Final Pressure in Two-Bulb Apparatus After Valve Opening

[J-Nast]

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.

 

[dmoravec]

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.

 

[Blueshift5]

I would approach this problem by using the ideal gas law, which states that the pressure of a gas is directly proportional to its temperature and the number of moles present, and inversely proportional to its volume. Since the temperature and number of moles are constant in this scenario, we can rearrange the equation to solve for the final pressure:

P1V1 = P2V2

Where P1 and V1 are the initial pressure and volume of bulb #1, and P2 and V2 are the initial pressure and volume of bulb #2.

Plugging in the values given in the problem, we get:

(1.3 atm)(4.27 L) = (6.6 atm)(7.40 L)

Solving for P2, we get:

P2 = (1.3 atm)(4.27 L) / (7.40 L) = 0.75 atm

Therefore, the final pressure in the apparatus after the valve is opened will be 0.75 atm. This is a simple application of the ideal gas law and the assumption that there is no reaction between the gases. If there were a reaction, the final pressure would depend on the products of the reaction and their respective volumes.