Get the mass of an argon atom and its molecular weight from Cv

[Ron_H]

Homework Statement

The mass of a gas molecule can be computed from the specific heat at constant volume. Take Cv=0.075 kcal/kg-K for argon and calculate (a) the mass of an argo atom and(b) the atomic weight of argon.

Relevant Equations

1/2 m <v^2> = 3/2 kT
1/2 M <v^2> = 3/2 RT
pV=NRT
U=(3/2)*NRT
Cv=(1/N)(dU/dT)

I don't see how isolate m or M.
That is, how to get rid of <v^2> which is how I wrote the average of v²

 

[Orodruin]

Your Cv in the relevant equations has the wrong physical dimension to be specific heat. What you have written has dimension energy per temperature per substance amount, not the correct energy per temperature per mass.

 

[Charles Link]

One thing the OP should know, or be able to compute is that $C_v=\frac{3}{2}R$. This will give the $C_v$ in calories/(mole K) which is calories per atomic weight per K. (You need to know $R$ in calories/(mole K ). It helps to memorize it. It's used a lot.).
They give $C_v$ to you in calories/(gram K). It's a simple ratio to determine the atomic weight.

 

[Ron_H]

Your Cv in the relevant equations has the wrong physical dimension to be specific heat. What you have written has dimension energy per temperature per substance amount, not the correct energy per temperature per mass.

Thank you Orodruin and Charles Link,

Orodruin, the book I took the problem from (Halliday & Resnick, 1966) says, in this context,

By definition of $C_v$ we have $\Delta Q = \mu C_{v} \Delta T$​

They are using $\mu$ for the number of moles.

This gives C_v dimensions ML²T^-3 or as you say 'energy per temp per quantity of substance amount'.

So, I could not agree with you that..

Your Cv in the relevant equations has the wrong physical dimension to be specific heat

Different yes, but not wrong.

But the statement of the problem gave $C_{v} = 0.075 kcal/kg K$.
And I complained I couldn't isolate the mass.
And your suggested 'energy per temperature per mass' had 'mass' in it.
Hmmm, your suggestion was not just different; it might be useful.

So, is this what you were suggesting?...

From kinetic theory and for a mole of a monoatomic gas like Ar,$$U=\frac 1 2 M\bar{ v^{2}} = \frac 3 2 RT$$ and now, with your suggestion$$U =MC_{v}T$$I got$$M=3R/2C_{v}$$

Plugging in the given $C_v$ and $R$ (converted to the same units as were given for $C_v$ gives about the molecular weight (in kg) for Ar.

The analogous calculation starting the average translational kinetic energy per molecule$$U=\frac 1 2 m\bar{ v^{2}} = \frac 3 2 kT$$gives$$m=3k/2C_v$$for the mass of the Ar atom. And plugging in $k$ in the same units as were given for $C_v$ gives a good value for the mass of the Ar atom.

(Could also divide the molecular weight by Avogadro's number to get the mass of the atom, but the topic is kinetic theory.)

 

[Charles Link]

Looking at post 3 again, $R=1.987$ cal/(mole K). With $C_v=\frac{3}{2}R$, the arithmetic is simple:

$C_v=3.0$ cal/(mole K) =$.075$ cal/(gram K).

This gives $1$ mole=3.0/.075=40 grams. The homework helper is not supposed to provide the solution, but it may be worthwhile to see what may be the easiest way to do it.

 

[Ron_H]

Hi Charles Link,
Thanks!

 

[Ron_H]

Hi Charles Link,

I've been thinking about your posts.
What was your thought process?
How do you know you are right?

You say

$C_{v} = 3.0 \newcommand\calmoleK{\operatorname{cal/mole-K}} \calmoleK= 0.075 \newcommand\calgmK{\operatorname{cal/gm-K}} \calgmK$.

And you also say

"1 mole=3.0/.075=40 grams".

In setting up my question, I'm going to re-write these statements.
Please correct me if I've misunderstood you.

The statement of the problem provided $\newcommand\calgmK{\operatorname{cal/gm-K}}C_{v} = 0.075 \calgmK$.
(I adopted your change of units. No issue there. BTW, I think of this as a measured value.)

I'll explicitly define $\newcommand\calmoleK{\operatorname{cal/mole-K}} C_{m} = 3.0 \calmoleK$.
This is a result of kinetic theory which you say we should either be able to remember or derive.
(This is only for a monatomic gas - which argon is.)
I think we're on the same page still.

I'll also define $M$=molecular weight of Ar.

You finish (in my paraphrase) by introducing this premise, $$M = \frac {C_{m}} { C_{v}} $$...plugging in the values for $C_{v}$ and $C_{m}$,$$ \frac {3.0 \calmoleK} {0.075g \calgmK} = 40 \frac {gm} {mole} $$... gettting the atomic weight of argon.

So here is my question:

What about that final premise?
Where did you get it?
Because it was not obvious to me, I proved it.
So now I know there is justification for getting the answer the way you describe.
I'm just wondering how you knew it was right to do that division.

 

[Orodruin]

If the heat capacity per mole is $C_m$, then the heat capacity per mass is $C_m/M$ because the mass per mole is $M$ by definition. The rest is basic algebra.

 

[Charles Link]

The way I like to look at it is the heat capacity is a type of vector, with a numerical result and units associated with it. Regardless of the units, the heat capacity "vector" is always the same, i.e. we can set the .075 with its units equal to the 3.0 and its units. The atomic mass was originally unknown, but appears when we set the two vectors equal, and cancel the other units leaving one mole=40 grams.

Incidentally, this trick comes in handy for doing conversion factors of all kinds:
e.g. $R=.08206$ liter-atm/(mole k)=$1.987$ cal/(mole K), with $1.0$ cal=$4.184$ joules.
( $1.0$ liter=$1.0$ E-3 m^3).
With a little algebra, you can determine that $1.0$ atm=$1.01$ E+5 Newton/m^2.

For an alternative approach, the heat $\Delta Q =C_{Vm} x_{moles} \Delta T=C_{Vg} x_{grams} \Delta T$. We can set $\Delta T= 1$, or simply cancel the $\Delta T$'s. When $x_{moles}=1$, we can solve for the corresponding $x_{grams}$.