Chemistry, 13.10.2020 20:01 therronvictorjr

What will happen when the valve between a 2.00-L bulb, in which the gas pressure is 2.00 atm, and a 3.00-L bulb, in which the gas pressure is 4.50 atm, is opened? Assume the temperature remains constant. (For this question, label each statement as True or False and give an explanation/justification for each of your answers to receive credit. No credit is given for only True or False answers with no adequate justification. To receive credit you must provide justification. You may have to give numerical answers followed by verbal explanations. You can use any reasoning that you see fit but be as thorough as possible.) The two gases will mix and react.
The two gases will remain separate and will not mix.
The two gases will occupy a volume of 5.0 L and the final pressure in the two bulbs will be 6.50 atm.
The two gases will occupy a volume of 5.0 L and the final pressure in the two bulbs will be 3.50 atm.
The two gases will occupy a volume of 5.0 L and the final pressure in the two bulbs will be 3.25 atm.

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Answer from: bunbun2913

- The two gases will mix and react.: FALSE.

- The two gases will remain separate and will not mix: FALSE.

- The two gases will occupy a volume of 5.0 L and the final pressure in the -two bulbs will be 6.50 atm.: FALSE.

- The two gases will occupy a volume of 5.0 L and the final pressure in the two bulbs will be 3.50 atm.: TRUE.

- The two gases will occupy a volume of 5.0 L and the final pressure in the two bulbs will be 3.25 atm: FALSE.

Explanation:

Hello.

In this case, given the options, since the total volume of the container includes the both of them, we find that:

- The two gases will mix and react: FALSE, since we do not know the identity of the gases which could be the same or two different inert gases.

- The two gases will remain separate and will not mix: FALSE, since as the valve is opened, the total gas will occupy the entire volume as the volume of a gas is the same to the container based on its constant molecules movements.

- The two gases will occupy a volume of 5.0 L and the final pressure in the two bulbs will be 6.50 atm: FALSE, since in this case, by using the Boyle's law for the first compartment we obtain the pressure of the gas there:

$P_2V_2=P_1V_1\\\\P_2=\frac{P_1V_1}{V_2} =\frac{2.00atm*2.00L}{5.00L} =0.8atm$

Now, we reuse it for the gas at the 3.00-L bulb to find its final pressure:

$P_2V_2=P_1V_1\\\\P_2=\frac{P_1V_1}{V_2} =\frac{2.00atm*3.00L}{5.00L} =2.7atm$

So the final pressure is:

$P_{TOTAL}=0.8atm+2.7atm\\\\P_{TOTAL}=3.5atm$

- The two gases will occupy a volume of 5.0 L and the final pressure in the two bulbs will be 3.50 atm: TRUE considering the total pressure computed above.

- The two gases will occupy a volume of 5.0 L and the final pressure in the two bulbs will be 3.25 atm: FALSE since the final pressure is 3.5 atm.

Regards.

Answer from: Quest

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.

Answer from: Quest

The answer is b metal denting

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What will happen when the valve between a 2.00-L bulb, in which the gas pressure is 2.00 atm, and a...