Calculating the Mean Free Path of Argon Atom

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Homework Help Overview

The discussion revolves around calculating the mean free path of argon atoms, specifically in the context of gaseous argon at a pressure of 1 atm and a temperature of 300K. The original poster presents a series of related questions involving both gaseous and liquid argon, including atomic density and root mean square calculations.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster questions which atomic density ('n') to use for the mean free path calculation, expressing uncertainty about whether to consider gaseous or liquid argon. Some participants suggest that the context implies a focus on gaseous argon. The poster also raises a related question about the root mean square calculation and its applicability to the different states of argon.

Discussion Status

Participants are exploring the implications of the problem's wording, particularly regarding the use of gaseous versus liquid argon for calculations. There is a recognition that the inclusion of both states in the question may lead to confusion, and some guidance has been offered regarding the likely intent of the problem.

Contextual Notes

The original poster notes the lack of explicit instructions on which state of argon to consider for certain calculations, leading to questions about the relevance of both states in the context of the homework problem.

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Homework Statement



Hi, I am not sure about the following question:

(Q) Argon (atomic weight 40) exists as a monatomic gas at room temperature and pressure. The density of liquid argon is 1784kg/m^3.

(a) Calculate the atomic density (atoms/m^3) in liquid argon *Done*

(b) Hence assumming the packing fraction in liquid argon is 0.7, determine the radius of an argon atom. *Done*

(c) calculate the atomic density in gaseous argon at a pressure of 1atm and a temperature of 300K *Done*

(d) hence calculate the mean free path of an argon atom at a pressure of 1 atm and a temperature of 300K?

Homework Equations



It is part (d) which I'm having trouble with

The mean free path formula is 1/[(4pi)nσ²]

The Attempt at a Solution



For part (d) which 'n' should I use the one for gas argon or for liquid argon, the question given doesn't seem to state which mean free path to calculate?
 
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The 'hence' suggests they want you to calculate it for gaseous argon.
 
Cyosis said:
The 'hence' suggests they want you to calculate it for gaseous argon.

Oh Hi! right! I actually forgot the next question which was:

(e) Calculate the root mean square of an argon atom at a temperature of 300K.

Again do you calculate it for gaseous or liquid argon? and why would they put both types of argon in the same question, its confusing..and they don't really ask any questions about comparing both types at the end.

I also found another question very similar to the above but how do I tackle this one?

(Q2.) The viscosity of argon gas (atomic weight 40) at a temperature of 273K and a pressure of 1atm is equal to 2.1 x 10^5 Pa s. The density of liquid argon is 1784kg/m^3 . Use these two pieces of data to determine the mean speed of an argon atom at 273k, its effective radius and Avogadro's number.
 
What is the boiling point of argon?
 
Cyosis said:
What is the boiling point of argon?

87.3K approx, how does this help me?
 
Seeing as the temperature of argon in question e) is 300K it would stand to reason that they are interested in gaseous argon.
 

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