# Homework Help: Change in entropy in a polytropic process

1. Nov 23, 2009

### ricof

1. The problem statement, all variables and given/known data

A piston cylinder device contains 1.2kg of nitrogen at 120kPa and 300K. Gas is compressed slowly in a polytropic process during which PV^1.3 = constant. The proces ends when the volume is reduced by one half. What is the entropy change?

2. Relevant equations

(P2/P1) = (V1/V2)^n

3. The attempt at a solution

Assuming V1 = 1 and therefore V2 = 0.5, I have worked out P2 as 295.5kPa but I am stuck on what to do next.

2. Nov 23, 2009

### ricof

I should add that I have absolutely no idea where or how the 1.2kg of nitrogen comes in to the equation, nor if V1 is in fact 1.

3. Nov 24, 2009

### ricof

anyone?

4. Nov 24, 2009

### Mapes

How many moles is 1.2kg of nitrogen?

5. Nov 24, 2009

### Redbelly98

Staff Emeritus
You can get V1 from the ideal gas equation. The 1.2 kg will come in handy there.

For entropy change, a useful relation is dQ/T = dS.

6. Nov 24, 2009

### ricof

Oh right ok so

120kPa x V = 1.2 x R x 300K

As the gas is Nitrogen do I still use R=8.314?

7. Nov 24, 2009

### Redbelly98

Staff Emeritus
This looks wrong. Where does this equation come from, and what does the 1.2 represent here? Include units with all quantities, and ask yourself if the units are what the should be.

No, use R = 8.314 Pa*m^3 / (mole*K). "8.314" is not the same thing.

8. Nov 24, 2009

### ricof

In your previous post you mentioned the ideal gas law which is PV=nRT no?

n is the number of moles so I think you use the mass of the nitrogen (1.2kg) in there somewhere?

9. Nov 24, 2009

### Redbelly98

Staff Emeritus
Yes, you use PV = nRT.

n is the number of moles of nitrogen, not the mass of nitrogen. You have to figure out how many moles there are. See Mapes's post #4

10. Nov 24, 2009

### ricof

The atomic weight of nitrogen is 7 but as it is diatomic, the molecular weight is 7x2 = 14.

So 1.2kgs represents 1200/14= 85.7 moles = n?

11. Nov 24, 2009

### Redbelly98

Staff Emeritus
That's the right idea, but you should look up the atomic weight of nitrogen.

12. Nov 25, 2009

### ricof

Thanks, have now got the answer