Final Pressure in Two-Bulb Apparatus After Valve Opening

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SUMMARY

The final pressure in a two-bulb apparatus after opening the valve can be calculated using the principles of partial pressures. Bulb #1 contains gas at 1.3 atm and 4.27 L, while Bulb #2 has gas at 6.6 atm and 7.40 L. Upon opening the valve, the total pressure is determined by the combined contributions of both bulbs, assuming constant temperature and no chemical reaction. The final pressure is the sum of the partial pressures from both bulbs.

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  • Understanding of gas laws, specifically Dalton's Law of Partial Pressures
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  • Basic algebra for calculations involving pressure and volume
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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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