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Equilibrium Conditions

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A container is partitioned by a porous, fixed, wall. Because the wall is porous, it allows particles to diffusive from one side to the other (note that by 'diffuse' I mean that there are tiny holes in the wall which allow the gas particles to make it through; there is not some big gaping hole). The wall also allows the flow of heat but it doesn't move. The container's total volume is 8 m3 and the small part of the container has a volume of 2 m3. The container is thermally isolated from the surroundings. Five moles of an ideal monatomic ideal gas are placed in the left side of the container (the small part of the container) - call this container A. Then two moles of an ideal monatomic gas are placed in the right side of the container - call this container B. Each gas is initially at a temperature of 300 K. The system will, of course, after awhile achieve some equilibrium state. So, in summary, the gases can exchange (i.e. change) internal energy and number of particles but their volumes are fixed.
a) What will be the condition(s) for equilibrium?
b) What is the equilibrium temperature?
c) What is the approximate change in entropy of this process?

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Solution Summary

The gas law suggests p1i*V1 = n1i*R*Ti and P2i*V2 = n2i*R*Ti.

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a) For equilibrium, the final temperature and pressure in two parts will be same i.e. if initial temperatures and pressures of two parts are T1i, T2i, p1i and p2i and finals are T1f, T2f, p1f and p2f. Then, for equilibrium, T1f = T2f and p1f = p2f. ...

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  • BEng, Allahabad University, India
  • MSc , Pune University, India
  • PhD (IP), Pune University, India
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