What Is the Final Pressure for a Monatomic Gas After Adiabatic Expansion?

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Discussion Overview

The discussion revolves around determining the final pressure of a monatomic ideal gas after it undergoes an adiabatic expansion and subsequent warming in a sealed container. Participants explore the implications of the process being irreversible and the relationship between initial and final pressures and temperatures.

Discussion Character

  • Homework-related
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes a two-step process involving the removal of a stopper from a bottle containing an ideal gas, leading to adiabatic expansion against an external pressure.
  • Another participant challenges the characterization of the process as reversible, arguing that the initial and external pressures cannot be equated during the adiabatic expansion.
  • A different participant clarifies the sequence of events, stating that the initial pressure is equal to the internal pressure before the stopper is removed, and the external pressure is only relevant when the stopper is replaced.
  • There is a discussion about using the ideal gas law to find the pressure after the temperature stabilizes at 323 K, emphasizing the need to calculate the number of moles per unit volume after the expansion.

Areas of Agreement / Disagreement

Participants do not reach consensus on whether the process can be considered reversible. There are competing views on how to approach the problem, particularly regarding the relationship between initial and final pressures and the implications of the adiabatic expansion.

Contextual Notes

Participants express uncertainty about the initial temperature after the stopper is removed and the implications of the irreversible nature of the expansion on the calculations. There are also unresolved mathematical steps related to the application of the ideal gas law and the temperature changes involved.

S.L.G.
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Homework Statement



A bottle at 323 K contains an ideal gas at a pressure of 1.667×10^5 Pa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against P(ext)=1.162×10^5 Pa, and some gas is expelled from the bottle in the process. When P=P(ext), the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 323 K.

I have to find the final pressure for a monatomic gas...that is, Cv,m=3R/2.

Homework Equations



T(final) = T(initial) ((Cv,m + (RP(ext)/P(initial)))/(Cv,m + (RP(ext)/P(final)))

The Attempt at a Solution



I know that P(initial) = P(external) (making this a reversible process), but I don't know how to find T (initial) (after the stopper is removed)...

Thank you!
 
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No way is this a reversible process! You are given p(initial) = 166.7 kPa and p(external) = 116.2 kPa for the adiabatic expansion.
For the second part of the process, warming in the sealed bottle, the initial pressure is equal to the external pressure, but the external pressure can play no part in a process inside a sealed bottle!
You are going to need to rethink your total approach to this problem, I fear.
 
I know that this is in two steps.

Step 1: The bottle is initially closed; p(initial) = 166.7 kPa. Then the bottle is opened.

Step 2: p(external) = 116.2 kPa. When P=P(external), the bottle is closed again, at which point p(initial) = 116.2 kPa. I have to find p(final)...
 
S.L.G. said:

Homework Statement



A bottle at 323 K contains an ideal gas at a pressure of 1.667×10^5 Pa. The rubber stopper closing the bottle is removed. The gas expands adiabatically against P(ext)=1.162×10^5 Pa, and some gas is expelled from the bottle in the process. When P=P(ext), the stopper is quickly replaced. The gas remaining in the bottle slowly warms up to 323 K.

I have to find the final pressure for a monatomic gas...that is, Cv,m=3R/2.


Homework Equations



T(final) = T(initial) ((Cv,m + (RP(ext)/P(initial)))/(Cv,m + (RP(ext)/P(final)))

The Attempt at a Solution



I know that P(initial) = P(external) (making this a reversible process), but I don't know how to find T (initial) (after the stopper is removed)...

Thank you!
pinitial doesn't equal pexternal. pfinal equals pexternal, where pfinal refers to the pressure within the bottle immediately after the irreversible adiabatic expansion, but before the temperature has had a chance to rise to 323K. "When p = pext, the stopper is replaced." The equation you give tells you the temperature in the bottle immediately after the irreversible adiabatic expansion has taken place. Knowing the pressure in the bottle (pexternal) and the temperature in the bottle (Tfinal) at the end of the irreversible adiabatic expansion gives you the information you need to calculate the number of moles per unit volume in the bottle at the end of the expansion. You can then use the ideal gas law to calculate the pressure when the temperature rises to 323K.
 

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