Adiabatic Expansion: Constant pressure and temperature

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Discussion Overview

The discussion revolves around the concept of adiabatic expansion in thermodynamics, specifically addressing the relationships between internal energy, work, and temperature in the context of ideal gases. Participants explore the implications of constant temperature and volume changes on internal energy and work done during adiabatic processes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant notes that for adiabatic expansion, the equation ΔU = q + w applies, with q = 0, leading to confusion about why ΔU = 0 when temperature is constant and volume changes.
  • Another participant introduces a relation between temperature and volume for adiabatic processes, suggesting that if temperature is constant, then volume must also remain constant, leading to zero work done.
  • A different viewpoint emphasizes that during adiabatic expansion, even though q = 0, temperature changes due to work done on the surroundings, indicating a change in internal energy.
  • One participant questions the assertion that work is zero, seeking clarification on the conditions under which work would be considered zero.
  • Participants discuss the implications of the equation ΔU = CvΔT, noting that if volume is constant, then dV = 0, which raises questions about the conditions under which ΔU can be zero.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between work, internal energy, and temperature during adiabatic expansion. There is no consensus on whether work is zero or how it relates to changes in internal energy under the given conditions.

Contextual Notes

Participants highlight the dependence of internal energy on temperature and the conditions under which work is calculated, indicating potential limitations in understanding the relationships involved in adiabatic processes.

Hpatps1
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I'm new to the forum, so please be kind.

I was reading through my pchem textbook, and I noticed something. We're given the equation:

ΔU = q + w

For an adiabatic expansion, we're told that q = 0. Fair enough, no heat transfer. But when there is a constant T and change in V, my book says:

ΔU = 0 (U is constant)

I don't understand. W = -∫pdV, doesn't it? Why is work zero?
The book also says that for a perfect gas, U isn't dependent on volume. When they use this fact, it makes sense why ΔU = 0, but it seems like a contradiction when using the definition of work.

Also, the book continues and says:

ΔU = CvΔT when V is constant.

Again, W = -∫pdV, right? So dV = 0, shouldn't ΔU = 0?

Thanks in advance!
 
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Hi Hpatps1,
For the adiabatic expansion we have the relation between T and V as T1V11-\gamma=T2V21-\gamma.
Where T1,V1 and T2,V2 are data of the system at two different states. If T is constant then T1=T2. Then by the above equation V1=V2. Hence dV is zero, and the W. That's why ΔU is zero.
Now for the next. By the above data it is clear that dV is zero iff T is constant (ΔT=0). So in the equation ΔU=CvΔT, ΔU certainly becomes zero when ΔT=0.
Regards.
 
Thanks! That makes sense.
 
As a gas expands or contracts adiabatically, although q=0, the temperature changes as a result of the particles doing work on the surroundings, so there is a change in internal energy equal to the work done by or to the system. U is only dependent on the kinetic energies of the molecules of the gas in an ideal system.
 
Hpatps1 said:
I'm new to the forum, so please be kind.

I was reading through my pchem textbook, and I noticed something. We're given the equation:

ΔU = q + w

For an adiabatic expansion, we're told that q = 0. Fair enough, no heat transfer. But when there is a constant T and change in V, my book says:

ΔU = 0 (U is constant)

I don't understand. W = -∫pdV, doesn't it? Why is work zero?
I am not sure why you say W = 0. Why do you think W = 0?

The book also says that for a perfect gas, U isn't dependent on volume. When they use this fact, it makes sense why ΔU = 0, but it seems like a contradiction when using the definition of work.

Also, the book continues and says:

ΔU = CvΔT when V is constant.

Again, W = -∫pdV, right? So dV = 0, shouldn't ΔU = 0?
Why would ΔU = 0?

AM
 

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