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

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Homework Help Overview

The discussion revolves around a thermodynamic system transitioning through various states (A, B, C) with specific processes: isochoric from A to B, isobaric from B to C, and a return to A. Participants are tasked with determining the signs of work and heat for these transitions based on the first law of thermodynamics.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants attempt to assign signs (Zero, Positive, Negative) to work and heat for each transition based on their understanding of thermodynamic principles.
  • Questions arise regarding the reasoning behind the signs assigned, particularly for the work done and heat flow in each process.
  • Some participants express confusion about the consistency of their results, especially regarding the transition from B to C.

Discussion Status

The discussion is ongoing, with participants seeking clarification on their reasoning and the application of thermodynamic conventions. Some guidance has been offered regarding the correct interpretation of work and heat signs, but no consensus has been reached on the specific assignments.

Contextual Notes

Participants are working under the constraints of homework rules that require them to analyze the problem without providing complete solutions. There is a focus on understanding the implications of the first law of thermodynamics in the context of the given processes.

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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?

(GRAPH ATTATCHED)
 

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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:

[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
 

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