Finding the efficiency of an ideal gas with adiabatic exponent 'γ'

[PhysicsEnthusiast123]

Homework Statement

What will be the efficiency of an ideal gas with adiabatic exponent 'γ' for a cyclic process as shown in the figure?
The answer in my book is [2ln(2) - 1](γ - 1) / γ, which I cannot reach as you will see in my attempt below.

Relevant Equations

Efficiency = work done /heat absorbed.
ΔQ = ΔU + W (first law of thermodynamics).

Here is what I did :
work done in going from A to C,
W₁ = 2nRT_oln(2) (isothermal process)

work done in going from C to B,
W₁ = pΔV = nRΔT = -nRT_o (isobaric process)

work done in going from B to A,
W₃ = 0 (isochoric process)

so, total work done = W₁ + W₂ + W₃
= nRT_o[2ln(2)-1]heat absorbed in going from A to C,
ΔQ₁ = W₁ = 2nRT_oln(2) ( ΔU = 0)

heat absorbed in going from B to A,
ΔQ₃ = nC_vΔT = nRT_o / (γ - 1) (isochoric)in going from C to B, heat is not absorbed but released (you can calculate it to be negative), so, it will not be included in the total heat absorbed.
So, total heat absorbed = nRT_o[2ln(2) - 1 / (γ - 1)]Now, you can see that efficiency that I would have calculated using all this would not have been equal to the solution provided.

 

[Chestermiller]

Homework Statement:: What will be the efficiency of an ideal gas with adiabatic exponent 'γ' for a cyclic process as shown in the figure?
The answer in my book is [2ln(2) - 1](γ - 1) / γ, which I cannot reach as you will see in my attempt below.
Relevant Equations:: Efficiency = work done /heat absorbed.
ΔQ = ΔU + W (first law of thermodynamics).

View attachment 264163
Here is what I did :
work done in going from A to C,
W₁ = 2nRT_oln(2) (isothermal process)

work done in going from C to B,
W₁ = pΔV = nRΔT = -nRT_o (isobaric process)

work done in going from B to A,
W₃ = 0 (isochoric process)

so, total work done = W₁ + W₂ + W₃
= nRT_o[2ln(2)-1]heat absorbed in going from A to C,
ΔQ₁ = W₁ = 2nRT_oln(2) ( ΔU = 0)

heat absorbed in going from B to A,
ΔQ₃ = nC_vΔT = nRT_o / (γ - 1) (isochoric)in going from C to B, heat is not absorbed but released (you can calculate it to be negative), so, it will not be included in the total heat absorbed.
So, total heat absorbed = nRT_o[2ln(2) - 1 / (γ - 1)]Now, you can see that efficiency that I would have calculated using all this would not have been equal to the solution provided.

I’ll be back a little later to help with this. Please be patient.

 

[Chestermiller]

I don't know what your book did, but I agree with your answer. As a check, if you add the three heat amounts, they sum, as they should, to the net work. So your answer is confirmed.

 

[PhysicsEnthusiast123]

I don't know what your book did, but I agree with your answer. As a check, if you add the three heat amounts, they sum, as they should, to the net work. So your answer is confirmed.

Thanks for all the help.
PS : Apparently, instead of calculating the heat absorbed, my book calculated the heat released, and then mistakenly put it in the efficiency formula, which resulted in a wrong answer.