Adiabatic Process in a heat engine

1. Jul 3, 2009

talaroue

1. The problem statement, all variables and given/known data
A heat engine uses a diatomic gas that follows the pV cycle shown in Figure.
Part 2→3 is adiabat, part 3→1 is isotherm, V=1040 cm3, P=100 kPa, T1=212 K.

Determine the pressure at point 2.

Determine the volume at point 2.

Determine the temperature at point 2.

Find Ws for process 1→2.

Find Q for process 1→2.

Tries 0/9
Find ∆E for process 1→2.

Tries 0/9
Find Ws for process 2→3.

Tries 0/9
Find Q for process 2→3.
0 J

Find ∆E for process 2→3.

Find Ws for process 3→1.

Find Q for process 3→1.

Tries 0/9
Find ∆E for process 3→1.

Tries 0/9
What is the thermal efficiency of this heat engine? (in percent)
2. Relevant equations

3. The attempt at a solution

I know how to do the rest if I can find the temperature at point 2. I tried 2 different ways.

1. I tried using the Ideal Gas law PV=nRT and solved for mols at pt 1. Then tried using that for point 2 and then used PV=nRT but that didn't work.

2. Then I tried just simply using PV/T=PV/T and came up with the same answer and they are both wrong.

What am I doing wrong?

2. Jul 3, 2009

talaroue

this would help out a lot!

3. Jul 3, 2009

talaroue

WHat is the equation i use to find the Work from point 2 to point 3?

4. Jul 3, 2009

Redbelly98

Staff Emeritus
It looks like there is a typo here. We know pv=constant for an isotherm, yet we have pv=2PV at 1, and pv=PV at 3. I suspect it should be p=0.5P at 3, judging from the scale of the p-axis.

For the adiabat, a useful equation is
pvγ = constant​

(Look up γ[/SUP] for an ideal diatomic gas in your textbook, if you're not sure what it is.)

5. Jul 6, 2009

talaroue

I understand what that symbol means for monatomic its 5/3 and for diatomic its 7/5. But for the W for 2-->3 I tried using the following....

W=nCv(delta T)
n-mols Cv-constant volume Delta T- Change in temp.

W=(.118 mol)(5/2)(8.31)(369.1-212)=138

6. Jul 6, 2009

Redbelly98

Staff Emeritus
I get a different number of moles than you did, calculated using p, v, and T at point 1. Perhaps you should reproduce that calculation.

I agree with the 369 K temperature at point 3.