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

Answer 1

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.

Answer 2

What pressure (in atm) would be exerted by 76 g of fluorine gas in a 1.50-liter vessel at -37oC