Molecular interpretation of temperature

Click For Summary
SUMMARY

The discussion focuses on the molecular interpretation of temperature in relation to the root mean square (rms) speed of gas molecules. The equation vrms = √(3kT/m) is central to the analysis, where vrms is the rms speed, k is the Boltzmann constant, T is the temperature in Kelvin, and m is the mass of a gas molecule. To raise the rms speed by 1%, the temperature must be increased to approximately 393 K, or about 6 degrees Kelvin from the initial 20 degrees Celsius. Additionally, the relationship vrms = √(P/ρ) is explored, where P is pressure and ρ is density, leading to a deeper understanding of gas behavior.

PREREQUISITES
  • Understanding of the ideal gas law and molecular theory
  • Familiarity with the equations for root mean square speed (vrms)
  • Knowledge of thermodynamic concepts including temperature and pressure
  • Basic algebra skills for manipulating equations
NEXT STEPS
  • Study the derivation of the rms speed equation vrms = √(3kT/m)
  • Learn about the implications of temperature changes on gas behavior
  • Explore the relationship between pressure, density, and temperature in gases
  • Investigate the Boltzmann constant and its role in statistical mechanics
USEFUL FOR

Students studying physics, particularly those focusing on thermodynamics and kinetic theory, as well as educators seeking to clarify concepts related to molecular speed and temperature relationships.

thattgirljoyy
Messages
2
Reaction score
0
i'm trying to do my ap physics summer work, but I've been having a hard time trying to teach this to myself. i do really well in class, but i need a little bit of guidance to help explain how to use the equations the right way.

i'm currently stuck on a question about vrms an it's relation to temp.

Homework Statement


the rms speed of molecules in a gas at 20degrees celsius is to be raised by 1%. To what temperature must it be raised?

Homework Equations


vrms= √3kT/m

The Attempt at a Solution


i don't have any clue where to start if it didn't give me the gas to find it's mass.-i was wondering if i could find someone who could maybe stay in touch for a day or two so i can reach out if i get stuck on anything
 
Last edited:
Physics news on Phys.org
What happens to constants when you need a ratio?
 
okay, the temperature would change by sqrt 2, right ?
essentially about 6 degrees kelvin.

i'm stuck on another problem, it seems to be more conceptual.
it says to prove how
vrms=√P/ρ
P, pressure
ρ, density of the gas

i started with the two equations
P=1/3Nmv^2/V
v^2=3KT/m

i assumed the goal was to get the V from the first equation into the denominator of the second to try to get the density (m/v)
some how all I've gotten to was
P/ρ=[3KT(1/3)N]/m

i'm trying to see if i can get the original 3KT/m to revert back to v^2, but i don't know how to get rid of the (1/3)N/m
 
Welcome to PF, thattgirljoyy! :smile:

thattgirljoyy said:
okay, the temperature would change by sqrt 2, right ?
essentially about 6 degrees kelvin.

It's not too far off 6 degrees... but there is no sqrt 2 involved that I can see... :confused:

Let's put it in formulas:
Before you have some unknown vrms,before at 20 degrees Celsius, or 393 K.
Afterward you have vrms,after=1.01vrms,before at some unknown temperature T.
Substitute and solve for T?
i'm stuck on another problem, it seems to be more conceptual.
it says to prove how
vrms=√P/ρ
P, pressure
ρ, density of the gas

i started with the two equations
P=1/3Nmv^2/V
v^2=3KT/m

i assumed the goal was to get the V from the first equation into the denominator of the second to try to get the density (m/v)
some how all I've gotten to was
P/ρ=[3KT(1/3)N]/m

i'm trying to see if i can get the original 3KT/m to revert back to v^2, but i don't know how to get rid of the (1/3)N/m

Did you consider that m is the mass of one molecule?
What is the mass of all molecules in the volume V?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 1 ·
Replies
1
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
Replies
5
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
5K
Replies
3
Views
9K
  • · Replies 1 ·
Replies
1
Views
2K