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Dry air near sea level has the following composition by volume: N2, 78.08 percent; O2, 20.94 percent; Ar, 0.93 percent; CO2, 0.05 percent. The atmosphere pressure is 1.00 atm. Calculate (a) the partial pressure of each gas in atmospheres and (b) the concentration of each gas in mol/L at 0 °C (Hint: Because volume is proportional to the number of moles present, mole fractions of gases can be expressed as ratios of volumes at the same temperature and pressure.)

Respuesta :

Explanation:

Suppose in 100 KL mixture of gas:

1) Percentage of volume of the nitrogen gas = 78.08 %

Mole fraction of nitrogen gas= 0.7808

Pressure of the mixture, P = 1 atm

Mole fraction of nitrogen : [tex]\chi_1[/tex]

Partial pressure of the nitrogen gas = [tex]P_1=P\times \chi_1[/tex] (Dalton's law)

[tex]1atm \times 0.7808 = 0.7809 atm[/tex]

[tex]P_1V_1=n_1RT[/tex] (ideal gas equation)

Concentration of nitrogen gas can be written as:

[tex]\frac{n_1}{V_1}=\frac{P_1}{RT}=\frac{0.7809 atm}{0.0821 atm L/mol K\times 273.15 K}=0.0348 mol/L[/tex]

2)Percentage of volume of the oxygen gas = 20.94%

Mole fraction of oxygen gas= 0.2094

Pressure of the mixture , P= 1 atm

Mole fraction of oxygen gas: [tex]\chi_2[/tex]

Partial pressure of the oxygen  gas = [tex]P_2=P\times \chi_2[/tex] (Dalton's law)

[tex]1atm \times 0.2094 = 0.2094 atm[/tex]

[tex]P_2V_2=n_2RT[/tex] (ideal gas equation)

Concentration of oxygen gas can be written as:

[tex]\frac{n_2}{V_2}=\frac{P_2}{RT}=\frac{0.2094 atm}{0.0821 atm L/mol K\times 273.15 K}=0.00934 mol/L[/tex]

3)Percentage of volume of the argon gas = 0.93%

Mole fraction of argon gas= 0.0093

Pressure of the mixture , P= 1 atm

Mole fraction of argon: [tex]\chi_3[/tex]

Partial pressure of the argon gas = [tex]P_3=P\times \chi_3[/tex] (Dalton's law)

[tex]1atm \times 0.0093= 0.0093 atm[/tex]

[tex]P_3V_3=n_3RT[/tex]  (ideal gas equation)

Concentration of argon gas can be written as:

[tex]\frac{n_3}{V_3}=\frac{P_3}{RT}=\frac{0.0093 atm}{0.0821 atm L/mol K\times 273.15 K}=0.000415 mol/L[/tex]

4)Percentage of volume of the carbon dioxide gas = 0.05%

Mole fraction of carbon dioxide gas= 0.0005

Pressure of the mixture, P = 1 atm

Mole fraction of carbon dioxide : [tex]\chi_4[/tex]

Partial pressure of the carbon dioxide gas = [tex]P_4=P\times \chi_4[/tex] (Dalton's law)

[tex]1atm \times 0.0005= 0.0005 atm[/tex]

[tex]P_4V_4=n_4RT[/tex]  (ideal gas equation)

Concentration of carbon dioxide gas can be written as:

[tex]\frac{n_4}{V_4}=\frac{P_4}{RT}=\frac{0.0005 atm}{0.0821 atm L/mol K\times 273.15 K}=2.23\times 10^{-5}mol/L[/tex]

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