Chapter 5, Problem 162CP

Interpretation Introduction

  $PV=nRT$

Interpretation:The diameter of circular hole in vessel is to be determined.

Concept introduction: The ideal gas equation can be expressed as follows,

Where,

• $P$is the pressure of gas.
• $V$is the volume of gas
• $T$is the temperature of gas.
• $n$is the mole of gas.
• $R$is the gas constant.

Answer to Problem 162CP

Diameter of hole is $1.18\times {10}^{-16}{\text{ m}}^2$ .

Explanation of Solution

The amount of gas is calculated as follows:

  $PV=nRT$

Rearrange above equation for $n$ .

  $n=\frac{PV}{RT}$

Where,

• $P$is pressure of gas.
• $V$is volume of gas.
• $n$isnumber of moles.
• $R$is universal gas constant.
• $T$is temperature.

Value of $P$ is $1.20\times {10}^{-4}\text{ atm}$ .

Value of $V$ is $1.00\text{ L}$ .

Value of $R$ is $0.0821\text{ atm L/mol K}$ .

Value of $T$ is $300\text{ K}$ .

Substitute values in above equation.

  $\begin{array}{c}n=\frac{PV}{RT}\\ =\frac{\left(1.20\times {10}^{-4}\text{ atm}\right)\left(1.00\text{ L}\right)}{\left(0.0821\text{ atm L/mol K}\right)\left(300\text{ K}\right)}\\ =4.87\times {10}^{-8}\text{ mol}\end{array}$

Number of molecules that collides with hole in unit time is expressed as follows:

  $Z_A=\frac{n}{24\text{ h}}$

Where,

• $Z_A$is number of collisions per unit area.
• $n$is number of molecules.

Value of $n$ is $4.87\times {10}^{-8}\text{ mol}$ .

Substitute value in above equation.

  $\begin{array}{c}{Z}_A=\frac{n}{24\text{ h}}\\ =\frac{4.87\times {10}^{-8}\text{ mol}}{24\text{ h}\left(\frac{3600\text{ s}}{1\text{ h}}\right)}\\ =3.39\times {10}^{11}\text{ molecules/s}\end{array}$

The amount of gas is calculated as follows:

  $PV=nRT$

Rearrange above equation for $\frac{n}{V}$ .

  $\frac{n}{V}=\frac{P}{RT}$

Where,

• $P$is pressure of gas.
• $V$is volume of gas.
• $n$isnumber of moles.
• $R$is universal gas constant.
• $T$is temperature.

Value of $P$ is $1\text{ atm}$ .

Value of $R$ is $0.0821\text{ atm L/mol K}$ .

Value of $T$ is $300\text{ K}$ .

Substitute values in above equation.

  $\begin{array}{c}\frac{n}{V}=\frac{P}{RT}\\ =\frac{\left(1\text{ atm}\right)}{\left(0.0821\text{ atm L/mol K}\right)\left(300\text{ K}\right)}\\ =4.06\times {10}^{-2}\text{ mol/L}\end{array}$

The formula to calculate molecules per volume is expressed as follows:

  $\begin{array}{c}\frac{N}{V}=\left(4.06\times {10}^{-2}\text{ mol/L}\right)\left(\frac{6.022\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)\left(\frac{{10}^3\text{ L}}{1{\text{ m}}^3}\right)\\ =2.45\times {10}^{25}{\text{ molecules m}}^{-3}\end{array}$

The expression for molar mass of gas is as follows:

  $\text{Molar mass of gas}=\left(\begin{array}{c}\left({\text{mass percent of N}}_2\right)\left({\text{molar mass of N}}_{\text{2}}\right)+\\ \left({\text{mass percent of O}}_2\right)\left({\text{molar mass of O}}_{\text{2}}\right)\end{array}\right)$

Value of mass percent of ${\text{N}}_2$ is $78\text{ \%}$ ,

Molar mass of ${\text{N}}_2$ is $28\text{ g/mol}$ .

Value of mass percent of ${\text{O}}_2$ is $22\text{ \%}$ ,

Molar mass of ${\text{O}}_2$ is $32\text{ g/mol}$ .

Substitute values in above equation.

  $\begin{array}{c}\text{Molar mass of gas}=\left(\begin{array}{c}\left({\text{mass percent of N}}_2\right)\left({\text{molar mass of N}}_{\text{2}}\right)+\\ \left({\text{mass percent of O}}_2\right)\left({\text{molar mass of O}}_{\text{2}}\right)\end{array}\right)\\ =\left(\begin{array}{c}\left(78\text{ \%}\right)\left(\frac{1}{100}\right)\left(28\text{ g/mol}\right)+\\ \left(22\text{ \%}\right)\left(\frac{1}{100}\right)\left(32\text{ g/mol}\right)\end{array}\right)\\ =28.9\text{ g/mol}\left(\frac{1\text{ kg}}{{10}^3\text{ g}}\right)\\ =2.89\times {10}^{-2}\text{ kg/mol}\end{array}$

The formula to calculate number of collisions per unit area is as follows:

  $Z_A=A\frac{N}{V}\sqrt{\frac{RT}{2\pi M}}$

Rearrange above equation for $A$ .

  $A=Z_A\frac{V}{N}\sqrt{\frac{2\pi M}{RT}}$

Where,

• $Z_A$is number of collisions per unit area.
• $A$is diameter.
• $\frac{N}{V}$is number of molecules per volume.
• $M$is molar mass of gas.
• $\pi$is constant term.
• $R$is universal gas constant.
• $T$is temperature.

Value of $Z_A$ is $3.39\times {10}^{11}\text{ molecules/s}$ .

Value of $\frac{N}{V}$ is $2.45\times {10}^{25}{\text{ molecules m}}^{-3}$ .

Value of $M$ is $2.89\times {10}^{-2}\text{ kg/mol}$ .

Value of $\pi$ is 3.14.

Value of $R$ is $8.314\text{ J/mol K}$ .

Value of $T$ is $300\text{ K}$ .

Substitute values in above equation.

  $\begin{array}{c}A=Z_A\frac{V}{N}\sqrt{\frac{2\pi M}{RT}}\\ =\left(\frac{3.39\times {10}^{11}\text{ molecules/s}}{2.45\times {10}^{25}{\text{ molecules m}}^{-3}}\right)\sqrt{\frac{2\left(3.14\right)\left(2.89\times {10}^{-2}\text{ kg/mol}\right)}{\left(8.314\text{ J/mol K}\right)\left(300\text{ K}\right)}}\\ =1.18\times {10}^{-16}{\text{ m}}^2\end{array}$

Hence, diameter of hole is $1.18\times {10}^{-16}{\text{ m}}^2$ .

Conclusion

Diameter of hole is $1.18\times {10}^{-16}{\text{ m}}^2$ .