Mastering Physics: Heat Engine Problem

[ricebowl07]

The figure shows the cycle for a heat engine that uses a gas having gamma =1.25. The initial temperature is T1=300K, and this engine operates at 20 cycles per second.

a.) What is the power output of the engine?

b.) What is the engine's thermal efficiency?

The link to this image is here:
http://session.masteringphysics.com/problemAsset/1074111/4/knight_Figure_19_54.jpg

This is how I attempted to solve this problem:

Given: T1 = T2 = 300K
P2 = P3
P1 = 1 atm = 1.013 * 10^(5) Pa
gamma = 1.25
V1 = V3 = 600 cm^(3)
V2 = 200 cm^(3)
R = 8.31 J/mol*K or 0.08206 atm *L/mol*K

Relevant equations:

PV = nRT

η = Wout/QH

Power = Wnet/time

Q_12 = W_12

Q_23 = nc_pΔT

Q_31 = nc_vΔTFinding number of moles

n = PV/RT

n = [1.013 * 10^(5) Pa(0.0006 m^(3))]/[(8.31J/(mol*K))(300K)]= 0.024 mol

Finding P2 and P3

P1V1 = P2V2

P2 = [P1V1]/V2 = [1atm(0.6L)]/0.2L

P2 = 3atm

P2 = P3

Finding T3

P3V3 = nRT3

T3 = P3V3/nR = 900K

Finding W_12

W_12 = nRT(ln(vf/vi)) = 0.024mol(8.31J/mol*K) (300K)(ln(0.0006m^(3)/0.0002m^(3))) = 65.73 J

Finding W_23

W_23 = P(vf-vi) = 3atm[(0.0006 - 0.0002)m^(3)] * 101325J = 121.59 J

W_31 = 0 because it is an isochoric process

Wnet = 187.32J

Power

P = Wnet/time = 187.32J(20(s^(-1))) = 3746.4 W = 3.746 kW

I got it wrong. Why was gamma given? I thought it was only used for adiabatic processes. I also don't know which c_p and c_v to use since the type of gas is not given.

 

[TSny]

Hello ricebowl07 and welcome to PF!

Did you consider the sign of the work for the isothermal part? (In other words, what is Vi and what is Vf for that part?)

Also, you might want to keep one more significant figure in your calculation of the number of moles.

[You should be able to find Cv amd Cp from gamma.]

 

[ricebowl07]

Thank you! I finally got the right answer.

 

[cluke95]

Hello, I have worked my way through this problem, except with a different value (16 cycles per second vs 20) and I am having trouble finding my thermal efficiency. You said find if from gamma=1.25, and I know that gamma=C_p/C_v, but I'm not sure where to go from there.

 

[TSny]

Hello, cluke95. Welcome to PF!

There is a simple relation between C_P and C_V for any ideal gas. It has the form: C_P = C_V + _ ? _ .

You can combine this equation with the equation γ = C_P/C_V to determine the values of C_P and C_V. Then you can calculate the heat Q_H to determine the efficiency.

 

[rude man]

My take:
You don't need either C_p or C_V.
(a): what is net work on a p-V diagram?
(b): use the 1st law and the ideal gas law.
Hint: what is ΔU for a complete cycle?