A box is separated by a partition into two parts of equal volume. The left side

Question

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500 molecules of nitrogen gas; the right side contains 100 molecules of oxygen gas. The two gases are at the same temperature. The partition is punctured, and equilibrium is eventually attained. Assume the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. You may also assume an ideal gas.

a.) On average, how many molecules of each type will there be in either half of the box?

b.) What is the change in entropy of the system when the partition is punctured?

c.) What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured, that is, 500 nitrogen molecules in the left half and 100 oxygen molecules in the right half?

Explanation / Answer

1) when the gases will attain equilibrium there will be a uniform distribution of molecules,

that is 500 +100 moles will be evenly distributed

hence one half will have 500/2 =250 moles and 100/2 =50 moles

i.e 250 N2 + 50 O2

2)Boltzmann entropy is defined by

S = k ln W

where k is theBoltzmann constant,

W called the thermodynamic probability .(in this case ,(( vinitial / vfinal )^n))

so when in eqilibrium

del(S)=?S = ? k ln W^600 = 600 k ? ln W = 600 k ln2 =
= 0.574 10-20 J/K

3) now to find the probabilty

w=( vinitial / vfinal )^n = (1/2)^600

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