Heat engine: moles/temp unknown

[joe_cool2]

Homework Statement

A heat engine uses the closed cycle shown in the diagram below. (Figure 1) The working substance is n moles of monatomic ideal gas. Find the efficiency of such a cycle. Use the values for pressure and volume shown in the diagram, and assume that the process between points 1 and 3 is isothermal.

Homework Equations

pV=nRT
efficiency = W_out/Q_H

The Attempt at a Solution

We know that T3 and T1 are on an isotherm, and so are both exactly 3 times T2. So, the absolute value of ΔT is 2 times T2. The reason I refer to the absolute value here is because the sign of ΔT for different processes will be different.

W_out = nRT*ln(V1/V3) - pΔV

So, W_out = nRT*ln(V1/V3) - nRΔT

Q_H= nRT*ln(V1/V3) + nC_vΔT

So the efficiency is:

3nRT₂*ln(V1/V3) + 2nRT₂ / 3nRT₂*ln(V1/V3) + 2nC_VT₂

Then the idea is to cancel out all the n's and T₂'s. All the rest are knowns. I get a value of 84.1%. There's definitely something very wrong about my approach.

 

[mfb]

Did you consider the heat involved in 1->2?

I don't understand the last, long expression and I think there are brackets missing.

 

[joe_cool2]

The heat involved in 1 -> 2 is actually leaving the system. So, Q_H is just the heat we put in during the isothermal and isochoric legs. So, the heat associated with the isobaric process, nC_PΔT, does not enter into the expression for Q_H.

Yes, there were brackets missing. Let me clarify as to where this expression comes from.

[3nRT₂*ln(V1/V3) + 2nRT₂] / [3nRT₂*ln(V1/V3) + 2nC_VT₂]

We know that T₃ is three times greater than T₂. This should help to explain the first term of the numerator and the first term of the denominator. The former is the heat intake during the isothermal process, and the latter is the work done during the isothermal process. These expressions are the same because in an isothermal process, W = Q_H.

The expression begins as nRT*ln(V1/V3). Now T can refer to the temeperature at point 1 or point 3, because we are dealing with an isothermal process. And we said in the last paragraph that 3T₂ = T₃. So this is where the 3T₂ comes from in both expressions.

The second terms in the numerator and the denominator contain the factor 2T₂. That's because ΔT = T₃ - T₂, so ΔT = 3T₂ - T₂.

Note the sign change on the second term in the numerator. That occurs because we lose temperature moving from 1 -> 2, so ΔT is negative. ΔT is not negative in the other place it occurs, moving from 2 -> 3.

EDIT: Let's not get trapped in the paradigm of my thinking. It hasn't gotten me anywhere with this problem.