Ideal gas PV diagram heat transfer process

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SUMMARY

The discussion centers on the energy changes in an ideal gas represented on a Pressure-Volume (PV) diagram, specifically analyzing the transitions between points A, B, C, and D. The professor concluded that the energy change from point A to B is equal to the energy change from point A to D, despite different work done in each process. This is confirmed by the fact that both points B and D are at the same temperature, leading to equal internal energy changes. The key equations referenced include deltaE = Q + W and the ideal gas law PV=nRT.

PREREQUISITES
  • Understanding of ideal gas laws, specifically PV=nRT
  • Knowledge of thermodynamic processes: isothermal and adiabatic
  • Familiarity with internal energy concepts and the first law of thermodynamics
  • Ability to interpret Pressure-Volume diagrams
NEXT STEPS
  • Study the implications of the first law of thermodynamics in various thermodynamic processes
  • Learn about the derivation and application of the ideal gas law in real-world scenarios
  • Examine the differences between isothermal and adiabatic processes in depth
  • Explore the concept of internal energy and its dependence on temperature in ideal gases
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Students of thermodynamics, physics educators, and anyone seeking to deepen their understanding of energy transfer processes in ideal gases.

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Homework Statement


There is a Pressure-Volume diagram of an ideal gas. The processes make up a (rough) triangle. At constant pressure, we have point A to point C to the right of it to point D to the right of C. Then above A we have point B. C goes to B and D goes to B.. thus forming something that resembles a triangle. C to B is adiabatic while D to B is isothermal.

Homework Equations


deltaE = Q + W

The Attempt at a Solution


My question is just on one part of a larger problem: The professor in class deduced that the energy change from point A to B (vertical) is equal to the energy change from point A to D (horizontal).

Now, why is that? I understand that AD is doing more work than AC, which does more work than AB (which does no work since it doesn't have a change in volume). From just that work statement, he concluded that the deltaE(AB) = deltaE (AD).

I believe this small part will help unravel the confusion in the rest of the problem. Thanks :)
 
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mbradar2 said:

Homework Statement


There is a Pressure-Volume diagram of an ideal gas. The processes make up a (rough) triangle. At constant pressure, we have point A to point C to the right of it to point D to the right of C. Then above A we have point B. C goes to B and D goes to B.. thus forming something that resembles a triangle. C to B is adiabatic while D to B is isothermal.

Homework Equations


deltaE = Q + W

The Attempt at a Solution


My question is just on one part of a larger problem: The professor in class deduced that the energy change from point A to B (vertical) is equal to the energy change from point A to D (horizontal).

Now, why is that? I understand that AD is doing more work than AC, which does more work than AB (which does no work since it doesn't have a change in volume). From just that work statement, he concluded that the deltaE(AB) = deltaE (AD).

I believe this small part will help unravel the confusion in the rest of the problem. Thanks :)
You seem to have one triangle: ABD the sides of which are straight lines and an adiabatic curve between B and C, C being a point between A and D.

I will assume you are looking at the change in internal energy. Do they give you the values of the changes in P from A to B and the changes in volume from A to C and A to D? If not, you cannot answer the question.

The internal energy is a function of temperature. So just work out the temperatures at B and at D using PV=nRT. If they are the same, then the internal energy is the same.

You know that the internal energy at B is greater than at A or C since the temperatures are lower at A and C (we know it is lower at C since AC is adiabatic, so there is work done by the gas in going from A to C but there is no heat flow into the gas).

AM
 
We are not given any numbers of any kind. It's meant to be a conceptual question and should be able to be answered conceptually, according to the professor.
 
AB is a constant volume process where delta U=Q=Cv(T2-T1)

AD is a constant pressure process where delta U=Q-W; Q=Cv(T2-T1)+R(T2-T1) and W=R(T2-T1)

R is a constant
 
mbradar2 said:
We are not given any numbers of any kind. It's meant to be a conceptual question and should be able to be answered conceptually, according to the professor.
Sorry, I was confused by your description of the graph. It is an L shaped graph with two curves from the base to the top point (B). I missed the part about the graph from D to B being isothermal.

Since B and D are at the same temperature, there is no difference in internal energy between D and B. So going from A to B or A to D results in the same change in internal energy.

AM
 

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