Calculation Involving Thermodynamics, Gravitation and Fluid mechanics

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SUMMARY

The discussion revolves around the application of Stokes' Law in the context of a monoatomic ideal gas in a gravitational field, specifically addressing the diffusion of gas molecules. The key equation referenced is Stokes' Law: F=6πNrν, where N is the coefficient of viscosity, r is the radius of the gas molecules, and ν is the velocity. Participants emphasize the importance of understanding the kinetic theory of gases, diffusion, and mean free path to solve the problem effectively, particularly in preparation for competitive exams like the JEE Advanced.

PREREQUISITES
  • Understanding of Stokes' Law and its application
  • Knowledge of kinetic theory of gases
  • Familiarity with diffusion concepts and mean free path
  • Basic principles of thermodynamics and fluid mechanics
NEXT STEPS
  • Research the Stokes-Einstein equation and its relevance to molecular diffusion
  • Study the concepts of random walks in the context of gas molecules
  • Explore the relationship between viscosity and kinetic energy in fluids
  • Review the principles of gas behavior under gravitational fields
USEFUL FOR

Students preparing for competitive exams such as JEE Advanced, physics enthusiasts, and professionals in thermodynamics and fluid mechanics seeking to deepen their understanding of gas behavior in various conditions.

Ayesha02
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Homework Statement
A monoatomic ideal gas is in gravitational field extended from the surface of the earth to a great height. Temperature of the gas is T and remains same at any height. At height 'h' number of molecules per unit volume 'n' is independent of time i.e equilibrium maintained. On the surface of the Earth n=##n_0##. Coefficient of viscosity of the gaseous medium is##N##. Stokes law is applicable. Radius of each spherical molecule is R. At height 'h' number of molecules diffusing upwards per unit area per second is given by -D dn/dh.
Find value of D. (boltzmann constant is K)
Relevant Equations
Stokes Law: F=6 pi ##N## r v
I figured out stokes law equation but I am unable to proceed further
 
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Ayesha02 said:
Homework Statement:: A monoatomic ideal gas is in gravitational field extended from the surface of the Earth to a great height. Temperature of the gas is T and remains same at any height. At height 'h' number of molecules per unit volume 'n' is independent of time i.e equilibrium maintained. On the surface of the Earth n=##n_0##. Coefficient of viscosity of the gaseous medium is##N##. Stokes law is applicable. Radius of each spherical molecule is R. At height 'h' number of molecules diffusing upwards per unit area per second is given by -D dn/dh.
Find value of D. (boltzmann constant is K)
Relevant Equations:: Stokes Law: F=6 pi ##N## r v

I figured out stokes law equation but I am unable to proceed further
I cannot understand what diffusion has to do with Stokes' law and viscosity. A trick question perhaps?
 
haruspex said:
I cannot understand what diffusion has to do with Stokes' law and viscosity. A trick question perhaps?

Probably. Its a competitive exam sum.

how to develop thought process in such sums?
 
Ayesha02 said:
Probably. Its a competitive exam sum.

how to develop thought process in such sums?
It helps if you can think through the physical processes involved.
Gas molecules diffuse because they travel at great speeds, typically, and bounce elastically off each other. The energy involved is thermal, so there is no mechanical energy to be lost to heat.
Viscosity involves the loss of KE in a much larger body moving through a fluid.

Next step is to think why an upwardly moving gas molecule would not always shoot up to the highest point its KE could take it. But unless you have actually studied this topic it's a bit tough to figure it all out in an exam.

What exam is this?

Google diffusion, mean free path.
 
haruspex said:
It helps if you can think through the physical processes involved.
Gas molecules diffuse because they travel at great speeds, typically, and bounce elastically off each other. The energy involved is thermal, so there is no mechanical energy to be lost to heat.
Viscosity involves the loss of KE in a much larger body moving through a fluid.

Next step is to think why an upwardly moving gas molecule would not always shoot up to the highest point its KE could take it. But unless you have actually studied this topic it's a bit tough to figure it all out in an exam.

What exam is this?

Its JEE Advanced Exam- for entry into undergrad engineering.

and the molecule won't go upto the highest point as per KE because of viscosity isn't it? Viscosity'll retard it..
 
Ayesha02 said:
and the molecule won't go upto the highest point as per KE because of viscosity isn't it? Viscosity'll retard it..
As I wrote, I would not say so. The kinetic theory of gases says they bounce off each other like perfectly elastic billiard balls.
Did you google diffusion and mean free path?
 
Try looking at the Stokes-Einstein equation
 
etotheipi said:
Try looking at the Stokes-Einstein equation
I still don't see the relevance. Those equations are for a relatively large particle moving through a fluid medium. Here we only have the molecules of the medium itself. Gaseous diffusion is governed by molecular velocity and mean free path.
 
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haruspex said:
I still don't see the relevance. Those equations are for a relatively large particle moving through a fluid medium. Here we only have the molecules of the medium itself. Gaseous diffusion is governed by molecular velocity and mean free path.

Welp yes you’re right! Didn’t read it carefully enough.

Random walks and mean free paths are the way to go here.
 

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