Ideal gas PV diagram heat transfer process

In summary: The professor said that the change in internal energy from point A to B is equal to the change in internal energy from point A to D.Since the temperatures are the same at B and D, the energy changes are also the same.
  • #1
accountkiller
121
0

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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  • #2
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
 
  • #3
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.
 
  • #4
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
 
  • #5
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
 

Related to Ideal gas PV diagram heat transfer process

1. What is an ideal gas PV diagram?

An ideal gas PV diagram is a graphical representation of the relationship between pressure (P) and volume (V) of an ideal gas at constant temperature, assuming that the gas follows the ideal gas law. It is a useful tool in understanding the behavior of gases under different conditions.

2. How is heat transfer represented on an ideal gas PV diagram?

Heat transfer is represented on an ideal gas PV diagram as a change in the internal energy of the gas. This can be seen as a change in the shape or position of the PV curve, as heat is either added to or removed from the system.

3. What happens to the PV diagram during an isothermal heat transfer process?

During an isothermal heat transfer process, the temperature of the gas remains constant. This means that the PV curve on the diagram will be a horizontal line, as the pressure and volume remain proportional to each other.

4. How does the PV diagram change during an adiabatic heat transfer process?

During an adiabatic heat transfer process, there is no heat exchange between the gas and its surroundings. This means that the PV curve on the diagram will be steeper, as the change in pressure and volume will be greater for the same change in temperature compared to an isothermal process.

5. What is the significance of the area under the PV curve on an ideal gas PV diagram?

The area under the PV curve on an ideal gas PV diagram represents the work done by the gas during a heat transfer process. This can be calculated by finding the area under the curve using integration, where the force is represented by the pressure and the displacement is represented by the change in volume.

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