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

Question

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.)

1.the two gases will mix and react

2. The two gases will remain separate and will not mix

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

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

5. The two gases will occupy a volume of 5.0 and the final pressure in the two bulbs will be 3.25 atm

Explanation / Answer

As per Boyles law PV = constant ( snce temperature constant)

For two liter container PV= 2*2= 4L.atm

For 3 liter container PV= 3*4.5= 13.5

when they are mixed total PV= 13.5+4= 17.5 L.atm

Volume after mixing= 2+3= 5L

P(after mixing )*5= 17.5

P= 17.5/5= 3.5 atm

So after mixing the pressure will be at 3.5 atm   ( 4 is correct)

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