Question

PLEASE HELPPP?

At 13.2°C and 103.8 kPa, the density of dry air is 1.26 g/L. What is the average "molar mass" of dry air at these conditions?

Answer 1

@satellite73 help?

Answer 2

density = \[\frac{ MP }{ RT }\]

Answer 3

Where M is the molar mass and R is the gas constant.

Answer 4

This formula can be derived from the ideal gas law.

Answer 5

R = 8.314 J/mol-K in this example I believe.

Answer 6

ya i knowthat but show me step by step?

Answer 7

Oh gosh, I don't wanna screw it up for you, but I'll try. So, we're solving for molar mass, M.

M = (dRT)/P

Answer 8

First, we need temperature in Kelvin. Temp(K) = Temp(C) + 273.15

Answer 9

no worries. ill catch it if you mess up i hope. I just dont know how to do it step by step. i know the equation but ya.

Answer 10

286.35k is the temp

Answer 11

M = [(1.26g/L)(8.314 J/mol-K)(286.35K)]/103.8kPa

Answer 12

That'll give you the molar mass in g/mol

Answer 13

so you just plugged these into the equation? and that's it?
btw howd you know the rate? that 8.314? like where does that come from? is it constant for everything?

Answer 14

Yes, it's one form of the gas constant.

Answer 15

theres many forms?

Answer 16

Yeah, but they are all derived from each other b/c of the different units that can be in the gas constant. Just depends on the problem.

Answer 17

wlp i guess i should study that. thanks!!!! i have another question, mind helping me on that also?

Answer 18

Perhaps lol.