# Thermodynamics energy balance

1. Jul 26, 2017

### mastermechanic

• Thread moved from the technical forums
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 lets 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,

Last edited: Jul 26, 2017
2. Jul 26, 2017

### Staff: Mentor

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

3. Jul 26, 2017

### mastermechanic

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?

4. Jul 26, 2017

### Staff: Mentor

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?

5. Jul 26, 2017

### mastermechanic

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?

6. Jul 27, 2017

### Staff: Mentor

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.

7. Jul 27, 2017

### mastermechanic

Last edited: Jul 27, 2017
8. Jul 27, 2017

### 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.

9. Jul 27, 2017

### Staff: Mentor

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

Last edited: Jul 27, 2017
10. Jul 27, 2017

### JBA

W = - mRT ln(P1/P2)

11. Jul 27, 2017

### mastermechanic

But I think the all compression work is already stored as potential energy. Actually we cant 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.

12. Jul 27, 2017

### JBA

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).

13. Jul 27, 2017

### Staff: Mentor

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.

Last edited: Jul 27, 2017
14. Jul 27, 2017

### JBA

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

15. Jul 27, 2017

### JBA

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.

16. Jul 27, 2017

### Staff: Mentor

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).

17. Jul 27, 2017

### 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.

18. Jul 27, 2017

### Staff: Mentor

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

19. Jul 27, 2017

### mastermechanic

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

20. Jul 27, 2017

### 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?