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

In summary: I DOING WRONG?In summary, the thermodynamic system changes isochorically from state A to state B, where the temperature is greater than state A. From state C the system returns to its original state A linearly. However, the work done by the gas from C to A is positive, and the work done on the gas from A to B is negative.
  • #1
yuvlevental
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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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  • #2
Determine ∆U for B → C, then find which one of your results is inconsistent regarding the B → C process.
 
  • #3
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
anyone??
 
  • #5
really, i need tp know the answer
 
  • #6
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
 

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

What do the letters and symbols mean in the thermodynamic equation?

The letters and symbols in the thermodynamic equation represent different variables and quantities related to energy and heat transfer. W represents work, which is the transfer of energy caused by a force acting on a system. Q represents heat, which is the transfer of thermal energy between systems. A and B represent two different systems, and the arrow pointing from A to B indicates the direction of energy or heat transfer.

How do you determine the sign of each term in the thermodynamic equation?

The sign of each term in the thermodynamic equation can be determined by considering the direction of energy or heat transfer. If energy or heat is being transferred from system A to system B, then the sign of the term will be positive. If the transfer is from system B to system A, then the sign will be negative. Additionally, work done by a system is considered positive, while work done on a system is considered negative.

What is the relationship between work and heat in a thermodynamic process?

The relationship between work and heat can be described by the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred between systems. In a thermodynamic process, work and heat can be exchanged between systems, but the total energy remains constant. This means that any work done by a system must be balanced by an equal amount of heat transfer.

How does the change in internal energy relate to the thermodynamic equation?

The change in internal energy of a system is related to the thermodynamic equation through the first law of thermodynamics. The internal energy of a system is the sum of its heat and work, so the equation W + Q = ΔU can be used to calculate the change in internal energy. In the thermodynamic equation, work and heat are represented as W and Q, respectively.

Can the thermodynamic equation be applied to all thermodynamic processes?

The thermodynamic equation can be applied to any process that involves the transfer of energy or heat between systems. However, it is important to note that the equation assumes certain ideal conditions, such as a closed system and no energy losses. In real-world scenarios, these conditions may not always hold, so the equation may not accurately represent the process. Other thermodynamic equations may need to be used in more complex or non-ideal situations.

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