The valve between a 2.00-l bulb, in which the gas pressure is 1.80 atm, and a 3.00-l bulb, in which the gas pressure is 3.40 atm, is opened. what is the final pressure in the two bulbs, the temperature remaining constant?

2.76 atm

Explanation:

Boyle's law states that, for an isothermic process (temperature remaining the same), the product of the pressure and volume is constant.

Dalton's law states that in a gas mixture, the total pressure is the sum of the partial pressure of the components.

So:

P1*V1 + P2*V2 = P*V

Where P1 is the pressure in the bulb 1, V1 is the volume in the bulb 1, P2 is the pressure in the bulb 2, V2 is the volume in the bulb 2, P is the pressure at the mixture after the valve was opened, and V is the final volume (5.00 L).

1.80*2.00 + 3.40*3.00 = P*5.00

5P = 13.80

P = 2.76 atm

We have that for the Question "The valve between a 2.00-L bulb, in which the gas pressure is 1.80 atm, and a 3.00-L bulb, in which the gas pressure is 3.40 atm, is opened. What is the final pressure in the two bulbs, the temperature remaining constant?" it can be said that the final pressure in the two bulbs, the temperature remaining constant

P_f=2.44atm

From the question we are told

The valve between a 2.00-L bulb, in which the gas pressure is 1.80 atm, and a 3.00-L bulb, in which the gas pressure is 3.40 atm, is opened. What is the final pressure in the two bulbs, the temperature remaining constant?

Generally the equation for the ideal gas  is mathematically given as

Where

Where

Therefore

P_f=2.44atm

Hence, the final pressure in the two bulbs, the temperature remaining constant

P_f=2.44atm

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