The stopcock connecting a 3.34 L bulb containing argon gas at a pressure of 8.95 atm, and a 8.00 L bulb containing oxygen gas at a pressure of 2.15 atm, is opened and the gases are allowed to mix. Assuming that the temperature remains constant, the final pressure in the system is atm.

The final pressure in the system is 4.153 atm

Explanation:

Given;

volume of Argon, V(Ar) = 3.34 L

pressure of Argon, P(Ar) = 8.95 atm

volume of oxygen, V(O₂) = 8.00 L

pressure of oxygen, P(O₂) = 2.15 atm

According to Boyle's law;

P₁V₁ = P₂V₂

Where;

P₁ is initial pressure of the gases

P₂ is the final pressure of the gases

V₁ is initial volume of the gases

V₂ is the final volume of the gases

We need to determine the final pressure of the gases;

Total volume of the gas mixtures = 3.34 L + 8.00 L = 11.34 L

Final pressure of Argon in the gas mixture = (8.95 atm) x ( 3.34 L / 11.34 L) = 2.636 atm

Final pressure of oxygen in the gas mixture = (2.15 atm) x (8.00 L / 11.34 L) = 1.517 atm

The final pressure in the system =  2.636 atm + 1.517 atm = 4.153 atm

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