compress the gas. If that's not it, then I give up.

I think this thread has come very close to running its course. I will be closing it soon.

 

[JBA]

First of all, I think the above post addresses my position correctly in more scientific terms than mine.

I am coming from an engineering perspective on all of this and I am not sure how to best name the energy contained in the compressed gas. I can call it potential energy because the work energy to compress the gas is conserved in the compressed gas in the cylinder and can be accessed. I might describe it simply as stored energy; either way, it is a source of accessible usable energy regardless of what it is named.

As to the relation of this to the thermodynamic balance equations, that is an issue far better addressed by Chestermiller, who is far more knowledgeable than I on that subject.

 

[Chestermiller]

First of all, I think the above post addresses my position correctly in more scientific terms than mine.

OK. I think we are pretty much on the same page now. I am hereby closing this thread. (Incidentally, I am an engineer (ChE) too.)