Basic thermodynamics, reversible engine process

AI Thread Summary
The discussion revolves around a thermodynamics homework problem involving a reversible engine process. The initial attempt at a solution incorrectly stated that the net work done in one cycle is zero due to dV being zero. However, participants clarified that the net work is not zero and guided the poster to calculate the work done in each process of the cycle. After determining the pressures, volumes, and temperatures at various states, the poster successfully calculated the work done and changes in internal energy. Ultimately, the poster expressed gratitude for the assistance received, indicating they were able to complete the problem confidently.
connorc234
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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?
 
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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 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?
 
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?
 
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?
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:
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.
ok i used pv=nRT taking R to be 8.31

p and V are known at all times so I just found the temperatures for each point.
a - pV/8.31
b - 3pV/8.31
c - pV/8.31
d - pV/24.93

then for the work since 2 are isochoric they are zero... the others
a-b W=2pV
c-d W= -2pv/3

adding them gives 4pV/3

is any of that right? haha
 
Chestermiller 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?
b to c and d to a both zero work since isochoric afaik

a to b W=2pV
c to d W= -2pV/3
 
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
 
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

I appreciate your help, but I was able to complete the question by myself after you guys got me rolling. I'm fairly confident I got the right answers.

Thanks again, Connor
 
Excellent. Congratulations!
 
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