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A mixture of 0.197 mol CO2 and 0.00278 mol H2O is held at 30.0 degrees celsius with a pressure of 2.50 atm. What is the partial pressure of each gas?

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CintiaSalazar

Considering the Dalton's partial pressure, the partial pressure of CO₂ is 2.465 atm and the partial pressure of H₂O is 0.035 atm.

Dalton's law

The pressure exerted by a particular gas in a mixture is known as its partial pressure.

So, Dalton's law states that the total pressure of a gas mixture is equal to the sum of the pressures that each gas would exert if it were alone:

[tex]P_{T} =P_{1} +P_{2} +...+P_{n}[/tex]

where n is the amount of gases in the mixture.

Dalton's partial pressure law can also be expressed in terms of the mole fraction of the gas in the mixture. The mole fraction is a dimensionless quantity that expresses the ratio of the number of moles of a component to the number of moles of all the components present.

So in a mixture of two or more gases, the partial pressure of gas A can be expressed as:

[tex]P_{A} =x_{A} P_{T}[/tex]

Partial pressure of CO₂ and H₂O

In this case, you know:

  • [tex]n_{CO_{2} }[/tex]= 0.197 moles
  • [tex]n_{H_{2}O }[/tex]= 0.00278 moles
  • [tex]n_{T }=n_{CO_{2} }+n_{H_{2}O }[/tex]= 0.197 moles + 0.00278 moles= 0.19978 moles
  • [tex]P_{T}= 2.50 atm[/tex]

Then:

  • [tex]x_{CO_{2}} =\frac{n_{CO_{2}} }{n_{T} }=\frac{0.197 moles}{0.19978 moles} = 0.986[/tex]
  • [tex]x_{H_{2}O} =\frac{n_{H_{2}O} }{n_{T} }=\frac{0.00278 moles}{0.19978 moles} =0.014[/tex]

Replacing in the Dalton's law:

[tex]P_{CO_{2} } =x_{CO_{2}} P_{T}[/tex]

[tex]P_{CO_{2} } =[/tex] 0.986× 2.50 atm

[tex]P_{CO_{2} } =[/tex] 2.465 atm

[tex]P_{H_{2}O } =x_{H_{2}O} P_{T}[/tex]

[tex]P_{H_{2}O } =[/tex] 0.014× 2.50 atm

[tex]P_{H_{2}O } =[/tex] 0.035 atm

Finally, the partial pressure of CO₂ is 2.465 atm and the partial pressure of H₂O is 0.035 atm.

Learn more about Dalton's partial pressure:

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