# Thermodynamics N2 gas Question

1. Dec 7, 2004

### Lakers08

Thermodynamics Question (What did I do wrong?)

Problem 19.46
Two moles of N2 gas undergo the cycle abcd. The pressure of the gas in each state is: pa = pd = 5700 Pa ;pb = pc 1700 . The volume of the gas in each state is: Va = Vb = 2.40 ;Vc= Vd 8.50 . The gas may be treated as ideal.

Find the total work done on (or by) the gas in the complete cycle?

from a to b volume is constant so therefore W = 0
from b to c pressure is constant , so W = p(Vc-Vb) = 10370 J
from c to d volume is constant so therefore W = 0
from d to a pressure is constant, so W = p(Va-Vd) = -34770 J

total W= 10370 J - 34770 J = -24400 J

somehow this awnser is wrong, someone please help me out, what am I doing wrong?

Last edited: Dec 7, 2004
2. Dec 7, 2004

### Andrew Mason

This is a kind of refrigerator cycle because of the direction abcda (expansion at low pressure, compression at high pressure). So work is done on the system.

The question asks for the total work done on the system in that cycle, so I am thinking that the answer they are looking for is just the work done on the compression cycle. That would be 5700 x 6.1 = 34770 J.

AM

3. Dec 8, 2004

### franznietzsche

Do you know what the right answer is?

I can't figure how that would be wrong, i keep getting that.

4. Dec 8, 2004

### Andrew Mason

Yes. The correct answer will help us to understand what the question is really asking.

The work done by the gas in expanding from b to c is due to absorption of heat from the cold reservoir. This does not lessen the amount of work that is required to be done on the system by the compressor from d to a. The work done by the gas from b to c relates to the efficiency, (ie. the amount of cooling effect the work done to the system will produce). So 34770 J. of work supplied from an external source has to be done on the gas in each cycle.

AM

5. Dec 8, 2004

### Lakers08

no, i actually dont know what the correct awnser is, but today after asking my professor he said that the correct awnser will be to graph those values on a pV graph and to get the area under the curve(or the area of the square it forms with those values) can anybody explain to me what is wrong with what i did, it makes a lot of sense to me.

6. Dec 8, 2004

### Andrew Mason

The PV diagram is a rectangle and you found the area : 24400 J. which represents 34770 J supplied by an external source on the gas less 10370 J done by the gas. So I don't see the problem. Ask your prof what his answer is.

AM

7. Dec 8, 2004

### franznietzsche

If you graph the cycle, and integrate, you get your answer, or the same answer with opposite sign (if you go in the other order)