# Gases #1

• October 4th 2009, 11:47 AM
jsu03
Gases #1

One mole of nitrogen and one mole of neon are combined in a closed container at STP.

How big is the container?

I did PV=nRT and V= nRT/P so V= ?(0.08206)(273)/1 atm... so I need to find n in moles but I dont know how. Thanks again!
• October 4th 2009, 11:59 AM
e^(i*pi)
Quote:

Originally Posted by jsu03

One mole of nitrogen and one mole of neon are combined in a closed container at STP.

How big is the container?

I did PV=nRT and V= nRT/P so V= ?(0.08206)(273)/1 atm... so I need to find n in moles but I dont know how. Thanks again!

What you need to do is multiply by n which is 2

Assuming both are ideal gases is a valid assumption so pV=nRT is the way to go. But because the gases are at STP we can use the fact that one mole of any gas at STP occupies 22.4L. (You can derive this from the ideal gas law)

As there are two moles of gas then the container has a volume of 44.8L

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This bit is unnecessary for the question at hand but it shows how we arrive at 22.4L.

$P_0 = 100\,kPa = 10^5\,Pa$

$T_0 = 273.15\,K$

$R = 8.314\,J\,mol^{-1}\,K^{-1}$

$n = 1$

$V = \frac{nRT_0}{P_0} = \frac{(1)(8.314)(273.15)}{10^5} = 0.0224\, m^3 = 22.4\,L$