Is ∆E = q always true at constant volume?

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    Thermodynamics
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The discussion revolves around the relationship between heat transfer and internal energy change in thermodynamics, specifically questioning whether the equation ∆E = q holds true at constant volume. Participants explore concepts related to the First Law of Thermodynamics and the implications of constant pressure versus constant volume conditions.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants examine the definitions of heat and internal energy, questioning how these relate under different thermodynamic conditions. There is a focus on the implications of constant volume and constant pressure on internal energy changes and the work done by the gas.

Discussion Status

Some participants have provided insights into the relationship between heat and internal energy at constant volume, while others express confusion regarding the differences in equations at constant pressure. The discussion reflects an exploration of these concepts without reaching a definitive consensus.

Contextual Notes

There is an ongoing examination of the assumptions related to the definitions of work and heat in thermodynamic processes, particularly in the context of constant volume and constant pressure scenarios. Participants reference the First Law of Thermodynamics as a foundational principle in their discussions.

tdod
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True or False:
"The Heat, q, at constant volume is equal to the change in internal energy, ∆E, for the process at constant pressure."

I know that this is true, but i can't for the life of me figure out why. Can anyone explain?

Thanks!
 
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Start from the First Law of Thermodynamics. You can change the internal energy by adding heat or/and doing work on the gas. How is work related to change of volume at constant pressure?

ehild
 
At constant volume, all the heat goes into internal energy - because it has nowhere else to go - but the pressure has to go up ... because of all that extra motion. So I'm a bit confused about what it is you are asserting is true. [edit]saw ehild's responce ... I think I get it.

The way to resolve confusions in thermodynamics is to concentrate on what each of the things are ... what is heat, and internal energy, and how does the gas give rise to pressure and temperature?
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cvpro.html
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/cppro.html
 
ehild said:
Start from the First Law of Thermodynamics. You can change the internal energy by adding heat or/and doing work on the gas. How is work related to change of volume at constant pressure?

ehild

So, I know that at constant volume ∆E = q.

However, at constant pressure ∆E = q + w. Therefore, shouldn't they be different?
 
tdod said:
... However, at constant pressure ∆E = q + w ...
Actually, that relation is true for any process, not just at constant pressure.

What do you know about w in a constant volume process?
 
At constant volume, the change in internal energy is equal to the amount of heat added.

At constant pressure, the change in enthalpy is equal to the amount of heat added.

[itex]\Delta[/itex]H= [itex]\Delta[/itex]E + [itex]\Delta(pV)[/itex]
 
At constant volume, the change in internal energy is equal to the amount of heat added.

At constant pressure, the change in enthalpy is equal to the amount of heat added.
 
tdod said:
So, I know that at constant volume ∆E = q.

However, at constant pressure ∆E = q + w. Therefore, shouldn't they be different?

In general, ∆E = q + w, the internal energy changes with heat transfer and with work. But the work done by the gas is w=-pΔV for a very small change when the pressure can be considered constant. In general, w=-∫pdV where the integration goes between the initial and final volume. At constant volume, the initial and final volumes are the same, ΔV=0, so w=0. The change of internal energy at constant volume is equal to the heat added: ∆E = q.

ehild
 
Last edited:

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