How Do Mass and Molar Velocities Differ in a Gas Mixture?

[gfd43tg]

Homework Statement

A gas mixture contains 50% He and 50% O2 by volume at 500 K and 1 bar. The absolute
mass fluxes of each species along the transportation direction z are $\hat J_{He} = -0.8 \frac {kg}{m^{2} s}$ and $\hat J_{O_{2}} = 0.8 \frac {kg}{m^{2} s}$. Determine the mass and molar average velocities. Also determine the mass and molar diffusion fluxes.

Homework Equations

The Attempt at a Solution

Hello, I am having considerable trouble solving this problem. The main reason I believe is because of all these definitions of terms and sorting them out. I will explain my thoughts. Firstly, I would like to mention that I will use the carat ''^" to denote mass quantities and ''~" for molar quantities.

So to begin, I figure I will calculate the mass concentration of oxygen, using the ideal gas law modified with the molar mass included,

$P_{O_{2}} V_{O_{2}}M_{O_{2}} = N_{O_{2}} M_{O_{2}}RT$
$$\hat C_{O_{2}} = \frac {P_{O_{2}}M_{O_{2}}}{RT}$$
where $P_{O_{2}} = y_{O_{2}}P = 0.5P$
So now I have the mass concentration of oxygen, and I divide the total mass flux given in the problem by the mass concentration to get the so called "average mass velocity".
$$\bar V_{O_{2}}^{mass} = \frac {\hat J_{O_{2}}}{\hat C_{O_{2}}}$$
I calculate 2.08 meters/sec

Now, if I do the ''molar average velocity"
$$\tilde J_{O_{2}} = \frac { \hat J_{O_{2}}}{M_{O_{2}}}$$
$$\tilde C_{O_{2}} = \frac {P_{O_{2}}}{RT}$$
$$\bar V_{O_{2}}^{molar} = \frac {\tilde J_{O_{2}}}{\tilde C_{O_{2}}}$$
Once again, I end up with 2.08 meters/sec.

So I do not know how to differentiate "average mass velocity'' from "average molar velocity". I thought velocity is velocity is velocity, period. I reckon this might have something to do with my partial pressure, one should be based on molar fraction, and the other on mass fraction, since I am using volume fraction right now. How can I go from a volume fraction to a mass fraction?

EDIT: I now know how to convert, I just use the ratio of the densities. I find that the mass fraction, $\hat y_{O_{2}} = 0.889$. Now, I go back to my ideal gas law,
$P_{O_{2}}V_{O_{2}}M_{O_{2}} = N_{O_{2}}M_{O_{2}}RT$
But I am unsure of where I should make the substitution in with the mass fraction. I guess it would be $P_{O_{2}} = \hat y_{O_{2}}P$?? If so, I find
$$ \hat C_{O_{2}} = \frac {\hat y_{O_{2}}PM_{O_{2}}}{RT} $$
and I calculate the mass concentration to be 0.684 kg/m^3

 

[Chestermiller]

Your calculation of the velocity of oxygen is correct. But, when they're asking for the mass average velocity, what they want is the mass average velocity of the combined stream, not of each species. The same goes for molar average velocity.

Chet

 

[gfd43tg]

I don't think I calculated the velocity of the oxygen correctly. I set $P_{O_{2}} = 0.5P$, but it is 50 vol% oxygen. I should be calculating the mol% oxygen. Is the vol% = mol% for an ideal gas by any chance?

For now I will assume they are the same. So I already calculated $V_{O_{2}}$ and $V_{He}$. So to find the mass average velocity, I should do
$$ V = \frac {\rho_{He}}{\rho}V_{He} + \frac {\rho_{O_{2}}}{\rho}V_{O_{2}} $$
Now I can get the density of both species, divide the absolute flux by the density to get the velocity. Now my question is, are these individual velocity terms the absolute velocity for both molar and mass average velocity? I calculate the mass average velocity to be 0.016 m/s
For the molar average velocity,
$$ \bar {V} = \frac {C_{He}}{C} \bar V_{He} + \frac {C_{O_{2}}}{C} \bar V_{O_{2}} $$
Is it the case that $V_{He} = \bar V_{He}$?

 

[Chestermiller]

I don't think I calculated the velocity of the oxygen correctly. I set $P_{O_{2}} = 0.5P$, but it is 50 vol% oxygen. I should be calculating the mol% oxygen. Is the vol% = mol% for an ideal gas by any chance?

For now I will assume they are the same. So I already calculated $V_{O_{2}}$ and $V_{He}$. So to find the mass average velocity, I should do
$$ V = \frac {\rho_{He}}{\rho}V_{He} + \frac {\rho_{O_{2}}}{\rho}V_{O_{2}} $$
Now I can get the density of both species, divide the absolute flux by the density to get the velocity. Now my question is, are these individual velocity terms the absolute velocity for both molar and mass average velocity?

Yes.

I calculate the mass average velocity to be 0.016 m/s

Shouldn't it be zero?

For the molar average velocity,
$$ \bar {V} = \frac {C_{He}}{C} \bar V_{He} + \frac {C_{O_{2}}}{C} \bar V_{O_{2}} $$
Is it the case that $V_{He} = \bar V_{He}$?

Yes.

 

[gfd43tg]

Well, I calculate
$$ \rho_{O_{2}} = \frac {P_{O_{2}}M_{O_{2}}}{RT} = \frac {(0.5)(1)(32)}{(0.08314)(500)} = 0.385 $$
$$ \rho_{He} = \frac {P_{He}M_{He}}{RT} = \frac {(0.5)(1)(4)}{(0.08314)(500)} = 0.048 $$
$$ \rho = \rho_{O_{2}} + \rho_{He} = 0.433 $$
$$V_{O_{2}} = \frac {J_{O_{2}}}{\rho_{O_{2}}} = \frac {0.8}{0.385} = 2.1 m/s $$
$$V_{He} = \frac {J_{He}}{\rho_{He}} = \frac {-0.8}{0.048} = -16.7 m/s $$
$$V = \frac {\rho_{He}}{\rho}V_{He} + \frac {\rho_{O_{2}}}{\rho}V_{O_{2}}$$
$$V = \frac {.048}{.433}(-16.7) + \frac {.385}{.433}(2.1) = 0.016 m/s $$
I mean, I guess it could be a rounding error, but there is no telling for sure at least right now.

 

[Chestermiller]

Well, I calculate
$$ \rho_{O_{2}} = \frac {P_{O_{2}}M_{O_{2}}}{RT} = \frac {(0.5)(1)(32)}{(0.08314)(500)} = 0.385 $$
$$ \rho_{He} = \frac {P_{He}M_{He}}{RT} = \frac {(0.5)(1)(4)}{(0.08314)(500)} = 0.048 $$
$$ \rho = \rho_{O_{2}} + \rho_{He} = 0.433 $$
$$V_{O_{2}} = \frac {J_{O_{2}}}{\rho_{O_{2}}} = \frac {0.8}{0.385} = 2.1 m/s $$
$$V_{He} = \frac {J_{He}}{\rho_{He}} = \frac {-0.8}{0.048} = -16.7 m/s $$
$$V = \frac {\rho_{He}}{\rho}V_{He} + \frac {\rho_{O_{2}}}{\rho}V_{O_{2}}$$
$$V = \frac {.048}{.433}(-16.7) + \frac {.385}{.433}(2.1) = 0.016 m/s $$
I mean, I guess it could be a rounding error, but there is no telling for sure at least right now.

It has to be roundoff error. You can show that immediately from your equations above.