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

In summary: You are on the right track. In summary, the problem involves a reversible cycle of a monatomic ideal gas with specific heat ratio of 1.66. The gas undergoes an isochoric process from a to b, followed by an adiabatic process from b to c, and finally a constant pressure process from c to a. The net work done by the gas over the cycle is equal to the heat exchanged over the isochoric and isobaric paths. Heat flows into the system over the isochoric path and out of the system over the isobaric path. The specific formulas for each type of process can be used to calculate the energy added to and leaving the gas, as well as the net work done and
  • #1
mindarson
64
0
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" [Broken]

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?
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
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?
 
  • #3
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?
 
  • #4
Yup!
 
  • #5

Thank you for providing the problem statement and your attempt at a solution. It seems like you have a good understanding of the relevant equations and concepts, but you are struggling with identifying when and how heat is entering and leaving the system.

In this problem, the key to determining the heat in and heat out is to remember that the process bc is adiabatic, meaning no heat is exchanged during this portion of the cycle. This means that the heat in and heat out only occur during the isothermal processes (ab and cd).

To find the heat in during process ab, you can use the following equation:

Q_in = nRTln(V_b/V_a)

where n is the number of moles, R is the gas constant, and T is the temperature at state b (which can be calculated using the ideal gas law).

Similarly, to find the heat out during process cd, you can use the following equation:

Q_out = nRTln(V_d/V_c)

where T is the temperature at state d.

Once you have calculated the heat in and heat out, you can use the First Law of Thermodynamics to find the net work done by the gas over the cycle. The efficiency of the cycle can then be calculated as the ratio of the net work done to the heat in.

I hope this helps clarify the problem for you. If you have any further questions, please don't hesitate to ask. Good luck with your solution!
 

1. What is work done in a reversible thermodynamic cycle?

In a reversible thermodynamic cycle, work is done when a system undergoes a change in its internal energy. This can occur through mechanical work, such as a piston moving in a cylinder, or through electrical work, such as a current flowing through a wire. The amount of work done can be calculated by the product of the force applied and the distance over which it is applied.

2. How is heat involved in a reversible thermodynamic cycle?

Heat is a form of energy that can be transferred between a system and its surroundings. In a reversible thermodynamic cycle, heat can be either added to the system (heat in) or removed from the system (heat out). This heat transfer occurs through a temperature difference between the system and its surroundings, and can be calculated using the specific heat capacity of the substances involved.

3. What is the significance of reversible thermodynamic cycles?

Reversible thermodynamic cycles are important in understanding the fundamental principles of thermodynamics and energy transfer. They allow scientists to analyze and predict the behavior of systems and processes, as well as optimize energy usage and efficiency. Additionally, many real-world systems are designed to operate on reversible cycles, making them an essential concept in engineering and technology.

4. How do you calculate the net work and heat transfer in a reversible thermodynamic cycle?

The net work done in a reversible thermodynamic cycle is equal to the difference between the heat added and the heat removed from the system. This can be calculated using the first law of thermodynamics, which states that the change in internal energy of a system is equal to the net heat transfer into the system minus the net work done by the system.

5. What factors affect the efficiency of a reversible thermodynamic cycle?

The efficiency of a reversible thermodynamic cycle depends on several factors, including the temperatures at which heat is added and removed, the type of heat transfer, and any inefficiencies or losses in the system. Generally, a higher temperature difference between the hot and cold reservoirs results in a higher efficiency, while any energy losses or irreversibilities decrease the efficiency.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
847
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
823
  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
649
  • Introductory Physics Homework Help
Replies
1
Views
841
  • Introductory Physics Homework Help
Replies
1
Views
893
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Back
Top