Solve Thermodynamic Signs Homework: W C → A, Q A → B, etc.

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The thermodynamic system transitions from state A to B isochorically, then expands isobarically to state C, and finally returns to state A. The user is confused about the signs for work and heat in various transitions, particularly questioning the positivity of work from C to A and the heat flow from A to B. Clarification is provided on the convention that work done by the gas is positive, while work done on the gas is negative, and emphasizes the need for reasoning behind each selection. The first law of thermodynamics is reiterated, highlighting the relationship between internal energy, heat, and work. The discussion ultimately seeks to resolve inconsistencies in the user's selections for the thermodynamic process.
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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?

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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...
 
anyone??
 
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:

\Delta Q = \Delta U + W or

\Delta U = \Delta Q - W

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
 
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