Question

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.

Answer 1

Answer: The partial pressure of oxygen gas is 2.76 bar

Step-by-step explanation:

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

`PV=nRT`

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 =
`0.0831\text{ L bar }mol^(-1)K^(-1)`

n = Total number of moles = ?

Putting values in above equation, we get:

`5.57bar* 2.19L=n* 0.0831\text{ L. bar }mol^(-1)K^(-1)* 298K\\\\n=(5.57* 2.19)/(0.0831* 298)=0.493mol`

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

`p_(A)=p_T* \chi_(A)` ........(1)

where,

`p_A` = partial pressure of carbon dioxide = 0.318 bar

`p_T` = total pressure = 5.57 bar

`\chi_A` = mole fraction of carbon dioxide = ?

Putting values in above equation, we get:

`0.318bar=5.57bar* \chi_(CO_2)\\\\\chi_(CO_2)=(0.381)/(5.57)=0.0571`

• Mole fraction of a substance is given by:

`\chi_A=(n_A)/(n_A+n_B)`

We are given:

Moles of nitrogen gas = 0.221 moles

Mole fraction of nitrogen gas,
`\chi_(N_2)=(0.221)/(0.493)=0.448`

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:

`p_(O_2)=5.57bar* 0.4949\\\\p_(O_2)=2.76bar`

Hence, the partial pressure of oxygen gas is 2.76 bar