Question

A mixture of$\mathbf{2}\mathbf{.00}\mathbf{}\mathbf{\text{mol}}$ nitrogen and$\mathbf{1}\mathbf{.00}\mathbf{}\mathbf{\text{mol}}$oxygen is in thermal equilibrium in a$\mathbf{100}\mathbf{\text{L}}$container at${\mathbf{25}}^{\mathbf{°}}\mathbf{\text{C}}$. Calculate the probability that at a given time all the nitrogen will be found in the left half of the container and all the oxygen in the right half.

Short answer

The probability of a mixture of nitrogen and oxygen is $5.44\times {10}^{23}$.

Step-by-step solution

Given information

The number of moles of nitrogen and oxygen are $2.00\text{mol}$and$1.00\text{mol}$

The volume of the container and temperature are$100\text{L}$ and ${25}^{°}\text{C}$.

Concept of probability

A probability is a number that expresses the likelihood or chance that a specific evenit will take place.

Probabilities can be stated as proportions with values between 0% and 100, or as

percentages with values between 0 and 100.

Calculate the number of molecules of nitrogen

First let's calculate the number of molecules as:

$\begin{array}{l}N(N_2)=N_A\times n(N_2)\\ N(N_2)=6.022\times {10}^{23}\times 2.00\\ N(N_2)=1.2044\times {10}^{24}\end{array}$

Calculate the probability of nitrogen

The probability for nitrogen is calculated as:

$\begin{array}{c}{\left(\frac{1}{2}\right)}^{N(N_2)}={\left(\frac{1}{10}\right)}^x\\ 1.2044\times {10}^{24}log\frac{1}{2}=xlog\frac{1}{10}\\ 1.2044\times {10}^{24}\times (-0.301)=x\times (-1)\\ x=3.63\times {10}^{23}\end{array}$

Where, x is the probability of nitrogen

Calculate the number of molecules of oxygen

First, let's calculate the number of molecules:

$\begin{array}{l}N\left(O_2\right)=N_A\times n\left(O_2\right)\\ N\left(O_2\right)=6.022\times {10}^{23}\times 1.00\\ N\left(O_2\right)=6.022\times {10}^{23}\end{array}$

Calculate the probability of oxygen

The probability for oxygen is calculated as:

$\begin{array}{c}{\left(\frac{1}{2}\right)}^{N\left(o_2\right)}={\left(\frac{1}{10}\right)}^y\\ 6.022\times {10}^{23}log\frac{1}{2}=ylog\frac{1}{10}\\ 6.022\times {10}^{23}\times (-0.301)=y\times (-1)\\ y=1.81\times {10}^{23}\end{array}$

Where, y represent the probability of oxygen

Calculate the probability of oxygen

The total probability is the sum of the probability of nitrogen and oxygen.

$\begin{array}{l}p=x+y\\ p=3.63\times {10}^{23}+1.81\times {10}^{23}\\ p=5.44\times {10}^{23}\end{array}$