Molar Mass and density gas relationship

[thomas49th]

Homework Statement

The molar mass of carbon dioxide is 0.045kg/mol

Calculate the density of the gas when thetemperature is 273 K and the pressure is 120 000 Pa

Homework Equations

Density = mass / volume
pV = nRT
n = mass/ molar mass

The Attempt at a Solution

I frist want to find n to use in pV = nRT to find V AND to find the mass from n = mass/ molar mass, then I want to use this in the Density = mass / volume to find the density

I feel I am missing an equation? THe one with that abigardo mumbo jumbo in it?

Thanks :)

 

[Redbelly98]

Try using PV=nRT to find the ratio (n/V), and work from there.

You cannot find n and V with the given information, you can only find their ratio (n/V). But, you can still find the density.

 

[thomas49th]

so would the right way be:

n = mass / molar mass

n = mass / 0.045

sub this into pv = nRT

V = massRT / 0.045p

sub this into density = mass / volume

density = mass /[(mass x 8.31 x 273)/(0.045 x 120000)]

mass cancels... yay

density = 8.31 x 273 / 45 x 120000

= 0.42 kg/m³

is that right?Thanks :)

 

[Redbelly98]

density = mass /[(mass x 8.31 x 273)/(0.045 x 120000)]

mass cancels... yay

Looks good up to this point.

density = 8.31 x 273 / 45 x 120000

= 0.42 kg/m³

is that right?

You've made an algebraic error. Try putting the units in with the numbers ... you won't get kg/m³ with the answer.

 

[thomas49th]

density = mass /[(mass x 8.31 x 273)/(0.045 x 120000)]

so are you saying that bits wrong?

i thought density would be kg/m³

density = mass / volume

mass per volume

kg per m³

Can you enlighten me? :)

Thanks :)

 

[Saladsamurai]

pV=nRT[/itex]<br /> <br /> \Rightarrow pV=\frac{m}{M}RT<br /> <br /> \Rightarrow \frac{m}{V}=\frac{pM}{RT}<br /> <br /> Do you see your mistake now?<br /> <br /> Like Redbelly said,<b>carry your units</b>. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> <br /> Casey

 

[thomas49th]

ahh yes, not both mass mass, one was molecular mass. so all i have to do is work out pM/RT

= 120000x0.045 / 8.31(273)

= 2.38

as for the units

wikipedia:

The SI unit for density is:

* kilograms per cubic metre (kg/m³)

and I've go the molecular mass in kg

I'm stumped :\ Any pointers

Thanks :)

 

[Saladsamurai]

I am not sure what you are asking. What exactly is your question now?

 

[thomas49th]

are my units for density correct
kg/m³

Thanks :)

 

[Saladsamurai]

Of course they are

What Redbelly and I were trying to say was this:

In post #3 you said density=.42 kg/m³

That is the reciprocal of the correct answer 2.38 kg/m³ which means that you had it upside-down.

If you had carried your units during that calculation, you would have gotten m³/kg which would have tipped you off that something was wrong.

Carrying your units is a good way off checking to see that you answer makes sense.

Hope that helps

Casey

 

[thomas49th]

ahh cheers :)

 

[Redbelly98]

ahh yes, not both mass mass, one was molecular mass. so all i have to do is work out pM/RT

= 120000x0.045 / 8.31(273)

= 2.38

as for the units

wikipedia:

The SI unit for density is:

* kilograms per cubic metre (kg/m³)

and I've go the molecular mass in kg

I'm stumped :\ Any pointers

Thanks :)

I'm not sure you quite understand what we are telling you to do with the units. Yes, you are aware that density units are kg/m³, but that doesn't mean to simply apply those units to your final calculated number.

You have to include them as part of the calculation:

120000x0.045 / 8.31(273)

should really be written as

(120000 Pa)x(0.045 kg/mol) / [(8.31 m³ Pa / (K mol)) (273 K)]

Many units will cancel out here, leaving you with 2.38 kg/m³. The units come from doing the algebra associated with the calculation, not because we know that's what they should be.

Had you done that earlier, you would have ended up with

0.42 m³ / kg

And, since m³ / kg is not a correct density unit, you would have realized that something had gone awry. That would alert you to check back over your work, and track down the error.

Working with units this way can often alert you that an error has been made somewhere -- a very useful thing during an exam.

Hopefully what I wrote is understandable. Good luck!