Asked Oct 15, 2019
Container A holds 742 mL of an ideal gas at 2.80 atm. Container B holds 174 mL of a different ideal gas at 4.50 atm.
If the gases are allowed to mix together, what is the resulting pressure?

Expert Answer

Step 1

Given, container A holds 742mL of an ideal gas at 2.80atm and container B holds 174mL of different ideal gas at 4.50atm.

The total volume of gases can be calculated as


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Total Volume 742 mL+174 mL 916 mL

Step 2

The pressure of ideal gas A in total container:


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P 2.80atm V742mL V916mL P.V P V2 2.80x742 916 2.268atm

Step 3

The pressure of ideal gas B...


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P 4.50atm P V 174mL V916mL PVPV P.V P 4.50x174 916 0.854 atm


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