Chapter 3: Q30E (page 678)

A box is separated by a partition into two parts of equal volume. The left side of the box contains 500molecules 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 that the volume of the box is large enough for each gas to undergo a free expansion and not change temperature. What is the probability that the molecules will be found in the same distribution as they were before the partition was punctured that is, 500nitrogen molecules in the left half and 100 oxygen molecules in the right half?

Short Answer

Expert verified

The probability that the molecules will be found in the same distribution as they were before the partition was punctured is $2.4 \times 10^{- 181}$ .

Step by step solution

The number of possible microscopic states

In a free expansion of N molecules in which the volumes doubles, the number of possible microscopic states increases by a factor of $2^{N}$ .

Therefore, the total probability that the molecules will be found in the same distribution as they were before the partition was punctured is

$P_{T} = {(\frac{1}{2})}^{500} . {(\frac{1}{2})}^{100} = {(\frac{1}{2})}^{600}$

Where, $P_{T}$ is the total probability of molecules that will be found in the same distribution as they were before the partition was punctured.

Calculation of total probability

_{$P_{T} = {(\frac{1}{2})}^{600}$}

Taking log on both sides

_{$In (P_{T}) = 600 In (\frac{1}{2}) = - 415.89 P_{T} = e^{- 415.89} = 2.4 \times 10^{- 181}$}

Thus, the probability that the molecules will be found in the same distribution as they were before the partition was punctured is $2.4 \times 10^{- 181}$ .

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Short Answer

Step by step solution

The number of possible microscopic states

Calculation of total probability

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