Calculating Heat Capacity of Krypton at 90K

[ibysaiyan]

Homework Statement

Atoms of the inert gas Krypton are adsorbed onto a smooth solid
surface at 90K. They can move freely over the surface but they cannot leave it. For a
sample with total surface area 2.5 m^2, and a surface density 3 nm-2, what is the heat
capacity of the krypton?

Homework Equations

$\Delta E$/ $\Delta T$ = Cv ( where Cv is specific heat @ constant volume)

$\Delta E$ = $\Delta W$ + $\DeltaQ$
U = 3/2nRT , U = 3/2NkbT

The Attempt at a Solution

Data given:

Krypton molecular weight : 84.8 /1000 = 0.0848 kg/mol.

I believe I have an idea on how to solve this. The fact that the gases can't escape implies work done = 0, however I am a little confused by the spatial dimensions provided.
Would I multiply 2.5 with 3 ? however that makes no sense to get mass. I was wondering.. could I use : Mass = Molecular weight / Avogadro's number and then plug in: Moles = mass/mr to give me 'n' which I could plug into U = 3/2nRT

Is my approach sensible ?

 

[gneill]

Hmm. Surface area and surface density give you the number of moles of gas. The fact that the gas is trapped on the surface (but moves freely thereon) implies that you have a 2d gas! That is, one less degree of freedom for kinetic energy distribution for internal energy than a 3d volume of gas.

Perhaps you need to investigate equipartition of energy for an ideal gas with a given temperature, and how the specific heat constants are based upon the available degrees of freedom for distributing the energy. Hot search terms would be "specific heat", "internal energy", "equipartition", "kinetic temperature".

 

[ibysaiyan]

Hmm. Surface area and surface density give you the number of moles of gas. The fact that the gas is trapped on the surface (but moves freely thereon) implies that you have a 2d gas! That is, one less degree of freedom for kinetic energy distribution for internal energy than a 3d volume of gas.

Perhaps you need to investigate equipartition of energy for an ideal gas with a given temperature, and how the specific heat constants are based upon the available degrees of freedom for distributing the energy. Hot search terms would be "specific heat", "internal energy", "equipartition", "kinetic temperature".

Thanks for your reply. I am presuming that surface density times surface area would give number of moles, right ?
Also I am not quite sure if I have come across the term '2d' gas... nor it's consequence. I haven't done any reading yet, I ought to. I will do that tomorrow morning.

 

[ibysaiyan]

I am so lost as to what to do with surface density and area... I tried to manipulate them to get something meaningful but to no avail. I mean if i divide or multiple the units.. they don't get anywhere sensible.. like for instance density has N/m^2.. which's kg/s^2 m...now if i multiply this with area... i get kgm/s^2... just a m factor away from getting energy.. any tips ?

 

[gneill]

The surface density is specified as 3 atoms of Krypton per square nanometer.

 

[ibysaiyan]

The surface density is specified as 3 atoms of Krypton per square nanometer.

Thanks for your reply.. the term 'surface density' is vague to me... so how does that relate to finding out mole of substance.

 

[gneill]

Thanks for your reply.. the term 'surface density' is vague to me... so how does that relate to finding out mole of substance.

It tells you how many atoms there are per square unit of surface area. What's the surface area? How many atoms does that make? How many atoms in a mole of atoms?

 

[ibysaiyan]

It tells you how many atoms there are per square unit of surface area. What's the surface area? How many atoms does that make? How many atoms in a mole of atoms?

edit: surface area is 2.5 m^2 , 3/ 2.5 = 1.2 so would n = 1.2 ?

 

[gneill]

Your surface area was given as 2.5 m². How many square nanometers is that?

 

[ibysaiyan]

your surface area was given as 2.5 m². How many square nanometers is that?

2.5 *10^(18)

 

[gneill]

2.5 *10^(18)

So if there are 3 atoms of Krypton for ever square nm, how many atoms is that?

 

[ibysaiyan]

So if there are 3 atoms of Krypton for ever square nm, how many atoms is that?

2.5 *10^18 / 1*10^-9 =2.5 *10^27... so I will divide with this by Avogadro's number, right ?

 

[gneill]

2.5 *10^18 / 1*10^-9 =2.5 *10^27... so I will divide with this by Avogadro's number, right ?

Why did you divide by 10^-9?

 

[ibysaiyan]

Why did you divide by 10^-9?

as you had said earlier.. there are three atoms per one nanometer so wouldn't i divide it by that ?

edit: it's about 4 am right now which may explain my incoherent thinking..

 

[gneill]

as you had said earlier.. there are three atoms per one nanometer so wouldn't i divide it by that ?

edit: it's about 4 am right now which may explain my incoherent thinking..

If you have 2 shoes per shoebox and you have a million shoeboxes, how many shoes do you have? Surely you don't divide a million by two!

Multiply the number of square nm by the number of atoms per nm to yield the total number of atoms. Once you have that, yes, you divide by Avogadro's number to find the number of moles.

 

[ibysaiyan]

If you have 2 shoes per shoebox and you have a million shoeboxes, how many shoes do you have? Surely you don't divide by a million by two!

Multiply the number of square nm by the number of atoms per nm to yield the total number of atoms. Once you have that, yes, you divide by Avogadro's number to find the number of moles.

Oh silly me...
so 2.5 *10^18 x 3 = 7.5*10^18 .. I hope i did this right...

 

[gneill]

Okay, so there are 7.5*10^18 atoms. Divide by N_A to yield moles.

 

[ibysaiyan]

Okay, so there are 7.5*10^18 atoms. Divide by N_A to yield moles.

I get approximately 1.25*10^-5 mol.
Now that gives me the 'n' value however I have to find specific heat @ 90k for krypton.
I know total energy of a system is given by : F/2 NkbT
Where f is the degrees of freedom but how do i figure out D.O.freedom for this gas, N , number of particles.
So if I find out the energy value and then divide that by T, then i should get my specific heat value, am I right ?
For number of particles I will multiply N_A with moles .

edit: is the degrees of freedom in this case 2 , instead of three ? if so why.. is it because of the three independent squared terms i.e x,y,z one freezes [translational]?

 

[ehild]

edit: is the degrees of freedom in this case 2 , instead of three ? if so why.. is it because of the three independent squared terms i.e x,y,z one freezes [translational]?

Yes, you have it correct: the atoms can not move away from the surface, so one from the usual 3 terms in the energy is zero.


ehild

 

[ibysaiyan]

Yes, you have it correct: the atoms can not move away from the surface, so one from the usual 3 terms in the energy is zero.


ehild

I get a value of 1.038 *10^-4, is that a reasonable value ? Since a quick google search shows it to be 0.247 J/g/C..

 

[ibysaiyan]

Anyone ?

 

[gneill]

Your value would appear to be the Joules per degree K for the given system. If you want a specific heat, it should be energy per unit mass (or per mole) per degree K.

Also, beware of comparing a specific heat value for a 3D gas to the one for a 2D gas. The kinetics are different (by one degree of freedom).

 

[ibysaiyan]

Thanks for your reply again.so is my answer right?! I have used E :f/2 Nkbt then cv by definition is e/t hm.. Also I have plugged 2 for f for reasons mentioned previously. Do I need convert my final answer or am I right.I appreciate your feedback / help. Sorry for my lack of latex usage I am at the trainstation ... On my celphone.

 

[gneill]

Your answer for the heat capacity looks okay to me. I used a value of 1.381 x 10^-23 J/K for k_b and got 1.035 x 10^-4 J/K for the heat capacity.

 

[ibysaiyan]

Oh that's brilliant. If I am to get the answer in the form of joules per mole per kelvin could I plug moles into energy equation I.e e= f/2 n times R T.is that doable

 

[gneill]

Actually it's simpler than that! For a monoatomic gas the 2D value of C_v will just be R. The degrees of freedom available are 2, so
$$U = \frac{2}{2} n R T ,~~~~~\text{so}~~~~C_v = \frac{U}{n T} = R$$

 

[ibysaiyan]

Actually it's simpler than that! For a monoatomic gas the 2D value of C_v will just be R. The degrees of freedom available are 2, so
$$U = \frac{2}{2} n R T ,~~~~~\text{so}~~~~C_v = \frac{U}{n T} = R$$

I am finally back on my laptop but shouldn't $C_{v}$ be nR although the unit of gas constant is same as specific heat. I say this because $C_{v}$ = $\Delta U$ /$\Delta T$ = nRT/ T = nr ?
edit: i have just googled this.. it seems my lecturer's notes are wrong.. since they show: Cv = Delta E/ delta T .. doesn't mention n/mole.

 

[gneill]

n is the number of moles. The specific heat should not depend upon the quantity of material.

The molar specific heat is defined by

$Q = n C_v \Delta T ~~~~~\text{(at constant volume)}$

and by the first law, $\Delta U + P \Delta V = n C_v \Delta T$, where ΔV = 0 in this case, so that $C_v = \frac{1}{n} \frac{\Delta U}{\Delta T}$.

For our 2D gas, $U = n R T$, so $\frac{\Delta U}{\Delta T} = n R$. Thus $C_v = \frac{1}{n} (n R) = R$.

 

[ibysaiyan]

n is the number of moles. The specific heat should not depend upon the quantity of material.

The molar specific heat is defined by

$Q = n C_v \Delta T ~~~~~\text{(at constant volume)}$

and by the first law, $\Delta U + P \Delta V = n C_v \Delta T$, where ΔV = 0 in this case, so that $C_v = \frac{1}{n} \frac{\Delta U}{\Delta T}$.

For our 2D gas, $U = n R T$, so $\frac{\Delta U}{\Delta T} = n R$. Thus $C_v = \frac{1}{n} (n R) = R$.

This makes sense. Thank you very much for all that you have done. I can't thank you enough.

-ibysaiyan

 

[davedavidson]

Sorry if I’ve missed something but why are you using specific heat at constant volume? The volume isn't constant; the question states that the gas is free to move along the surface. Wouldn’t it be c at constant pressure?

 

[gneill]

Sorry if I’ve missed something but why are you using specific heat at constant volume? The volume isn't constant; the question states that the gas is free to move along the surface. Wouldn’t it be c at constant pressure?

Well, the surface area is of a fixed size. The atoms are constrained to that fixed-size surface. It's the 2D equivalent to the 3D constant volume case.

I suppose that, notationally, the constant should be C_A, for constant area!

 

[ehild]

Sorry if I’ve missed something but why are you using specific heat at constant volume? The volume isn't constant; the question states that the gas is free to move along the surface. Wouldn’t it be c at constant pressure?

The problem states that the atoms of the 2D gas are free to move on the surface.

The particles of a normal gas in 3 dimension move freely inside the container. The volume of the gas is defined as the volume of the container.

The particles of the 2D gas move freely on a certain surface. The area of the surface plays the role of volume.


ehild