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.

Respuesta :

Answer:

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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