Atomic density of argon in liquid and gas form

[EmmaLemming]

Homework Statement

Argon (atomic weight 40) exists as a monatomic gas at room temperature and
pressure. The density of liquid argon is 1784 kg m⁻³.

(a) Calculate the atomic density (atoms m⁻³) in liquid argon.

(b) Calculate the atomic density (atoms m⁻³) in gaseous argon at a pressure of 1 atm and a temperature of 300K.

The attempt at a solution

(a)

PV = NkT
where,
P = 1.01 x 10⁵
k = 1.38 x 10^-23
T = 300

N/V = P/kT = 2.44 x 10²⁵ atoms m⁻³

(b)

I have no idea. I know there should be a difference but I don't know what to do.

I think perhaps my answer to (a) is in fact the answer to (b) in which case, how do I answer (a)?

 

[I like Serena]

Hi EmmaLemming!

Yep. Your answer to (a) is the answer to (b).

That leaves you with 2 quantities that you have not used yet: atomic mass of argon and its density.

Do you know what that number 40 for the atomic mass represents and how to use it in a formula?
And what is density (as a formula)?

Btw, the formula PV = NkT only applies to ideal gasses.
Luckily gaseous argon does behave like an ideal gas.

 

[EmmaLemming]

Hello :)

Thanks for you're help

So for part (a) could I just divide the atomic density by the mass?
I've seen before that atomic mass, 40, can be written as 0.04kg however I thought,

mass = M/avagadro = 6.64 x 10^-23kg

Which mass do I use? :s

 

[I like Serena]

Yes, you can divide the mass density by the mass of 1 atom, to get the atomic density.The atomic mass number 40 means that 1 mol of argon has a mass of 40 grams.

And 1 mol of argon is a number of atoms that is equal to the number of avogadro.I'm not sure which masses you calculated with just now, however.

 

[EmmaLemming]

I did,

40/(6.02 x 10²³) = 6.64 x 10^-23 kg,

to get the mass of one argon atom. Now that I think about it though should it be,

0.04/(6.02 x 10²³) = 6.64 x 10^-26 kg.

 

[I like Serena]

Ah okay.
That looks better. :)

 

[EmmaLemming]

Yaaay :) Thanks very much for your help.