A student has a 2.19 L bottle that contains a mixture of O 2 , N 2 , and CO 2 with a total pressure of 5.57 bar at 298 K . She knows that the mixture contains 0.221 mol N 2 and that the partial pressure of CO 2 is 0.318 bar . Calculate the partial pressure of O 2.

The partial pressure of oxygen gas is 2.76 bar

Explanation:

To calculate the number of moles, we use the equation given by ideal gas which follows:

where,

P = pressure of the gas = 5.57 bar

V = Volume of the gas = 2.19 L

T = Temperature of the gas = 298 K

R = Gas constant =

n = Total number of moles = ?

Putting values in above equation, we get:

To calculate the mole fraction of carbon dioxide, we use the equation given by Raoult's law, which is:

        ........(1)

where,

= partial pressure of carbon dioxide = 0.318 bar

= total pressure = 5.57 bar

= mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas,

Calculating the partial pressure of oxygen gas by using equation 1, we get:

Mole fraction of oxygen gas = (1 - 0.0571 - 0.448) = 0.4949

Total pressure of the system = 5.57 bar

Putting values in equation 1, we get:

Hence, the partial pressure of oxygen gas is 2.76 bar

the answer for sodium is 0.186

the answer for rubidium is 0.251

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