Chapter 5, Problem 121E

Interpretation Introduction

Interpretation: For the given data, partial pressure, number of molecules per cubic meter and number of molecules per cubic centimeter of carbon monoxide should be determined.

Concept introduction:

• Mixing ratio is used to express the concentration of trace compounds in air. Mixing ratios are often expressed as ppmv (parts per million volume);

  $ppmvofX=\frac{volumeofXatSTP}{totalvolumeofairatSTP}\times 10^6$

• A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture.

  Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

  $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

  Partial pressure of a gas in terms of its mole fraction ( ${\text{χ}}_{\text{A}}$ ) and total pressure is,

${\text{P}}_{\text{A}}\text{=}{\text{χ}}_{\text{A}}\text{×}{\text{P}}_{\text{TOTAL}}$

• By combining the three gaseous laws namely Boyle’s law, Charles’s law and Avogadro’s law a combined gaseous equation is obtained. This combined gaseous equation is called Ideal gas law.

According to ideal gas law,

$PV=nRT$

Where,

P = pressure in atmospheres

V= volumes in liters

n = number of moles

R =universal gas constant ( $\text{0}\text{.08206L×atm/K×mol}$)

T = temperature in kelvins

Total number of moles of gases in the mixture of gases can be determined by using ideal According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\text{=}\frac{\text{Total pressure}\times \text{Volume}}{\text{R}\text{×}\text{Temperature}}$

•   $\text{Number}\text{of}\text{molecules}\text{=Number}\text{of}\text{moles}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{mol}}$

Answer to Problem 121E

Answer

• Partial pressure of carbon monoxide $P_{CO}=0.19torr$
• Number of molecules per cubic meter of carbon monoxides $=\frac{6.6\times 10^{21}molecules}{m^3}$
• Number of molecules per cubic  centimeter of carbon monoxides $\frac{6.6\times 10^{15}molecules}{c{m}^3}$

Explanation of Solution

Explanation

• To determine: The partial pressure of carbon monoxide

Partial pressure of carbon monoxide $P_{CO}=0.19torr$

Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone. Partial pressure of a gas in terms of its mole fraction ( ${\text{χ}}_{\text{CO}}$) and total pressure is,

` ${\text{P}}_{\text{CO}}\text{=}{\text{χ}}_{\text{CO}}\text{×}{\text{P}}_{\text{TOTAL}}$ (1)

A mole fraction of a molecule in a mixture is the ratio of number of moles of particular molecule to the sum of number of moles of all molecules in the mixture.

$\begin{array}{l}\text{To}\text{calculate}\text{the}\text{partial}\text{pressure,molefraction}\text{and}\text{volume}\text{of}\text{carbon}\text{monoxide}\text{should}\text{be}\text{determined}\\ \text{mole}\text{fraction}\text{of}\text{CO,}\text{(}{\text{χ}}_{\text{CO}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{ CO}{\text{(n}}_{\text{CO}}\text{)}}{\text{total number of moles}\text{of}\text{mixture}\left({\text{n}}_{\text{total}}\right)}\\ \text{Since,volume}\text{of}\text{a}\text{gas}\text{is}\text{directly}\text{proportional}\text{to}\text{the}\text{number}\text{of}\text{moles}\text{of}\text{gas,}\\ \text{That}\text{is,}\text{Vµ}\text{n}\\ \text{So,mole}\text{fraction}\text{in}\text{terms}\text{of}\text{volume}\text{can}\text{be}\text{written}\text{as:}\\ {\text{χ}}_{\text{CO}}\text{=}\frac{\text{volume}\text{of}\text{carbonmonoxide}\left({\text{V}}_{\text{CO}}\right)}{\text{Total}\text{volume}\text{of}\text{gas}\left({\text{V}}_{\text{total}}\right)}\mathrm{...}\left(\text{2}\right)\\ \text{From}\text{the}\text{given}\text{data,}\\ \text{Concentration}\text{of}\text{carbon}\text{monoxide}\text{inthe}\text{air}\text{is}\text{3}{\text{.0×10}}^{\text{2}}\text{ppmv}\\ \text{ppmv}\text{of}\text{X}\text{=}\frac{\text{volume}\text{of}\text{X}\text{at}\text{STP}}{\text{total}\text{volume}\text{of}\text{air}\text{at}\text{STP}}{\text{×10}}^{\text{6}}\\ \text{If}\text{we}\text{have}\text{1}{\text{.0×10}}^{\text{6}}\text{L}\text{of}\text{air}\left({\text{V}}_{\text{total}}\right)\\ \text{3}{\text{.0×10}}^{\text{2}}\text{ppmv}\text{=}\frac{{\text{V}}_{\text{CO}}}{\text{1}{\text{.0×10}}^{\text{6}}\text{L}}{\text{×10}}^{\text{6}}\\ V_{CO}=\frac{3.0\times 10^{2}ppmv\times 1.0\times 10^{6}L}{10^6}=3.0\times 10^{2}L\\ \text{By}\text{adding}\text{this}\text{values}\text{to}\text{the}\text{equation}\left(\text{2}\right)\\ {\chi }_{CO}=\frac{3.0\times 10^{2}L}{1.0\times 10^{6}L}=3.0\times 10^{-4}\\ \text{By}\text{adding}\text{the}\text{value}\text{of}\text{mole}\text{fraction}\text{to}\text{the}\text{equation}\left(\text{1}\right)\\ {\text{P}}_{\text{CO}}\text{=}{\text{χ}}_{\text{CO}}\text{×}{\text{P}}_{\text{TOTAL}}\\ {\text{χ}}_{\text{CO}}\text{=}\text{3}{\text{.0×10}}^{\text{-4}}\\ \text{From}\text{the}\text{given}\text{data}\\ {\text{P}}_{\text{TOTAL}}\text{=}\text{628}\text{torr}\\ {\text{P}}_{\text{CO}}\text{=}\text{3}{\text{.0×10}}^{\text{-4}}\text{×628}\text{torr=}\text{0}\text{.19torr}\end{array}$

• To determine: Number of molecules per cubic meter of carbon monoxides.

Number of molecules per cubic meter of carbon monoxides $=\frac{6.6\times 10^{21}molecules}{m^3}$

Total number of moles of gases in the mixture of gases can be determined by using ideal gas equation.

According to ideal gas equation,

$\text{Total number}\text{of}\text{moles}\left(\text{n}\right)\text{=}\frac{\text{Total pressure}\left(\text{P}\right)\text{×}\text{Volume}\left(\text{V}\right)}{\text{R}\text{×}\text{Temperature}\left(\text{T}\right)}$

Number of molecules per cubic meter of carbon monoxides, ${\text{n}}_{\text{CO}}\text{=}\frac{{\text{P}}_{\text{CO}}\times \text{V}}{\text{R}\times \text{T}}$

$\begin{array}{l}\text{Here,}\\ \text{Assuming}\text{1}{\text{.0m}}^{\text{3}}\text{air,}\text{1}{\text{m}}^{\text{3}}\text{=}\text{1000}\text{L}\\ \text{V=}\text{1}{\text{.0×10}}^{\text{3}}\text{L}\\ {\text{P}}_{\text{CO}}\text{=0}\text{.19}\text{torr}\text{=}\text{0}\text{.00025}\text{atm}\text{Since,}\text{1}\text{torr}\text{=}\text{760}\text{atm}\\ \text{0}\text{.19torr=}\frac{\text{0}\text{.19}}{\text{760}}\text{atm}\\ \text{=}\text{0}\text{.00025}\text{atm}\\ \text{R}\text{=}\frac{\text{0}\text{.08206}\text{L}\text{atm}}{\text{K mol}}\\ \text{T}\text{=0°C}\text{=}\text{273K}\text{Since,Kelvin=}\text{°C+273}\\ \text{=}\text{0°C+273}\\ \text{=}\text{273K}\\ {\text{n}}_{\text{CO}}\text{=}\frac{\text{0}\text{.00025atm×1}{\text{.0×10}}^{\text{3}}\text{L}}{\frac{\text{0}\text{.08206}\text{L}\text{atm}}{\text{K mol}}\text{×273K}}\\ \text{=}\text{1}\text{.1}\text{×}{\text{10}}^{\text{-2}}\text{mol}\text{CO}\\ \text{Number}\text{of}\text{molecules}\text{=Number}\text{of}\text{moles}\text{×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{mol}}\\ \text{=}\text{1}\text{.1}\text{×}{\text{10}}^{\text{-2}}\text{mol×}\frac{\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}}{\text{mol}}\\ \text{=}\text{6}{\text{.6×10}}^{\text{21}}\text{CO}\text{molecules}\text{in}\text{1}{\text{.0m}}^{\text{3}}\text{of}\text{air}\end{array}$

• To determine: Number of molecules per cubic centimeter of carbon monoxides

Number of molecules per cubic centimeter of carbon monoxides $\frac{6.6\times 10^{15}molecules}{c{m}^3}$

$\begin{array}{l}\text{Since,}\\ \text{1}{\text{m}}^{\text{3}}{\text{=10}}^{\text{6}}{\text{cm}}^{\text{3}}\text{and}\\ \text{Number of molecules per cubic meter of carbon monoxides=}\frac{\text{6}{\text{.6×10}}^{\text{21}}\text{molecules}}{{\text{m}}^{\text{3}}}\\ \\ \text{Number of molecules per cubic centimeter of carbon monoxides}\text{=}\frac{\frac{\text{6}{\text{.6×10}}^{\text{21}}\text{molecules}}{{\text{m}}^{\text{3}}}}{{\text{10}}^{\text{6}}{\text{cm}}^{\text{3}}}\\ \\ \text{=}\frac{\text{6}{\text{.6×10}}^{\text{15}}\text{molecules}\text{CO}}{{\text{cm}}^{\text{3}}}\end{array}$

Conclusion

Conclusion

Partial pressure, number of molecules per cubic meter and number of molecules per cubic centimeter of carbon monoxide is determined on the basis of respective equations.