Answer:

The total pressure is 61.4 kPa, and the volume is 1.31 L.

Explanation:

Based on the given information, a gas mixture comprising 320 mg or 0.320 grams of CH4, 175 mg or 0.175 grams of Ar, and 225 or 0.225 grams of N. The number of moles of the gases presents within the mixture can be determined by using the formula,

Number of moles = Mass/ molecular mass

The molecular mass of methane is 16.04 grams per mole, the molecular mass of Argon is 40 grams per mole, and the molecular mass of Nitrogen is 28.02 grams per mole.

Now, the number of moles of CH4 is,

= 0.320 grams/ 16.04 grams per mole

= 0.0199 moles

The number of moles of Ar is,

= 0.175 grams/40 grams per mole

= 0.0044 moles

The number of moles of N2 is,

= 0.225 grams/28.02 grams per mole

= 0.0080 moles

The partial pressure of nitrogen given is 15.2 kPa or 0.15 atm. Thus, the partial pressure of other two gases will be,

CH4 = (15.2 kPa) (0.0199 moles)/(0.0080 moles)

= 37.8 kPa

Ar = (15.2 kPa) (0.0044 moles)/(0.0080 moles)

= 8.36 kPa

Therefore, the total pressure is 15.2 + 37.8 + 8.36 = 61.4 kPa or 0.606 atm

The total volume can be determined by using the formula,

V = nRT/P

Here n is the total number of moles of the gas, which is 0.0323 moles.

Now putting the values we get,

V = (0.0323 moles) (0.0826 atm*L/mol*K)(300 K)/(0.606 atm)

V = 1.31 L