Finding velocity of gas molecule

[Miike012]

Equations:
1. v_rms = √(3RT/M)
2. PV = nRT
3.M = mN_A

I want to answer part a.
I am given T,P, and the density of gas molecules (d).


From equation 3. I want to solve for m.
m = dV = d(nRT/P)
m[N_A] = M = d(nRT/P)[N_A] = dNRT/P; N = nN_A.

From equation 1.
v_rms = √(3RTP/(dNRT)) = √(3P/dN).
I get stuck at this part because I don't know the value of N.

 

[DrClaude]

M = mN_A

What is that equation?

 

[Miike012]

What is that equation?

Is M molar mass or mass?
I thought it was molar mass...

 

[DrClaude]

Is M molar mass or mass?
I thought it was molar mass...

I'm asking you! The equation doesn't make sense to me. I think it is wrong and that this is where your problem lies.

Please detail what are M, m and N_A.

 

[Miike012]

I'm asking you! The equation doesn't make sense to me. I think it is wrong and that this is where your problem lies.

Please detail what are M, m and N_A.

I added the equations in the attachment from the book

 

[Miike012]

I think I figured it out. M=mNa and m = M/Na = d*v, so M = d*v*Na

 

[DrClaude]

I think I figured it out. M=mNa and m = M/Na = d*v, so M = d*v*Na

If $m$ is the mass of one molecule, then how can $m = d \times V$?

 

[Miike012]

If $m$ is the mass of one molecule, then how can $m = d \times V$?

Well it could be the integral of a single mass through volume V. But I am not sure. Do you have a suggestion?

 

[Miike012]

And it didn't work because the variable V from equation 2 contains the value n which is unkown

 

[DrClaude]

What about $M_\mathrm{sam}$?

 

[Miike012]

What about $M_\mathrm{sam}$?

M_sam = M*n

 

[Miike012]

And M = dRT/P

 

[Miike012]

Thank you!

 

[Miike012]

I am now solving for M for part b.

M = 3RT/(V_rms)²

The units of this RHS is Joul/(mol*Kelvin)*Kelvin/(meter/second)² = Joul*Second²/(mol*meter²)

The LHS has units of

kilograms/mole.

How does the RHS turn into kilograms/mole?

 

[DrClaude]

I didn't understand $M = m N_\mathrm{A}$ because you had written $m = d \times V$, so I had taken $m$ to be the total mass of the gas. Using the notation of the book, $m$ is the mass of one molecule and $M_\mathrm{sam}$ the total mass of the gas, therefore the correct relation is $M_\mathrm{sam} = d \times V$, which I guess is what you now have figured out, which you used along with $n = M_\mathrm{sam} / M$.

I am now solving for M for part b.

M = 3RT/(V_rms)²

The units of this RHS is Joul/(mol*Kelvin)*Kelvin/(meter/second)² = Joul*Second²/(mol*meter²)

The LHS has units of

kilograms/mole.

How does the RHS turn into kilograms/mole?

Go back to the definition of the joule in terms of base units:

J = kg m² s^-2

 

[Miike012]

I didn't understand $M = m N_\mathrm{A}$ because you had written $m = d \times V$, so I had taken $m$ to be the total mass of the gas. Using the notation of the book, $m$ is the mass of one molecule and $M_\mathrm{sam}$ the total mass of the gas, therefore the correct relation is $M_\mathrm{sam} = d \times V$, which I guess is what you now have figured out, which you used along with $n = M_\mathrm{sam} / M$.



Go back to the definition of the joule in terms of base units:

J = kg m² s^-2

Thank you for all your help.