# Compressing ideal gas

1. Oct 12, 2012

### Sunfire

Hi,

trying to find the error here

compressing ideal gas from p1 to p2 adiabatically means the invested energy in the gas is negative and equal to

E = -αnRT1[(p2/p1)γ-1/γ-1] = -cvm(T2-T1)

If then the gas is allowed to cool down isochorically back to T1, it would lose heat

Q =cvm(T2-T1)

This is equal to the invested energy, therefore in the end the gas is back where it started. Is this correct?

Does this mean if one needs compressed gas at temperature T1, the only way to achieve it is to compress the gas starting at some T3<T1?

Thank you

2. Oct 13, 2012

### Andrew Mason

Since it is an ideal gas, its internal energy is a function of temperature only. So it has the same temperature and internal energy as it did originally.
No. One could compress the gas isothermally starting at temperature T1. You would just have to put it in thermal contact with reservoir at T1 and compress it slowly so that the temperature remained constant.

AM

3. Oct 13, 2012

### 256bits

The gas does not have the same state as before the adiabatic compression and constant volume cooling. While the internal energy, which is a function of temperature, is the same in bothe states, the pressure and volume are different.

Note that if you had not cooled the gas after the adiabatic compression, you could have retrieved the work put into the system and the internal energy would have returned to its initial value.

If, with the adiabatic compression and isochoric cooling, you could now attempt to have the gas perform some work at the expense of its internal energy, which means the temperature will decrease to below T1.

4. Oct 13, 2012

### Staff: Mentor

This is a very interesting problem you have described. You have compressed a gas by an adiabatic reversible path, and then cooled it back down to its original temperature. The final state is the same as if you had compressed the gas isothermally in a single step operation. But the amount of heat you removed in the second step and the amount of work you did in the first step are higher than a single step reversible isothermal compression to the final state. Yet the cooling in the second step could have been carried out reversibly. So this problem illustrates that, even for two different overall reversible paths from an initial state to a final state, in which (in this case) the overall internal energy change is zero, the amount of work done can differ between the reversible paths, and the amount of heat transfer can also differ. But, as you can readily verify, the change in entropy for the two different overall reversible paths will be the same.

5. Oct 14, 2012

### Sunfire

Chestermiller,

could you elaborate?... I am trying to understand how to minimize the work lost as heat and still end up with
-same temperature as the initial state
- higher pressure than the initial state

Thank you

6. Oct 14, 2012

### Staff: Mentor

From your original post, I did not recognize that this is what you were trying to do. I recommend that you compare (a) reversible adiabatic compression followed by cooling, with (b) reversible isothermal compression. To carry out this comparison, it will be easier to work with the ratio of the volumes than the ratio of the pressures. The ratio of the volumes for the two processes will be the same, and there will be no change in volume during the cooling phase that follows the adiabatic compression. Take the ratio of the works for the two processes, which will be the same as the ratio of the heats. All the other parameters will cancel out except for the volume ratio. Also, the volume ratio will be the inverse of the pressure ratio, since the final and initial temperatures are the same.

7. Oct 15, 2012

### Sunfire

Perhaps I am missing something...

It seems to me that
(1) Adiabatic compression followed by isohoric cooling
is different than
(2) Isothermal compression

in that (1) invests work in a gas at initial conditions (p,T) and then returns to the same point (p,T) by rejecting all invested work as heat

(2), on the other hand, starts at (p,T) but ends at (p',T) where p'>p

or am I missing something

8. Oct 15, 2012

### Sunfire

The correction should probably be:

in process (1), the final state has smaller volume, but same T as the starting condition. The conclusion then is that the final pressure is p'>p

This means, even though the rejected heat equals the amount of invested work, still portion of the invested work remains in a form of pressure in the gas at its final state

9. Oct 15, 2012

### Staff: Mentor

Yes, you are missing something. Read over the first paragraph of 256bits reply #3 carefully. Any questions?

You are going to do an analysis of the two processes I described, which have identical initial states and identical final states. You are going to be comparing work and heat involved for the two processes.

10. Oct 16, 2012

### Sunfire

Yes, you are right, the rejected heat in the case of isothermal compression is smaller, e.g. this is preferable, since it's less wasteful.

Do you think that isobaric cooling would require more heat rejection than isothermal compression?

In isobaric cooling, I would imagine a pressure rise from ambient, then cooling while pumping gas in the system to maintain pressure

11. Oct 16, 2012

### Staff: Mentor

What I think really doesn't matter. You can resolve the answers to these questions on your own by doing some modeling of the physical systems. It doesn't pay to speculate. Why don't you define some specific problems, using actual gas parameter values (initial temperature, initial pressure), and analyze the problems. You are the one who needs to get the experience, not I. When I am learning a new subject, I often make up my own problems and solve them.

Chet

12. Oct 17, 2012

### Sunfire

Thanks to all for your valuable input

-Sunfire

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