The net work done is not zero. Let us go through each part step by step.connorc234 said:Homework Statement
http://i.imgur.com/jmLqca9.jpg
pic of question
Homework Equations
W = p x dV, Q = dU + W etc
The Attempt at a Solution
I know what the different stages are, a-b and c-d are isobaric, d-a and b-c are isochoric
and I believe the answer to a, the net work done in one cycle, is zero as dV is zero.
but the further questions I'm lost. can anyone help or point me in the right direction?
The net work is not zero. How much work is done from a to b? from b to c? from c to d? from d to a?connorc234 said:Homework Statement
http://i.imgur.com/jmLqca9.jpg
pic of question
Homework Equations
W = p x dV, Q = dU + W etc
The Attempt at a Solution
I know what the different stages are, a-b and c-d are isobaric, d-a and b-c are isochoric
and I believe the answer to a, the net work done in one cycle, is zero as dV is zero.
but the further questions I'm lost. can anyone help or point me in the right direction?
Continuing from there, you should first calculate the work done in each process in the cycle, and then add them to give you the answer to a. Let me know if you are comfortable with that.Chandra Prayaga said:The net work done is not zero. Let us go through each part step by step.
1, First, since the given values are p0, V0, we must find the values of pressure volume and temperature of each of the corner states, in terms of these values. Can you do that, and write them down?
Chandra Prayaga said:The net work done is not zero. Let us go through each part step by step.
1, First, since the given values are p0, V0, we must find the values of pressure volume and temperature of each of the corner states, in terms of these values. Can you do that, and write them down?
ok i used pv=nRT taking R to be 8.31Chandra Prayaga said:Continuing from there, you should first calculate the work done in each process in the cycle, and then add them to give you the answer to a. Let me know if you are comfortable with that.
b to c and d to a both zero work since isochoric afaikChestermiller said:The net work is not zero. How much work is done from a to b? from b to c? from c to d? from d to a?
Chestermiller said:This is all correct so far. In your equations for the temperatures, leave the R in (for now), instead of substituting the 8.31. Now, from your results for the temperatures, what is the change in internal energy ΔU from a to b? from b to c? from c to d? from d to a?
Chet
Basic thermodynamics is the study of energy and its transformations. It is important because it helps us understand how energy is used and conserved in various systems, from engines to power plants to the human body.
A reversible engine process is a theoretical concept in thermodynamics where the engine operates in a forward and reverse direction, with no energy loss, and returns to its original state. It is used as a benchmark for comparing the efficiency of real engines.
In a reversible engine process, the system goes through a series of changes and then returns to its original state, while in an irreversible process, the system is unable to return to its initial state. This is due to energy loss and friction in the irreversible process, making it less efficient.
No, it is not possible for all engines to operate on a reversible process. This is because real-world engines are subject to energy loss and friction, making them unable to return to their original state without external intervention.
The efficiency of a reversible engine process is calculated by dividing the work output of the engine by the heat input. This theoretical maximum efficiency is known as the Carnot efficiency and is used as a benchmark for comparing the efficiency of real engines.