Thermodynamics - Heat Engine Cycle

[azukibean]

Homework Statement

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?

Homework Equations

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

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
ΔQ_{1 to 2} = (65520-21840) + 29120 = 72800 (correct)
ΔQ_{2 to 3} = (57796.88-65520) + 0 = -7723.12 (correct: -7020)
ΔQ_{3 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?

 

[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 ?!

 

[azukibean]

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

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 ΔQ_1→2 I equated it to Q_1→2 because I assumed at "step 0" there was no heat energy.

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 ?

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

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 ?!

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.

 

[SteamKing]

What happened to the units in your calculations?

 

[azukibean]

What happened to the units in your calculations?

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

 

[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

 

[azukibean]

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

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