Finding the original pressure, thermodynamics

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Discussion Overview

The discussion revolves around determining the original pressure of Freon 12 contained in a sealed glass container as it is cooled from 50°C to 20°C. The context includes thermodynamic principles and the interpretation of vapor pressures at different temperatures.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant calculates the original pressure by subtracting pressures from the Freon 12 table at 50°C and 20°C, arriving at 652 kPa, but expresses uncertainty about the method.
  • Another participant argues that at 50°C, the Freon is fully vaporized and suggests that the original pressure should be slightly above the vapor pressure at 20°C, which they state is 566 kPa.
  • A later post questions the validity of the initial calculation and suggests that the first temperature could have been any value above 20°C, indicating that the approach may be flawed.
  • Two participants inquire about the pressure of the gas at 50°C, assuming constant volume, but do not provide a definitive answer.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate method to determine the original pressure, with no consensus reached on the correct approach or final answer.

Contextual Notes

There are unresolved assumptions regarding the behavior of Freon 12 at different temperatures and the implications of vapor pressure calculations. The discussion reflects varying interpretations of thermodynamic principles.

jweie29nh
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Homework Statement


Freon 12 is contained in a sealed glass container at 50°C. As it is cooled, vapor droplets arc noted
condensing on the sidewalls at 20°C. Find the original pressure in the container.

Homework Equations

The Attempt at a Solution


Looking at the Freon 12 table I subtracted the given pressures from 50 degrees and 20 degrees to get 652 kpa. I'm pretty sure the this is an appropriate solution. More likely there is another approach to solving this problem.
 
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At ## T=50^{\circ} ## C , it is all vaporized so the pressure is less than the vapor pressure at that temperature. I'm presuming at ## T=21^{\circ} ## C there still were no droplets forming. Would not the approximate answer be the vapor pressure at ## T=20^{\circ} ## C which is ## P=566 ## kPa, or just slightly more than this? ## \\ ## As Homework Helpers we are not supposed to supply the answer, but this one involves an interpretation, and I think it has a very simple answer, but the OP still needs to determine if what I have proposed is indeed correct. ## \\ ## See also: https://edisciplinas.usp.br/pluginfile.php/283793/mod_resource/content/2/R12 Properties.pdf ## \\ ## And meanwhile, I think the solution that you have proposed is incorrect. The first temperature they chose to present could have been any temperature above ## T=20^{\circ} ## C , including ## T=21^{\circ} ## C . Taking ## P(21^{\circ} \, C)-P(20^{\circ} \, C) ## would be incorrect.
 
Last edited:
If the vapor pressure of an ideal gas is 566 kPa at 20 C, what is the pressure of the gas at 50 C (assuming the volume is constant)?
 
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Chestermiller said:
If the vapor pressure of an ideal gas is 566 kPa at 20 C, what is the pressure of the gas at 50 C (assuming the volume is constant)?
Thank you @Chestermiller :smile: . I overlooked this second part of the problem. Hopefully the OP returns to also figure out how to work it.
 

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