Asked by ana
When the valve between the 2.00-L bulb, in which the gas pressure is 2.00 atm, and the 3.00-L bulb, in which the gas pressure is 4.50 atm, is opened, what will be the final pressure in the two bulbs? Assume the temperature remains constan
Answers
Answered by
DrBob222
One way to do this is to use PV = nRT and solve for n = PV/RT.
Calculate n for one bulb and n for the other bulb, add the two and calculate P from PV = nRT using the sum of the bulbs for n. I get 3.5. What do you use for R and T. Whatever values are convenient because they end up canceling.
Another way is to look at the equation for n = PV/RT. Since T is a constant and R is a constant, we can just leave them out of the equation.
Then n1 = PV = 2*2 = 4
n2 = 3.5*4 = 13.5
Total n = 17.5
17.5 = PV. V is the total of 3L + 2L = 5L. Solve for P.
Calculate n for one bulb and n for the other bulb, add the two and calculate P from PV = nRT using the sum of the bulbs for n. I get 3.5. What do you use for R and T. Whatever values are convenient because they end up canceling.
Another way is to look at the equation for n = PV/RT. Since T is a constant and R is a constant, we can just leave them out of the equation.
Then n1 = PV = 2*2 = 4
n2 = 3.5*4 = 13.5
Total n = 17.5
17.5 = PV. V is the total of 3L + 2L = 5L. Solve for P.
Answered by
T
What pressure (in atm) would be exerted by 76 g of fluorine gas in a 1.50-liter vessel at -37oC
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