posted by Cliff on .
Please let me know how I'm doing with the following chemistry problems. I appreciate your imput.
Here's problem 1:
A sample of Helium gas occupies a volume of 29.4mL AT 18deg.celcius.
(a)Would the volume of this gas sample be doubled at 36deg.celcius?
(b)At what temperature(degree celcius)would the volume of the gas be cut in half?
Solution (a): finding V2
=29.4mL * 309k/291k = 31.2mL
If 3.25 mol of Argon gas occupies a volume of 100L,what volume does 14.15 mol occupy?
Since 1 mol of a gas at STP= 22.4L
14.15 mol Ar * (39.95/1 mol Ag)*(22.4L/1 mol Ar)= 12,226L
The way I've solved question 2 looks strange.
Note the correct spelling of celsius.
1a. The work you did is ok but you didn't answer the question. The answer is no, the volume will not be doubled by doubling the celsius T.
2b. You halved the volume but the temperature went up. Something wrong here.
I would use 29.4 mL as V1 and 1/2 that for V2.
3.You can do this one of two ways but I don't think your way is one of them.
You can use PV = nRT (assume a value for T or V an calculate the other, then redo PV = nRT and substitute 14.15 for n and recalculate the volume.
The easier way is
100 L x (14.15 mols/3.25 mols) = ?? L
You have been really helpful..may I have your suggestion.. as to the approach to solving these 3 chemistry questions please.I'm new to the topic area:PARTIAL PRESSURES
1.A gas mixture consists of 6.91g of N2 gas,4.71g of O2 gas and 2.95g of He.What volume does the mixture occupy at 28degrees celsius and 1.05 atm?
2.A 50.0L tank contains 5.21kg of N2,4.49kg of O2. What is the pressure of the tank at 24degrees celsius?
3.A tank contains a mixture of 3.0 mol of N2,2.0mol of O2,and 1.0 CO2 at 25degrees celsius and a total pressure of 10.0atm. Calculate the partial pressure of each gas in the mixture.