Ideal gas PV diagram heat transfer process

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Homework Help Overview

The discussion revolves around a Pressure-Volume diagram of an ideal gas, where the processes form a triangular shape. The original poster seeks clarification on the energy changes between specific points, particularly from point A to B and from point A to D, in the context of constant pressure and adiabatic processes.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between energy changes at different points, questioning why the energy change from A to B is considered equal to that from A to D. There is discussion on the work done during various processes and the implications of constant volume and constant pressure on internal energy.

Discussion Status

Some participants have offered insights into the internal energy changes and the conditions under which they occur, while others have noted the lack of numerical values, emphasizing the conceptual nature of the question. Multiple interpretations of the diagram and processes are being explored, particularly regarding the temperature relationships at points B and D.

Contextual Notes

Participants indicate that the problem is intended to be conceptual, without specific numerical data provided. There is an acknowledgment of the need to understand the implications of the processes described in the diagram.

accountkiller
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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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