# Thermodynamics - Heat Engine Cycle

1. Feb 3, 2014

### azukibean

1. The problem statement, all variables and given/known data
n = 5 moles, monatomic, ideal gas
p1 = 7280 Pa; v1 = 2 m3
p2 = 7280; v2 = 6
p3 = 6500; v3 = 6

What is the change in heat, from 2 to 3, and from 3 to 1?

2. Relevant equations
PV = nRT
U = (3/2)nRT
Q = du + dw

3. The attempt at a solution

PV = nRT
(7280)(2) = (5)(8.32)T1

T1 = 350 (correct)
T2= 1050 (correct)
T3 = 937.5 (correct)

U = (3/2)nRT
U1 = (3/2)(5)(8.32)(350) = 21840
U2 = 65520
U3 = 57796.88

W1 = 29120
W2 = 0
W3 = -27560
(all correct)

Q = du + dw
ΔQ1 to 2 = (65520-21840) + 29120 = 72800 (correct)
ΔQ2 to 3 = (57796.88-65520) + 0 = -7723.12 (correct: -7020)
ΔQ3 to 1 (21840-57796.88) - 27560 = -63516.88 (correct: -64220)

This is driving me nuts. I can get the first heat quantity right, but not the next two. I've triple checked my work (though that doesn't totally eliminate the possibility of an arithmetic mistake). Where am I going wrong?

Last edited: Feb 3, 2014
2. Feb 3, 2014

### BvU

Is Q = du + dw or is dQ = du + dw ?

How come I get T3 = 312.5 from pV = nRT ? Any chance p3 = 19500 ?
Any (other) typos or omissions (adiabatic, isothermal?) in the problem statement ?

You say W1, do you mean W$_{1\rightarrow2}$ ? There I get 29120 for the work pΔV the gas is doing.
If so, why is W$_{2\rightarrow3}$ = 0 ?
If so, why is W$_{3\rightarrow1}$ = -27560 ? The volume does not change ?!

Last edited: Feb 3, 2014
3. Feb 3, 2014

### azukibean

The first one is correct. The heat added or subtracted is equal to the change in internal energy and work.
I'm unsure how to work the problem to find dQ; in ΔQ1→2 I equated it to Q1→2 because I assumed at "step 0" there was no heat energy.

There are no omissions. V3 should actual be 6 instead of 2. I fixed it. Sorry about that.

Right.
W$_{2\rightarrow3}$ = 0 because the volume does not change.
W$_{3\rightarrow1}$ = -27560 because I made a mistake in the original post; V3 should be 6 instead of 2.

4. Feb 3, 2014

### SteamKing

Staff Emeritus
What happened to the units in your calculations?

5. Feb 3, 2014

### azukibean

I added the units to the original problem. There was no unit conversion involved, so I didn't want to clutter my work.

6. Feb 3, 2014

### Andrew Mason

Your calculation of U3 is a bit low. I get 58500 J.

You can use ΔU = 3nRΔT/2

ΔU from 2-3 is (3)(5)(8.32)(112.5)/(2) = 7020.

AM

7. Feb 3, 2014

### azukibean

Ah, that's it. Thank you so much.