What is the pressure in mm of Hg, of a gas mixture that contains 1g of H2, and 8.0 g of Ar in a 3.0 L container at 27°C.

4362.4 mmHg

Explanation:

Step 1:

Determination of the number of mole of H2 and the number of mole of Ar.

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 = 1g

Number of mole of H2 = ?

Number of mole = Mass/Molar Mass

Number of mole of H2 = 1/2 = 0.5 mole

Molar Mass of Ar = 40g/mol

Mass of Ar = 8g

Number of mole of Ar = ?

Number of mole = Mass/Molar Mass

Number of mole of Ar = 8/40 = 0.2 mole

Step 2:

Data obtained from the question.

Volume (V) of the mixture = 3L

Temperature (T) = 27°C = 27°C + 273 = 300K

Number of mole (n) of the mixture = 0.5 + 0.2 = 0.7 mole

Gas constant (R) = 0.082atm. L/Kmol

Pressure (P) of the mixture = ?

Step 3:

Determination of the pressure of the mixture.

The pressure of the mixture can be obtained as follow:

PV = nRT

P x 3 = 0.7 x 0.082 x 300

Divide both side by 3

P = (0.7 x 0.082 x 300) / 3

P = 5.74 atm

Step 4:

Conversion of the pressure obtained from atm to mmHg.

1 atm = 760mmHg

Therefore, 5.74 atm = 5.74 x 760 = 4362.4 mmHg