A 3.00-L container with O₂ at 4.00 atm is connected to a 2.00-L container with Ne at 2.00 atm, resulting in a final pressure of 3.20 atm in the container.
We have 2 containers connected by a valve.
One container has a volume of 3.00 L and contains O₂ at 4.00 atm. The other container has a volume of 2.00 L and contains Ne at 2.00 atm.
When the valve is opened, the total volume is the sum of the 2 containers.
[tex]V_2 = 3.00 L + 2.00 L = 5.00 L[/tex]
Assuming constant temperature and ideal behavior, we can calculate the final pressure of each gas using Boyle's law.
O₂
[tex]P_1 \times V_1 = P_2 \times V_2\\\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{4.00atm \times 3.00L}{5.00L} = 2.40atm[/tex]
Ne
[tex]P_1 \times V_1 = P_2 \times V_2\\\\P_2 = \frac{P_1 \times V_1}{V_2} = \frac{2.00atm \times 2.00L}{5.00L} = 0.800atm[/tex]
The final pressure in the container will be the sum of the final pressures of the gases.
[tex]P = 2.40 atm + 0.800 atm = 3.20 atm[/tex]
A 3.00-L container with O₂ at 4.00 atm is connected to a 2.00-L container with Ne at 2.00 atm, resulting in a final pressure of 3.20 atm in the container.
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