What is the Temperature of Vessel 2 After Effusion?

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SUMMARY

The discussion centers on calculating the temperature of vessel 2 after effusion from vessel 1, utilizing the principles of kinetic theory. The temperature of vessel 1 is determined to be 400 K based on the average kinetic energy of the gas molecules, while vessel 2 reaches a final temperature of 533 K. The flux of particles effusing from vessel 1 is expressed through the equation φ = (1/4)n⟨v₁⟩, and the discussion highlights the importance of understanding the average kinetic energy of diatomic gases, specifically nitrogen, in these calculations.

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  • Understanding of kinetic theory of gases
  • Familiarity with the Maxwell-Boltzmann speed distribution
  • Knowledge of thermodynamic principles related to temperature and kinetic energy
  • Basic calculus for integrating functions related to energy distributions
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  • Study the Maxwell-Boltzmann distribution for diatomic gases
  • Learn about the relationship between kinetic energy and temperature in thermodynamics
  • Explore the concept of effusion and its mathematical modeling
  • Investigate the properties of nitrogen gas and its behavior under varying temperatures
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Students in physics or chemistry, particularly those studying thermodynamics and gas behavior, as well as educators seeking to explain the principles of effusion and temperature calculations in gases.

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Homework Statement



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Part (a): Explain what A and f means. Describe an experiment that shows this relation is true.
Part (b):Find the temperature of vessel 1.
Part (c): Why is temperature of vessel initially the same as vessel 1? Calculate the final temperature of vessel 2. Find the time taken for vessel 2 to reach this temperature.

Homework Equations


The Attempt at a Solution



Part (b)
Letting ##T_0 = 300 K## be room temperature, and ##\alpha = \frac{m}{2kT}##,
the speed distribution of effusing molecules from 1 into 2 is:

g_{(v)} = 2 \alpha^2 v^3 \space exp\left(-\alpha v^2\right)

The average energy of these particles is:

\langle E'\rangle = \frac{1}{2}m \int_0^{\infty} 2\alpha^2 v^5 \space exp\left(-\alpha v^2\right)
= 2kT_0

Now, in general the average KE of a system with temperature ##T## is ##\frac{3}{2}kT##.

\langle E' \rangle = \frac{3}{2}k\left(\frac{4}{3}T_0\right)
= \frac{3}{2}kT_1

This implies that temperature of box 1 is ##T_1 = \frac{4}{3}T_0 = 400 K##.

Part (c)
In the beginning, the average KE of molecules entering box 2 should be higher than the average KE of particles in box 1, since particles effusing have higher KE? I'm not sure why it would be the same.

Now the same as above,
T_2 = \frac{4}{3}T_1 = 533 K

The flux of particles effusing from box 1 is given by:
\phi = \frac{1}{4}n \langle v_1\rangle = \frac{1}{4} \frac{P}{kT} \sqrt{\frac{8kT_1}{\pi m}}

Thus number of particles per unit time is ##\phi A = \pi r^2## where ##r^2## is the area of the hole.

Not sure how to find the time taken.
 
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Your result for part (b) looks correct for a monatomic gas where KEavg = (3/2)kT. (But note that at the end of the statement of the problem, it mentions nitrogen which is diatomic.)

For part (c), you are looking for the initial temperature in box 2 a short time after hole 1 is opened. Almost all of the molecules that make it through hole 2 initially are those molecules that travel directly from hole 1 through hole 2 without any collisions on the way. See if you can show that the average translational KE of such a molecule is 2kT0.
 
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TSny said:
Your result for part (b) looks correct for a monatomic gas where KEavg = (3/2)kT. (But note that at the end of the statement of the problem, it mentions nitrogen which is diatomic.)

For part (c), you are looking for the initial temperature in box 2 a short time after hole 1 is opened. Almost all of the molecules that make it through hole 2 initially are those molecules that travel directly from hole 1 through hole 2 without any collisions on the way. See if you can show that the average translational KE of such a molecule is 2kT0.

Yes, I worked through the question and I got the right answers. Thanks a lot!
 

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