A company that produces footballs uses a proprietary mixture of ideal gases to inflate their footballs. If the temperature of 230 grams [g] of gas mixture in a 15-liter [L] tank is maintained at 465 degrees Rankine [°R] and the tank is pressurized to 135 pound-force per square inch [psi], what is the molecular weight of the gas mixture in units of grams per mole

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whitneytr12 whitneytr12 · Answer 1 · 2020-09-30T02:52:29+02:00

Answer:

The molecular weight of the gas mixture is 35.38 g/mol.

Explanation:

The molecular weight of the gas can be found using the following equation:

[tex] M = \frac{m}{n} [/tex]

Where:

m: is the mass = 230 g

n: is the number of moles

First, we need to find the number of moles using Ideal Gas Law:

[tex] PV = nRT [/tex]

Where:

P: is the pressure = 135 psi

V: is the volume = 15 L

R: is the gas constant = 0.082 L*atm/(K*mol)

T: is the temperature = 465 °R (K = R*5/9)

[tex]n = \frac{PV}{RT} = \frac{135 psi*\frac{1 atm}{14.6959 psi}*15 L}{0.082 L*atm/(K*mol)*465*(5/9) K} = 6.50 moles[/tex]

Finally, the molecular weight of the gas is:

[tex] M = \frac{m}{n} = \frac{230 g}{6.50 moles} = 35.38 g/mol [/tex]

Therefore, the molecular weight of the gas mixture is 35.38 g/mol.

I hope it helps you!