Question

Two tanks (tank A and tank B) of gas are connected by a closed valve. Tank A is 5 liters and contains O2 gas at a pressure of 24 atm. Tank B is 3 liters and contains N2 gas at a pressure of 32 atm. Both tanks are held at the same temperature. The valve between the two tanks is opened and the gases are allowed to mix. After the gases have had time to mix, what is the partial pressure of the oxygen gas?

Answer 1

Answer:

15 atm

Explanation:

Tank A is V₁ = 5 liters and contains O₂ gas at a pressure (P₁) of 24 atm. When it is connected to tank B, the total volume V₂ is 5 L + 3 L = 8 L. We can calculate the final pressure (P₂) of O₂ using Boyle's law (assuming ideal behavior and constant temperature).

$P_{1}.V_{1}=P_{2}.V_{2}\\P_{2}=\frac{P_{1}.V_{1}}{V_{2}} =\frac{24atm.5L}{8L} =15atm$

The pressure of O₂ is independent of the pressure of N₂.

Answer 2

The partial pressure of O₂ after mixing of the two gase is 15 atm.

How does a change in volume affect gas pressure?

The relationship between gas volume and pressure at constant temperature is given by Boyle's law which statesthat the volume is inversely proportional to pressure.

From the data provided;

Volume of Tank A is V₁ = 5 liters

Tank A contains O₂ gas at a pressure (P₁) = 24 atm. When it is connected to tank B, total volume V₂ is 5 L + 3 L = 8 L.

The pressure of O₂ is independent of the pressure of N₂.

Using Boyle's law:

• P1V1 = P2V2

Then;

P2 = P1V1/V2

P2 = 24 × 5/8

P2 =15atm

Therefore, the partial pressure of O₂ is 15 atm.

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