Help to understand fast adiabatic expansion

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SUMMARY

The discussion centers on the concept of fast adiabatic expansion in thermal physics, specifically addressing a scenario where a gas in one flask is allowed to expand into a vacuum upon opening a valve. Participants clarify that while the ideal gas law suggests a decrease in temperature with an increase in volume, this is not applicable in an adiabatic process where pressure also decreases. The first law of thermodynamics is emphasized, indicating that in an adiabatic process, the change in internal energy (dU) equals the work done (dW), leading to the conclusion that the final temperature remains the same as the initial temperature despite the absence of heat transfer (Q=0).

PREREQUISITES
  • Understanding of the ideal gas law and its implications.
  • Familiarity with the first law of thermodynamics (dU = dQ + dW).
  • Knowledge of adiabatic processes and their characteristics.
  • Concept of irreversible processes in thermodynamics.
NEXT STEPS
  • Study the implications of the first law of thermodynamics in various thermodynamic processes.
  • Explore the concept of adiabatic processes in greater detail, including real-world applications.
  • Investigate the differences between isobaric and adiabatic processes in thermodynamics.
  • Learn about the behavior of gases during rapid expansions and compressions.
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Students and professionals in physics, particularly those studying thermodynamics, as well as engineers and researchers involved in gas dynamics and energy systems.

juanedilio
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Hi,

I am going through a book on thermal physics (specifically on a section on the 2nd Law)

It talks about an extremely fast adiabatic expression (as a gas in one flask separated by a valve to another flask in vacuum and then the valve is opened).

What I fail to understand is that once the valve is opened and the gas is allowed to reach equilibrium then the final temperature will be the same as the initial temperature (even though Q=0). The ideal gas law tells me that for an adiabatic process when volume increases the temperature must decrease.

Please help me in understanding why the temperature stays the same,

Thanks
 
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juanedilio said:
Hi,

I am going through a book on thermal physics (specifically on a section on the 2nd Law)

It talks about an extremely fast adiabatic expression (as a gas in one flask separated by a valve to another flask in vacuum and then the valve is opened).

What I fail to understand is that once the valve is opened and the gas is allowed to reach equilibrium then the final temperature will be the same as the initial temperature (even though Q=0). The ideal gas law tells me that for an adiabatic process when volume increases the temperature must decrease.

Please help me in understanding why the temperature stays the same,

Thanks
You mention that the ideal gas law states that temperature is proportional to volume, which is indeed correct. However, you should also note that temperature is proportional to pressure. So to say that an increase in volume results in a decrease in pressure is not technically correct, this is only the case during an isobaric process. In your thought experiment, the volume does indeed increase, but that pressure also decreased (i.e. the process is no isobaric). An important point to note here is that a rapid adiabatic expansion is irreversible, and as such, whilst we can apply the ideal gas law to the initial and final equilibrium states, we cannot do the same for any intermediate states since these states do not have very well defined state variables.

Noting that for an ideal gas, it's temperature is proportional to it's internal energy, it is perhaps more useful to move away from the ideal gas law and instead consider the first law of thermodynamics:

[tex]dU = dQ+dW[/tex]

The problem states that the process is adiabatic and therefore the first term is zero, leaving us with:

[tex]dU = dW[/tex]

So now what you have to ask yourself, is that whether or not the gas does any work on the surroundings?
 
Thanks for the response.

I guess I got to caught up in the ideal gas law before checking if the 1st law was met.
 

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