Atank contains a mixture of 52.5g oxygen gas and 65.1g carbon dioxide gas at 27c. the total pressure in the tank is 9.21 atm calculate the partial pressure of each gas in the containter

Taking into account the Dalton's partial pressure law, the partial pressure of oxygen and carbon dioxide in the containter is 4.84 atm and 4.37 atm  respectively.

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:

PT = PA + PB

This relationship is due to the assumption that there are no attractive forces between the gases.

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.

The mole fraction of a gas in a gas mixture is given by:

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

PA = xA× PT

In this case, you know that a tank contains a mixture of 52.5g oxygen gas and 65.1g carbon dioxide gas at 27 °C. Considering the molar mass of each compound, that is, the amount of mass that a substance contains in one mole, it is possible to calculate the number of moles present in the tank as:

Then, the total number of moles of all the components present is:

Total moles = (1.641 + 1.479) mol = 3.120 mol

So, the mole fraction of each gas is

Finally, the partial pressure of each gas in the containter can be calculated as:

0.5259× 9.21 atm= 4.84 atm

0.4741× 9.21 atm= 4.37 atm

In summary, the partial pressure of oxygen and carbon dioxide in the containter is 4.84 atm and 4.37 atm  respectively.

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The partial pressure of oxygen is 4.84 atm, and the partial pressure of carbon dioxide is 4.37 atm.

Moles of O₂ = 52.5 g O₂ × (1 mol O₂/32.00 g O₂) = 1.641 mol O₂

Moles of CO₂ = 65.1 g CO₂ × (1 mol CO₂/44.01 g CO₂) = 1.479 mol CO₂

Total moles = (1.641 + 1.479) mol = 3.120 mol

Let O₂ be Gas 1 and CO₂ be Gas 2.

χ₁= 1.641/3.120 = 0.5259

χ₂= 1.479/3.120 = 0.4741

p₁ = χ₁p_tot = 0.5259 × 9.21 atm = 4.84 atm

p₂ = χ₂p_tot = 0.4741 × 9.21 atm = 4.37 atm

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