Gas Laws -- why calculate the mean square speed at 273K?

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Discussion Overview

The discussion revolves around the application of the formula ##p = \frac{1}{3}\rho## for calculating the mean square speed of gas molecules at a specific temperature of 273K. Participants explore the reasons for using this temperature and the implications of the ideal gas assumptions in the context of a homework problem.

Discussion Character

  • Homework-related
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant questions why the formula is specifically applied at 273K and not at other temperatures like 546K.
  • Another participant emphasizes that the formula can be used at any temperature, seeking more context about the problem.
  • A participant provides specific values for density and pressure at 273K, suggesting that the calculations for root mean square (rms) speed should be based on these values.
  • Concerns are raised about the applicability of the formula at different temperatures, with one participant noting that the pressure values indicate the temperature for which the rms speed should be calculated.
  • There is a request for a more complete problem statement to clarify the confusion surrounding the application of the formula.
  • Another participant confirms that no information is missing from the problem statement and explains how the ideal gas assumption allows for the use of the ideal gas law in conjunction with the provided equation.
  • One participant mentions the derivation of the formula from other equations related to gas behavior.
  • A later reply indicates that the initial confusion has been resolved, thanking others for their help.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the formula at various temperatures, with some asserting it can be used universally while others believe it is limited to specific conditions. The discussion remains unresolved regarding the implications of using the formula at temperatures other than 273K.

Contextual Notes

Some participants highlight the importance of understanding the context of the problem, including the assumptions made about the gas being ideal and the specific conditions under which the formula is applied. There are references to missing details in the problem statement that could clarify the situation further.

Faiq
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Homework Statement


Why is the formula ##p = \frac{1}{3}\rho<c^2>## used to calculate the mean square speed at 273K?
Why 273K?
 
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It works at any temperature. Could you give a bit more context ? Is this a section in your textbook you have a question about or is it in an exercise ?
 
The density of a gas at a temperature of ##273~K## and a pressure of ##1.02*10^5~ Pa## is ##0.9kgm^{-3}##. It may be asssumed to be an ideal gas.
Calculate rms speed at 273K and 546K.

For 273, we were supposed to use the given equation and for second we were supposed to used the ratio of mean square speed and temperature.
My thinking is the equation only applies for those temperature for which we are using the pressure values.
 
You are given pressure and density at 273 K and asked to calculate rms at exactly this temperature, I don't see where is the problem?
 
Borek said:
You are given pressure and density at 273 K and asked to calculate rms at exactly this temperature, I don't see where is the problem?
The confusion I had was why is that formula only applicable for the 273K. That's to say why I can't just put that value down for 546K? Later I realized, the pressure inserted in those equations are the indicator of what temperature should the RMS value be for.
 
We're still missing context. Please be a bit more complete in formulating problem statements. This way we spend way too much time dragging out what is troubling you.
Faiq said:

Homework Statement

Missing:

Homework Equations



The Attempt at a Solution


Accidentally ? Or did you erase them ? That's a no-no in PF !

And anyway, 1. should probably have been
Faiq said:
The density of a gas at a temperature of ##273~K## and a pressure of ##1.02*10^5~ Pa## is ##0.9kgm^{-3}##. It may be asssumed to be an ideal gas.
Calculate rms speed at 273K and 546K.
Any word or sentence missing from this full problem statement ? If no, then a list of all variables is good to have at hand. For you too.
Faiq said:
For 273, we were supposed to use the given equation and for second we were supposed to used the ratio of mean square speed and temperature.
What given equation ?

Did you do anything at all with the information that the gas can be considered an ideal gas ?

Where did the
Faiq said:
formula ##p = \frac{1}{3}\rho<c^2>##
come from ? From the textbook or from the problem statement ? (or perhaps from the solution manual :rolleyes: ?)
 
1. No words or sentences are missing.
2. The given equation is ##p = \frac{1}{3}\rho<c^2>##
3. By telling us that it could be considered an ideal gas, permission was given to use the ideal gas law and the provided equation.
4. The formula was supposed to be derived by using the equation ##pV = \frac{1}{3}Nm<c^2>## and ##\rho = \frac{Nm}{V}##

Btw my confusion has been sorted out. Thanks for the help though
 
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