Chapter 19, Problem 26P

(a)

To determine

The average time before molecules reach in left half of the box.

Answer to Problem 26P

The average time before molecules reach in left half of the box is $5\text{sec}$ .

Explanation of Solution

Given:

The number of molecules is $10$ .

The volume of the box is $V=1.0\text{L}$ .

Formula used:

The expression for average time is given as,

  $t=\frac{2^n}{2\left(100{\text{s}}^{-1}\right)}$

Here, $n$ is the number of molecules.

Calculation:

The average time for $n=10$ molecules can be calculated as,

  $\begin{array}{c}t=\frac{2^n}{2\left(100{\text{s}}^{-1}\right)}\\ =\frac{2^{10}}{2\left(100{\text{s}}^{-1}\right)}\\ =5.12\text{s}\\ \approx 5\text{s}\end{array}$

Conclusion:

Therefore, the average time before molecules reach in left half of the box is $5\text{sec}$ .

(b)

To determine

The average time before molecules reach in left half of the box.

Answer to Problem 26P

The average time before molecules reach in left half of the box is $2\times {10}^{20}\text{yr}$ .

Explanation of Solution

Given:

The number of molecules is $100$ .

Calculation:

The average time for $n=100$ molecules can be calculated as,

  $\begin{array}{c}t=\frac{2^n}{2\left(100{\text{s}}^{-1}\right)}\\ =\frac{2^{100}}{2\left(100{\text{s}}^{-1}\right)}\\ =6.34\times {10}^{27}\text{s}\times \frac{1\text{yr}}{3.156\times {10}^7\text{s}}\\ \approx 2\times {10}^{20}\text{yr}\end{array}$

Conclusion:

Therefore, the average time before molecules reach in left half of the box is $2\times {10}^{20}\text{yr}$ .

(c)

To determine

The average time before molecules reach in left half of the box.

Answer to Problem 26P

The average time before molecules reach in left half of the box is $1.5\times {10}^{291}\text{yr}$ .

Explanation of Solution

Given:

The number of molecules is $1000$ .

Calculation:

The average time for $n=1000$ molecules can be calculated as,

  $\begin{array}{c}t=\frac{2^n}{2\left(100{\text{s}}^{-1}\right)}\\ =\frac{2^{1000}}{2\left(100{\text{s}}^{-1}\right)}\end{array}$ ........(a)

The conversion of $2^{1000}$ in power of 10 can be calculated as,

  $\begin{array}{l}{10}^x=2^{1000}\\ x\ln 10=1000\times \ln 2\\ x=301\end{array}$

Substitution of the conversion in equation (a) can be given as,

  $\begin{array}{c}t=\frac{{10}^{301}}{2\left(100{\text{s}}^1\right)}\\ =0.5\times {10}^{299}\times \frac{1\text{yr}}{3.156\times {10}^7\text{s}}\\ \approx 1.5\times {10}^{291}\text{yr}\end{array}$

Conclusion:

Therefore, the average time before molecules reach in left half of the box is $1.5\times {10}^{291}\text{yr}$ .

(d)

To determine

The average time before molecules reach in left half of the box.

Answer to Problem 26P

The average time before molecules reach in left half of the box is $1.5\times {10}^{15}\text{yr}$ .

Explanation of Solution

Given:

The number of molecules is $1\text{mol}$ .

Calculation:

The average time for $n=1\text{mol}$ , molecules can be calculated as,

  $\begin{array}{c}t=\frac{2^n}{2\left(100{\text{s}}^{-1}\right)}\\ =\frac{2^{1\text{mol}}}{2\left(100{\text{s}}^{-1}\right)}\\ =\frac{2^{6.022\times {10}^{23}}}{2\left(100{\text{s}}^{-1}\right)}\end{array}$ ........(b)

The conversion of $2^{6.022\times {10}^{23}}$ in power of 10 can be calculated as,

  $\begin{array}{l}{10}^x=2^{6.022\times {10}^{23}}\\ x\ln 10=6.022\times {10}^{23}\times \ln 2\\ x\approx {10}^{23}\end{array}$

Substitution of the conversion in equation (2) can be given as,

  $\begin{array}{c}t=0.5\times {10}^{23}\text{s}\times \frac{1\text{yr}}{3.156\times {10}^7\text{s}}\\ \approx 1.5\times {10}^{15}\text{yr}\end{array}$

Conclusion:

Therefore, the average time before molecules reach in left half of the box is $1.5\times {10}^{15}\text{yr}$ .

(e)

To determine

The physicist waiting time before all of the gas molecules in the vacuum chamber occupy only the left half of chamber and comparison with expected lifetime of the universe.

Answer to Problem 26P

The average time is ${10}^{3\times {10}^9}\text{yr}$ and it is $0.3$ times of expected lifetime of universe.

Explanation of Solution

Given:

The pressure at which best vacuum created is $P={10}^{-12}\text{torr}$ .

The expected lifetime of the universe is ${10}^{10}\text{yr}$ .

Formula used:

The expression for the ideal gas is given as,

  $PV=NKT$

Here, $K$ is the Boltzmann constant, $T$ is the temperature and $N$ is the number of molecules.

The expression for the comparison is given as,

  $\frac{t}{t_{universe}}$

Calculation:

The number of moles in vacuum condition at $300\text{K}$ and ${10}^{-12}\text{torr}$ can be calculated as,

  $\begin{array}{l}PV=NKT\\ N=\end{array}$

  $\begin{array}{c}PV=NKT\\ N=\frac{PV}{KT}\\ =\frac{\left({10}^{-12}\text{torr}\times \frac{133.32\text{Pa}}{1\text{torr}}\right)\left(1.0\text{L}\right)}{1.381\times {10}^{-23}\text{J}/\text{K}\left(330\text{K}\right)}\\ =\frac{1.33\times {10}^{-10}\text{Pa}\cdot \text{L}\times \frac{{10}^{-5}\text{bar}}{1\text{Pa}}}{1.381\times {10}^{-23}\text{J}}\end{array}$

On further solving the above equation,

  $\begin{array}{c}N=\frac{1.33\times {10}^{-15}\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}}{1.381\times {10}^{-23}\text{J}}\\ =\frac{1.33\times {10}^{-15}\text{bar}\cdot \text{L}\times \frac{100\text{J}}{1\text{bar}\cdot \text{L}}}{1.381\times {10}^{-23}\text{J}}\\ =9.63\times {10}^9\end{array}$

The average time of molecules in vacuum to occupy left half of the chamber can be calculated as,

  $t=\frac{2^{9.63\times {10}^9}}{2\left(100{\text{s}}^1\right)}$

The conversion of $2^{9.63\times {10}^9}$ in power of 10 can be calculated as,

  $\begin{array}{l}{10}^x=2^{9.63\times {10}^9}\\ x\ln 10=9.63\times {10}^9\times \ln 2\\ x\approx 3\times {10}^9\end{array}$

Substitution of the conversion in equation (2) can be given as,

  $\begin{array}{c}t=\frac{{10}^{3\times {10}^9}}{2\left(100{\text{s}}^1\right)}\\ =0.5\times {10}^{3\times {10}^9}\times \frac{1\text{yr}}{3.156\times {10}^7\text{s}}\\ \approx {10}^{3\times {10}^9}\text{yr}\end{array}$

The comparison can be given as,

  $\begin{array}{c}\frac{t}{t_{universe}}=\frac{{10}^{3\times {10}^9}\text{yr}}{{10}^{{10}^{10}}\text{yr}}\\ \approx 0.3t_{universe}\end{array}$

Conclusion:

Therefore, the average time is ${10}^{3\times {10}^9}\text{yr}$ and it is $0.3$ times of expected lifetime of universe.