Thermodynamic Signs

  • #1
yuvlevental
44
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: 396
Last edited:

Answers and Replies

  • #2
Mapes
Science Advisor
Homework Helper
Gold Member
2,593
20
Determine ∆U for B → C, then find which one of your results is inconsistent regarding the B → C process.
 
  • #3
yuvlevental
44
0
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...
 
  • #4
yuvlevental
44
0
anyone??
 
  • #5
yuvlevental
44
0
really, i need tp know the answer
 
  • #6
Andrew Mason
Science Advisor
Homework Helper
7,736
437

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
 

Suggested for: Thermodynamic Signs

  • Last Post
Replies
2
Views
340
Replies
9
Views
246
Replies
5
Views
286
  • Last Post
Replies
1
Views
340
Replies
3
Views
218
Replies
2
Views
394
Replies
10
Views
377
Replies
19
Views
631
  • Last Post
Replies
2
Views
354
Replies
19
Views
815
Top