Final Pressure in Two-Bulb Apparatus After Valve Opening

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To find the final pressure in a two-bulb apparatus after opening the valve, calculate the partial pressures contributed by each bulb. Bulb #1, with a pressure of 1.3 atm and a volume of 4.27 L, and Bulb #2, with a pressure of 6.6 atm and a volume of 7.40 L, will mix without any reactions at constant temperature. The final pressure is determined by adding the partial pressures from both bulbs once the valve is opened. This approach assumes that the gases behave ideally and that the total volume available for the gases is the sum of the volumes of both bulbs. The final pressure can be calculated using the ideal gas law principles.
J-Nast
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Consider an apparatus where there are two rigid glass bulbs separated from one another by a valve that is closed initially. Bulb #1 has a pressure of 1.3 atm and a volume of 4.27 L and bulb #2 has a pressure of 6.6 atm and a volume of 7.40L. What is the final pressure (in atm) in the apparatus after the valve is opened so that the gases from each side can mix? Assume that the T remains constant and there is no reaction between gases.

I really have no clue where to start with this one, any help would be wonderful. :wink:
 
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you could do it using partial pressures. how much pressure does the gas on bulb 1 put on the entire system and how much does the gas in bulb 2 put in the system? (basically assume that bulb 2 is empty when opened for the first part and the opposite for the second). The pressures will then add to be the final pressure.
 
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