Thermodynamic cycle with isochoric, isobaric and linear processes

  • Thread starter Thread starter yuvlevental
  • Start date Start date
  • Tags Tags
    Thermodynamic
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
5 replies · 5K views
yuvlevental
Messages
44
Reaction score
0

Homework Statement


A thermodynamic system changes isochorically from state A to state B where the temperature is greater than state A. The system then expands isobarically to state C where the temperature is greater than state B. From state C the system returns to its original state A linearly. Select Zero, Positive, or Negative for all selections.
W C → A
Q A → B
W B → C
∆U C → A
Q B → C
∆U A → B

Homework Equations


∆U = Q + W
W = WORK ON GAS
Q = HEAT ADDED TO GAS
W = -P∆V

The Attempt at a Solution


W C → A +
Q A → B +
W B → C -
∆U C → A 0
Q B → C 0
∆U A → B +

WHAT AM I DOING WRONG?

(GRAPH ATTATCHED)
 

Attachments

  • ISO.JPG.jpg
    ISO.JPG.jpg
    3.6 KB · Views: 508
Last edited:

Discussion

Physics news on Phys.org
Determine ∆U for B → C, then find which one of your results is inconsistent regarding the B → C process.
 
It's W = -P∆V. The volume is increasing because the temperature is increasing. So the work is negative. So I can't find anything inconsistent regarding the B → C process...
 
really, i need tp know the answer
 
yuvlevental said:

Homework Statement


A thermodynamic system changes isochorically from state A to state B where the temperature is greater than state A. The system then expands isobarically to state C where the temperature is greater than state B. From state C the system returns to its original state A linearly. Select Zero, Positive, or Negative for all selections.
W C → A
Q A → B
W B → C
∆U C → A
Q B → C
∆U A → B


Homework Equations


∆U = Q + W
W = WORK ON GAS
Q = HEAT ADDED TO GAS
W = -P∆V

The Attempt at a Solution


W C → A +
Q A → B +
W B → C -
∆U C → A 0
Q B → C 0
∆U A → B +

WHAT AM I DOING WRONG?

(GRAPH ATTATCHED)
Make sure that you are using the right convention for W. The convention is to use W done by the gas as positive and W done on the gas as negative. The first law is:

[tex]\Delta Q = \Delta U + W[/tex] or

[tex]\Delta U = \Delta Q - W[/tex]

Also, we can't tell what you are doing wrong without knowing your reasoning.

ie. What is your reason for saying that:

1. the work done by the gas from C to A is positive?

2. the heat flow into the gas from A to B is positive?

3. the work done by the gas from B to C is negative?

4. the change in internal energy of the gas from C to A is 0?

5. the heat flow into the gas from B to C is 0?

6. the change in internal energy of the gas from A to B is positive?

AM