Question

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

Answer 1

Answer:

$P_{T}$ = 2.94 atm

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:

$n_{O_{2}} = \frac{m}{M}$

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

$n_{O_{2}} = \frac{32.00 g}{31.998 g/mol} = 1.00 moles$  

And the number of moles of helium is:

$n_{He} = \frac{8.00 g}{4.0026 g/mol} = 2.00 moles$

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

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

$P_{He} = \frac{nRT}{V} = \frac{2.00 moles*0.082 Latm/(Kmol)*298 K}{25.0 L} = 1.96 atm$

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!