Finding work done, heat in, and heat out over a reversible thermodynamic cycle

[mindarson]

1. Statement of the Problem

One mole of a monatomic ideal gas is taken through the reversible cycle shown. Process bc is an adiabatic expansion, with P_b = 10.0 atm and V_b = 1.00E-3 m^3. Find a) the energy added to the gas as heat, b) the energy leaving the gas as heat, c) the net work done by the gas, and d) the efficiency of the cycle.

Here is a picture of the problem, with the figure (it is problem #7):

http://www.pa.msu.edu/courses/PHY215/handouts/HW3.pdf"

Homework Equations

I honestly don't even know which equations are relevant at this point, beyond the First Law of Thermodynamics and the Ideal Gas Law. I believe I have incorporated every equation in lectures and the book and gotten nowhere.

The Attempt at a Solution

Since it is a monatomic ideal gas, I'm assuming that the ratio of specific heats (gamma) is 1.66. Then I use the fact that P*V^gamma = constant for adiabatic processes to solve for P_c (since I already know the values of P_b, V_b, and V_c). From here I can calculate P, V, and T for each of the 3 states of the problem. I'm not sure I need all this information, but I have it in case I do need it.

To find the net work done by the gas over the cycle, I used the following equation for net work done over an adiabatic process:

http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/imgheat/adiab.gif

By the First Law of Thermodynamics this would also give me the opposite of the internal energy change on the path bc.

Beyond this, I have no idea. My trouble is that I don't know when/how heat is entering the system and when/how it is leaving the system, except that I know that no heat is exchanged during the adiabatic portion of the cycle (path bc).

This problem seems straightforward but there is a key piece of logic that I'm missing! Can somebody help me out?

 

[vela]

You're doing fine. The key is to identify the type of process for each leg. You should be able to find the necessary formulas for each type of process.

From a to b, the gas undergoes an isochoric process, that is, the volume doesn't change. So how much work is done? What's the change in internal energy?

From c to a, you have a constant pressure process. Again, what's the work done and what's the change in internal energy?

 

[mindarson]

Thanks for the encouragement!

Here is the approach I'm taking now. I'm forgetting about the total work done for now and concentrating on the heat (delta Q) exchanged over each of the 3 paths. For the adiabatic path, delta Q = 0. For the isochoric path, delta Q = delta U. For the isobaric path, delta Q = (3/2)R(delta T) + P(delta V).

For one cycle, delta U = 0, and so the work done by the gas over the whole cycle is delta W = delta Q. And delta Q for the whole cycle is just the sum of the heat exchanged over the isochoric and isobaric paths.

My only remaining confusion is which path - isochoric or isobaric - represents input of heat and which represents outflow. I'm thinking that heat flows in over the isochoric path, since according to the ideal gas law temperature must rise with pressure, and I already know that internal energy rises with temperature, and for constant volume heat must then be positive. Similar reasoning leads me to believe that heat flows out over the isobaric path.

Am I on the right track here?

 

[vela]

Yup!