Question

A 25.0 liter rigid container has a mixture of 32.00 grams of oxygen gas and 1 point

8.00 grams of helium gas at 298K. What is the total pressure in the

container?

Answer 1

`P_(T)` = 2.94 atm

Step-by-step explanation:

The total pressure (
`P_(T)`) in the container is given by:

`P_(T) = P_{O_(2)} + P_(He)`

The pressure of the oxygen (
`P_{O_(2)}`) and the pressure of the helium (
`P_(He)`) can be calculated using the ideal gas law:

`PV = nRT`

Where:

V: is the volume = 25.0 L

n: is the number of moles of the gases

R: is the gas constant = 0.082 Latm/(Kmol)

T: is the temperature = 298 K

First, we need to find the number of moles of the oxygen and the helium:

`n_{O_(2)} = (m)/(M)`

Where m is the mass of the gas and M is the molar mass

`n_{O_(2)} = (32.00 g)/(31.998 g/mol) = 1.00 moles`

And the number of moles of helium is:

`n_(He) = (8.00 g)/(4.0026 g/mol) = 2.00 moles`

Now, we can find the pressure of the oxygen and the pressure of the helium:

`P_{O_(2)} = (nRT)/(V) = (1.00 moles*0.082 Latm/(Kmol)*298 K)/(25.0 L) = 0.98 atm`

`P_(He) = (nRT)/(V) = (2.00 moles*0.082 Latm/(Kmol)*298 K)/(25.0 L) = 1.96 atm`

Finally, the total pressure in the container is:

`P_(T) = P_{O_(2)} + P_(He) = 0.98 atm + 1.96 atm = 2.94 atm`

Therefore, the total pressure in the container is 2.94 atm.

I hope it helps you!