Number of Molecules vs Velocity: Finding N for v > v_0

[alexmahone]

Suppose I have a graph with velocity of a gas v on the x-axis and number of molecules N on the y-axis, how would I find the number of molecules with v>v_0, where v_0 is a given velocity?

 

[Ackbach]

How about

$$\int_{v_{0}}^{\infty}N(v)\,dv?$$

 

[alexmahone]

How about

$$\int_{v_{0}}^{\infty}N(v)\,dv?$$

But that has dimensions vxN.

As stated, the answer to the problem is infinity since we have to consider infinitely many velocities. (Note that I have a continuous distribution.) So I guess we cannot have such a graph.

 

[Ackbach]

You're right about the units. I'll have to rethink that. However, you're not correct about infinities. What if $N(v)=e^{-v^{2}}?$ Note that $\int_{-\infty}^{\infty}N(v)\,dv=\sqrt{\pi}$, in that case. You can definitely have a finite integral, if the number of molecules with a given velocity drops off fast enough.

 

[Ackbach]

I have it, I think. Your function $N$ does not have units of [number of molecules]. It has units of [number of molecules]/[v]. So the [v]'s will cancel out in the integration. You're integrating a density function. So, I think that
$$\int_{v_{0}}^{\infty}N(v)\,dv$$
still works. And if your $N(v)$ is to be at all realistic, it must die off fast enough for the integral to converge, since we believe there are a finite number of molecules in the universe!

 

[alexmahone]

I have it, I think. Your function $N$ does not have units of [number of molecules]. It has units of [number of molecules]/[v]. So the [v]'s will cancel out in the integration. You're integrating a density function. So, I think that
$$\int_{v_{0}}^{\infty}N(v)\,dv$$
still works. And if your $N(v)$ is to be at all realistic, it must die off fast enough for the integral to converge, since we believe there are a finite number of molecules in the universe!

But I told you that N has units of [number of molecules]. With all due respect, N having units of [number of molecules]/[v] doesn't make any sense. Pretty sure you're mistaken.

 

[CaptainBlack]

But I told you that N has units of [number of molecules]. With all due respect, N having units of [number of molecules]/[v] doesn't make any sense. Pretty sure you're mistaken.

On a continuous graph N cannot be a number it must be a number density.

CB

 

[alexmahone]

On a continuous graph N cannot be a number it must be a number density.

CB

What do you mean by number density?

 

[Jameson]

If N is continuous then P[N=a] for any a is 0, so it can only be considered between a range of values.

As I understand the situation you are trying to create, the x-axis is continuous and the y-axis is discrete so for any $N_i$ there should be a corresponding range of velocities \(\displaystyle v_a<v_i<v_b=N_i\). However, N can't realistically be non-integer though unless molecule means something different than I've ever seen.

 

[CaptainBlack]

What do you mean by number density?

Number of molecule in a unit speed interval (or rather \( \lim_{\Delta v\to 0} N(v-\Delta v/2, v+\Delta v/2)/ \Delta v \) where \( N(v-\Delta v/2, v+\Delta v/2) \) denotes the number in the speed range \((v-\Delta v/2, v+\Delta v/2) \) )

CB

 

[CaptainBlack]

An alternative type of data presentation where the vertical scale is number is a cumulative number plot, where against speed v we plot the total number (or fraction) with speed less than or equal to v.

This gives a plot that is independedent of assumptions about discrete or continuous (or for that matter mixed continuous and discrete) distributions (one has steps, the other does not).

CB

 

[chisigma]

Suppose I have a graph with velocity of a gas v on the x-axis and number of molecules N on the y-axis, how would I find the number of molecules with v>v_0, where v_0 is a given velocity?

The speed distribution of the molecules of a gas with absolute temperature T has been found by Maxwell and Boltzmann about hundred and fifty years ago...

$\displaystyle f(v)= \sqrt{\frac{2}{\pi}\ (\frac{m}{K\ T})^{3}}\ v^{2}\ e^{- \frac{m\ v^{2}}{2\ K\ T}}$ (1)

... where m is the mass of a single molecule and K is the 'Boltzmann's constant'. If N is the global number of molecules, the the number of molecules with $\displaystyle v> v_{0}$ is given by...

$\displaystyle N_{0}= N\ \int_{v_{0}}^{\infty} f(v)\ dv$ (2)

Kind regards

$\chi$ $\sigma$

 

[chisigma]

The speed distribution of the molecules of a gas with absolute temperature T has been found by Maxwell and Boltzmann about hundred and fifty years ago...

$\displaystyle f(v)= \sqrt{\frac{2}{\pi}\ (\frac{m}{K\ T})^{3}}\ v^{2}\ e^{- \frac{m\ v^{2}}{2\ K\ T}}$ (1)

... where m is the mass of a single molecule and K is the 'Boltzmann's constant'. If N is the global number of molecules, the the number of molecules with $\displaystyle v> v_{0}$ is given by...

$\displaystyle N_{0}= N\ \int_{v_{0}}^{\infty} f(v)\ dv$ (2)

Of course in the formula (2) You consider $N_{0}$ as continuous variable and is $\displaystyle N_{0}= N\ P \{v>v_{0}\}$. In case You intend to consider $N_{0}$ as an integer, then the x-y diagram must be an histogram where on the X axis is reported an integer k and on the Y axis is reported the number of molecules $N_{k}$ having speed $\displaystyle k\ v_{u} \le v< (k+1)\ v_{u}$. In this case the number of molecules having speed $\displaystyle > v_{0}$ is...

$\displaystyle N_{0}= \sum_{k= \text{int} \frac{v_{0}}{v_{u}}} ^ {\infty} N_{k}$

Kind regards

$\chi$ $\sigma$