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Finding velocity of gas molecule

  • Thread starter Miike012
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  • #1
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Equations:
1. vrms = √(3RT/M)
2. PV = nRT
3.M = mNA

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[NA] = M = d(nRT/P)[NA] = dNRT/P; N = nNA.

From equation 1.
vrms = √(3RTP/(dNRT)) = √(3P/dN).
I get stuck at this part because I dont know the value of N.
 

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  • #2
DrClaude
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  • #3
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What is that equation?
Is M molar mass or mass?
I thought it was molar mass...
 
  • #4
DrClaude
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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 NA.
 
  • #5
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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 NA.
I added the equations in the attachment from the book
 

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  • #6
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I think I figured it out. M=mNa and m = M/Na = d*v, so M = d*v*Na
 
  • #7
DrClaude
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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##?
 
  • #8
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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?
 
  • #9
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And it didn't work because the variable V from equation 2 contains the value n which is unkown
 
  • #10
DrClaude
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What about ##M_\mathrm{sam}##?
 
  • #11
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  • #12
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And M = dRT/P
 
  • #13
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Thank you!
 
  • #14
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I am now solving for M for part b.

M = 3RT/(Vrms)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?
 
  • #15
DrClaude
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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/(Vrms)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
 
  • #16
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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.
 

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