Reading a P-V Diagram for heat in/out and work in/out

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SUMMARY

This discussion focuses on interpreting P-V diagrams to determine heat transfer and work done by or on a gas. Key conclusions include that during adiabatic processes, there is no heat exchange, while isovolumic processes involve heat transfer without work. Isobaric processes result in both heat transfer and work, and isothermal processes involve heat exchange equal to work done. The discussion emphasizes the importance of understanding these relationships for accurate thermodynamic analysis.

PREREQUISITES
  • Understanding of thermodynamic processes: adiabatic, isovolumic, isobaric, and isothermal.
  • Familiarity with the first law of thermodynamics: Q = ΔU + W.
  • Ability to interpret P-V diagrams and their significance in thermodynamics.
  • Knowledge of heat transfer concepts: positive and negative Q values.
NEXT STEPS
  • Study the first law of thermodynamics in detail, focusing on the relationships between heat, work, and internal energy.
  • Learn how to calculate work done during different thermodynamic processes using P-V diagrams.
  • Explore real-world applications of P-V diagrams in engines and refrigeration cycles.
  • Investigate the implications of heat transfer in various thermodynamic systems and processes.
USEFUL FOR

This discussion is beneficial for students studying thermodynamics, engineers working with heat engines, and anyone interested in understanding the principles of energy transfer in gases.

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Homework Statement


Predict whether heat is lost or gained by a gas, and whether work is done by the gas or the gas does work on its surroundings, given a P-V diagram.


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The Attempt at a Solution



I'm trying to understand how to read a P-V diagram and am wondering whether there are any shortcut ways of determining whether heat has been added to the gas (which I think is positive Q) or left the gas (which I think is negative Q), and whether the gas has done work (positive) or had work done to it (negative). Is it ok to say the following?

Adiabatic steps moving from left to right and downward on the diagram: 0 heat change and gas does (positive) work.

Adiabatic steps moving from right to left and upward on the diagram: 0 heat change and environment does work on gas (negative work).

Isovolumic step moving from bottom to top on the diagram: heat added to the gas (positive Q) and 0 W

Isovolumic step moving from top to bottom on the diagram: heat taken from the gas (negative Q) and 0 W

Isobaric step moving from left to right on the diagram: heat added to the gas (positive Q) and the gas does (positive) work.

Isobaric step moving from right to left on the diagram: heat taken from the gas (negative Q) and the environment does work on the gas (work is negative)

Isothermal step moving from left to right and downward on the diagram: heat added to the gas (negative Q) and the gas does (positive) work.

Isothermal step moving from right to left and upward on the diagram: heat taken from the gas (negative Q) and the gas does (negative) work.
 
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Pretty good. I would make the following changes (in bold):

Adiabatic steps moving from left to right and downward on the diagram: 0 heat flow and gas does (positive) work: W = ΔU

Adiabatic steps moving from right to left and upward on the diagram: 0 heat flow and environment does work on gas (negative work):W = ΔU

Isovolumic step moving from bottom to top on the diagram: heat flow into the gas (positive Q) and 0 W: Q = ΔU

Isovolumic step moving from top to bottom on the diagram: heat flow out of the gas (negative Q) and 0 W: Q = ΔU

Isobaric step moving from left to right on the diagram: heat flow into the gas (positive Q) and the gas does (positive) work.

Isobaric step moving from right to left on the diagram: heat flow out of the gas (negative Q) and the environment does work on the gas (work is negative)

Isothermal step moving from left to right and downward on the diagram: heat flow into the gas (negative Q) and the gas does (positive) work: Q = W

Isothermal step moving from right to left and upward on the diagram: heat flow out of the gas (negative Q) and the gas does (negative) work: Q = W

AM
 

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