# Electron concentration as a function of pressure

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1. Feb 4, 2017

### Farang

1. The problem statement, all variables and given/known data
Determine electron concentration for both Helium and Nitrogen as a function of pressure at 300K.

This is from preliminary task for Energy loss of alpha particles in gases laboratories. It's too late to submit this, I'd just like to understand. Can post more info from the lab script if needed. If someone could point out what I'm doing wrong, that'd be great.

2. Relevant equations
$$n(p,T)=\frac{p}{T}\frac{kN_{A}}{V_{m}}$$

3. The attempt at a solution
Not sure if the value I use for Vm is correct. It's $22.4\times10^{-3}m^{3}mol^{-1}$ for all gases under standard conditions ($273K$). Does it change with temperature?

$$n(p,300K)=\frac{p}{300K}\frac{2(6.022\times10^{23}mol^{-1})}{(22.4\times10^{-3}m^{3}mol^{-1})}=(1.792\times10^{23}m^{-3}K^{-1})p$$

So that gives units of electron concentration = $m^{-4}s^{-2}K^{-1}kg$, which doesn't make sense. I'd expect $m^{-3}$

I've got a lot of catching up to do on thermodynamics, I'm sure you can tell haha

Andre

PS. I wonder if there's a way to write the Dirac equation in natural units without the slashed package(see sig)

2. Feb 4, 2017

### kuruman

Can you explain how you got the relevant equation that you quoted? Also it seems that k = 2 in that equation from your substitution. What does k stand for?

3. Feb 4, 2017

### Farang

k is the number of electrons per molecule. The lab script says to use that equation :) It follows after:

The electron density that is responsible for the stopping power can be estimated for a given material of atomic mass $A$, with $k$ electrons per molecule and a density of $\rho$ which can be expressed as
$$n=\frac{k \rho}{A u}$$
where the atomic mass, $u=1.6605 \times 10^{-27} kg$

For example, for dry air under standard conditions (which is taken to be $79\% N_{2}$ and $21\% O_{2}$), $k = 14.4$ and $A = 28.8$. This allows the calculation of the density of air
$$\rho(\rho,T)=\frac{p A u N_{A}}{T V_{m}}$$
where Avogadro's number is $N_{A} = 6.022 \times 10^{23} mol^{-1}$, the volume of gas under standard conditions ($0^{\circ} C$) is $22.7 l mol^{-1}$. This gives a value of
$$\rho(\rho,T)=1.269 kg m^{-3} \frac{p}{1000 hPa}\frac{273}{T}$$
where the pressure is determined in hectoPascals, $hPa$, with $1 hPa \equiv 100 Pa$ and the temperature in Kelvin.

Similarly, the calculation of the electron density can be completed to show the expected variation with both temperature and pressure
$$n(p,T)=\frac{p}{T}\frac{kN_{A}}{V_{m}}=3.825 \times 10^{26} m^{-3} \frac{p}{1000 hPa}\frac{273}{T}$$
In this experiment, the pressure dependence of the energy loss will be determined for air and helium at room temperature (typically 294K) in order to determine if the predictions made above are correct.

Last edited: Feb 5, 2017
4. Feb 5, 2017

### Farang

Why are there $1/1000hPa$ and $273$(must be missing K) in the 3rd equation in my previous post? Seriously, I'd like to know.

5. Feb 5, 2017

### haruspex

Because Avogadro's number pertains to "standard conditions", i.e. 1atm (1000hPa, or maybe 1023?) and 0C (273K). To adjust to p and T, need to multiply/divide by the pressure and temperature ratios.

6. Feb 5, 2017

### kuruman

This equation makes sense. The next one
does not make sense and does not have the correct dimensions. You can write the density of a mole of gas under standard p and T as$$\rho=\frac{m}{V}=\frac{N_A Au}{V}$$. Then use the ideal gas law $pV=N_Ak_BT$ ($k_B$ is the Boltzmann constant) to find
$$\rho(p,T)=\frac{pAu}{k_BT}$$

7. Feb 5, 2017

### Farang

Ratios are dimensionless, so there should be no $hPa$ in there. This lab script needs correcting...

8. Feb 5, 2017

### haruspex

The ratio p:1000hPa is dimensionless. The hPa is necessary.

9. Feb 5, 2017

### Farang

In that case the units won't cancel... Is this incorrect then:
$$n(p,T) = \frac{k p}{k_{b} T} = (1.4486 \times 10^{23} m^{-3} Pa^{-1} K) (\frac{273}{1.0 \times 10^{5}}) (\frac{p}{T})$$
For 300K:
$$n(p,300 K) = (1.3182 \times 10^{18} m^{-3} Pa^{-1})p$$
Looks good to me :)

10. Feb 5, 2017

### haruspex

Not sure what you mean by that. Can you elaborate?

Because "hPa" appeared in the equation, it is correct whatever units p is in. The units of p should be included when the substitution is made. If p is replaced by "2000 hPa" then the two "hPa"s cancel. If it is replace by "200,000 Pa" then the "Pa" and the "hPa" cancel to leave a constant 1/100.
You are right that the 273 should similarly have been qualified with "K". If it is not then the temperature value must be plugged in expressed as so many Kelvin.

Your equations in post #9 look ok to me with regard to units and dimensions.

11. Feb 5, 2017

### Farang

I take that back. I think I get it now, kind of... Is this the way to express density as a function of pressure and temperature? Multiply by ratios? How does this even work?
$$\rho=\frac{m}{V}=\frac{N_A Au}{V}$$
$$\rho(p, T)=\frac{AuN_A}{V_m} (\frac{p}{10^{5} Pa}\frac{273K}{T})$$
Why multiply by $p$ and divide by $T$ and not the other way around then?

12. Feb 5, 2017

### haruspex

Because it is all based on pV=nRT. ρ(p, T)=m/V(p, T)=mp/(nRT).
Using the subscript s for standard conditions, ρs(p, T)=mps/(nRTs).
So ρ/ρs=(p/ps)(Ts/T)

13. Feb 5, 2017

### Farang

Wow, this place is way more effective than tutorial sessions. Thanks very much :)