Chapter 5, Problem 164CP

Interpretation Introduction

Interpretation: The pressure in each given chamber of $\text{He and Rn}$ needs to be calculated after $\text{10}\text{.0 h}$ has passed.

Concept Introduction:The expression relating the pressure, P, volume, V with number of moles, n and temperature, T of an ideal gas is:

  $\text{PV = nRT}$ (obtained by combining three gaseous laws: Charles’s law, Boyle’s law and Avogadro’s law)

Where R is Universal gas constant.

Answer to Problem 164CP

  ${\text{P}}_{\text{He container}}=\text{ 1}\text{.94}\times {\text{10}}^{-6}\text{ atm}$ and ${\text{P}}_{\text{Rn container}}=\text{ 2}\text{.06}\times {\text{10}}^{-6}\text{ atm}$ .

Explanation of Solution

Given:

Two sample of gases, one of pure helium and the other is pure radon, separated in two rectangular 1.00 L chambers by thin metal wall.

Temperature for both the samples = ${\text{27 }}^{\text{o}}\text{C}$ and pressure = $\text{2}\times {\text{10}}^{-6}\text{ atm}$

A circular hole is developed of radius $\text{1}\text{.0}\times {\text{10}}^{-6}\text{ m}$ on the metal wall separating the gases.

Converting temperature from degree Celsius to Kelvin as:

  $\text{27 + 273 = 300 K}$

The ratio of number of moles to volume (which is equal for both the gases as temperature is same) is calculated using ideal gas equation as:

  $\begin{array}{l}\frac{\text{n}}{\text{v}}=\frac{\text{P}}{\text{RT}}\\ \frac{\text{n}}{\text{v}}=\frac{\text{(2}\text{.00}\times {\text{10}}^{-6}\text{ atm)}}{\left(\text{0}\text{.08206 }\frac{\text{L}\text{.atm}}{\text{K}\text{.mol}}\right)(300.\text{ K)}}\\ \frac{\text{n}}{\text{v}}=\text{ 8}\text{.12}\times {\text{10}}^{-8}{\text{ mol L}}^{-1}\end{array}$

Converting this value from ${\text{mol L}}^{-1}$ to ${\text{molecules m}}^{-3}$ as:

  $\begin{array}{l}\frac{\text{N}}{\text{V}}=(8.12\times {10}^{-8}{\text{ mol L}}^{-1})\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}\text{ molecules}}{\text{1 mol}}\right)\left(\frac{{\text{10}}^3\text{ L}}{{\text{1 m}}^3}\right)\\ \frac{\text{N}}{\text{V}}=\text{ 4}\text{.89}\times {\text{10}}^{19}{\text{ molecules m}}^{-3}\end{array}$

Calculating the area, $\text{A}$ of hole developed using formula:

  $\text{A=}\pi {\text{r}}^2$

Where r is radius.

Substituting the values:

  $\begin{array}{l}\text{A = }\pi {\text{(1}}^{\text{.00}}\\ \text{A = 3}\text{.14}\times {\text{10}}^{-12}{\text{ m}}^2\end{array}$

Now, the number of collisions is calculated using formula:

  ${\text{Z}}_{\text{A}}=\text{A}\frac{\text{N}}{\text{V}}\sqrt{\frac{\text{RT}}{\text{2}\pi \text{M}}}$

Where, ${\text{Z}}_{\text{A}}$ is the number of collisions per unit area, A is area in which collisions are taking place, $\frac{\text{N}}{\text{V}}$ is the number of molecules per unit volume, R is universal gas constant, T is temperature and M is molar mass.

The number of collisions for He will be:

  $\begin{array}{l}{\text{Z}}_{\text{A,He}}=\text{A}\frac{\text{N}}{\text{V}}\sqrt{\frac{\text{RT}}{\text{2}\pi \text{M}}}\\ {\text{Z}}_{\text{A,He}}=\text{(3}\text{.14}\times {\text{10}}^{-12}{\text{ m}}^2)(4.89\times {10}^{19}{\text{ molecules m}}^{-3})\sqrt{\frac{\text{(8}{\text{.3145 J K}}^{-1}{\text{ mol}}^{-1})(300.\text{ K)}}{\text{2}\pi \text{(4}\text{.003}\times {10}^{-3}{\text{ kg mol}}^{-1})}}\\ {\text{Z}}_{\text{A,He}}=\text{ 4}\text{.84}\times {\text{10}}^{10}{\text{ molecules s}}^{-1}\end{array}$

The number of collisions for Rn will be:

  $\begin{array}{l}{\text{Z}}_{\text{A,Rn}}=\text{A}\frac{\text{N}}{\text{V}}\sqrt{\frac{\text{RT}}{\text{2}\pi \text{M}}}\\ {\text{Z}}_{\text{A,Rn}}=(3.14\times {10}^{-12}{\text{ m}}^2)(4.89\times {10}^{19}{\text{ molecules m}}^{-3})\sqrt{\frac{(8.3145{\text{ J K}}^{-1}{\text{ mol}}^{-1})(300.\text{ K)}}{2\pi (2.22\times {10}^{-1}{\text{ kg mol}}^{-1})}}\\ {\text{Z}}_{\text{A,Rn}}=\text{6}\text{.49}\times {\text{10}}^9{\text{ molecules s}}^{-1}\end{array}$

The rate of collisions is assumed to be constant, so, the number of molecules transferred into another chamber is:

  $\begin{array}{l}{\text{N}}_{\text{He,transferred}}\text{= (4}\text{.84}\times {\text{10}}^{10}{\text{ molecules s}}^{-1})(10.0\text{ h)}\left(\frac{3600\text{ s}}{\text{1 h}}\right)\\ {\text{N}}_{\text{He,transferred}}\text{ = }(1.74\times {10}^{15}\text{ molecules)}\left(\frac{\text{1 mol}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{ molecules}}\right)\\ {\text{N}}_{\text{He,transferred}}=\text{ 2}\text{.89}\times {\text{10}}^{-9}\text{ mol}\end{array}$

  $\begin{array}{l}{\text{N}}_{\text{Rn,transferred}}\text{ = (6}\text{.49}\times {\text{10}}^9{\text{ molecules s}}^{-1})(10.0\text{ h)}\left(\frac{3600\text{ s}}{\text{1 h}}\right)\\ {\text{N}}_{\text{Rn,transferred}}\text{ =}(2.34\times {10}^{14}\text{ molecules)}\left(\frac{\text{1 mol}}{\text{6}\text{.022}\times {\text{10}}^{23}\text{ molecules}}\right)\\ {\text{N}}_{\text{Rn,transferred}}\text{ = 3}\text{.89}\times {\text{10}}^{-10}\text{ mol}\end{array}$

Now, the original amount of gas is calculated using ideal gas equation as:

  $\begin{array}{l}\text{n = }\frac{\text{PV}}{\text{RT}}\\ \text{n = }\frac{\text{(2}\text{.00}\times {\text{10}}^{-6}\text{ atm)(1}\text{.00 L)}}{\left(\text{0}\text{.08206 }\frac{\text{L}\text{.atm}}{\text{K}\text{.mol}}\right)\text{(300}\text{. K)}}\\ \text{n = 8}\text{.12}\times {\text{10}}^{-8}\text{ mol}\end{array}$

Side that began with He now contains:

  $\text{8}\text{.12}\times {\text{10}}^{-8}\text{ mol - 2}\text{.89}\times {\text{10}}^{-9}\text{ mol = 7}\text{.83}\times {\text{10}}^{-8}\text{ mol}$

Total moles are:

  $\text{7}\text{.83}\times {\text{10}}^{-8}\text{ mol He + 3}\text{.89}\times {\text{10}}^{-10}\text{ mol Rn = 7}\text{.87}\times {\text{10}}^{-10}\text{ mol}$

So, the final pressure of He in container is:

  $\begin{array}{l}{\text{P}}_{\text{He container}}=\frac{\text{nRT}}{\text{V}}\\ {\text{P}}_{\text{He container}}=\frac{\text{(7}\text{.87}\times {\text{10}}^{-8}\text{ mol)}\left(0.8206\frac{\text{L}\text{.atm}}{\text{K}\text{.mol}}\right)(300.\text{ K)}}{\text{(1}\text{.00 L)}}\\ {\text{P}}_{\text{He container}}=\text{ 1}\text{.94}\times {\text{10}}^{-6}\text{ atm}\end{array}$

Side that began with Rn now contains:

  $\text{8}\text{.12}\times {\text{10}}^{-8}\text{ mol - 3}\text{.89}\times {\text{10}}^{-10}\text{ mol = 8}\text{.08}\times {\text{10}}^{-8}\text{ mol}$

Total moles are:

  $\text{8}\text{.08}\times {\text{10}}^{-8}\text{ mol Rn + 2}\text{.89}\times {\text{10}}^{-9}\text{ mol He = 8}\text{.37}\times {\text{10}}^{-8}\text{ mol}$

So, the final pressure of Rn in container is:

  $\begin{array}{l}{\text{P}}_{\text{Rn container}}=\frac{\text{nRT}}{\text{V}}\\ {\text{P}}_{\text{Rn container}}=\frac{\text{(8}\text{.37}\times {\text{10}}^{-8}\text{ mol)}\left(0.8206\frac{\text{L}\text{.atm}}{\text{K}\text{.mol}}\right)(300.\text{ K)}}{\text{(1}\text{.00 L)}}\\ {\text{P}}_{\text{Rn container}}=\text{ 2}\text{.06}\times {\text{10}}^{-6}\text{ atm}\end{array}$