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Thermodynamics - RMS speed

  • #1

Homework Statement


So this is the final section of a problem I had posted earlier, and I'm kinda stumped again.

The rms speed of the molecules in 1.2 g of hydrogen gas is 1800 m/s.
300 J of work are done to compress the gas while, in the same process, 1500 J of heat energy are transferred from the gas to the environment. Afterward, what is the rms speed of the molecules?


Homework Equations



E = N(1/2)mc^2

m = Mass of H molecule = 1.67 x 10^-27 kg
N = Total # of Molecules = 7.19 x 10^23 Molecules
c = RMS speed

E(total) = E1 (initial energy) + E2 (300J of work) + E3 (1500 J lost to environment)

The Attempt at a Solution




initial Energy = N x (1/2)mC2
Which equates to 1.9 KJ earlier in the assignment.

E = (1900J) + (300J - the work done on the gas) - (1500J subtracted because it is lost to the environment)
= 700 J

Now plug this back in to solve for C^2

700J = N x (1/2)mc^2

700J = (7.19 x 10^23 Molecules)(1/2)(1.67 x 10^-27 kg)(C^2)

(c^2) = 1,166,083.6

c = 1,825 m/s

It is telling me I am wrong, so can someone shed some light on my mistake?

Thank you.
 

Answers and Replies

  • #2
anyone?
 
  • #3
collinsmark
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(c^2) = 1,166,083.6

c = 1,825 m/s

It is telling me I am wrong, so can someone shed some light on my mistake?
I haven't gone through all the numbers. But for starters you might want to check the square root function on your calculator. :wink:
 
  • #4
collinsmark
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Couple of other things.

I'm not sure where you got the mass of an H2 molecule. But as I remember, there are two hydrogen atoms in a single hydrogen molecule. (It seems to me that you might have assumed one H atom per molecule.)

Beyond that, you might want to make sure you use a sufficient number of significant digits in your calculations.
 
  • #5
Haha, oops.

Well, I tried this equation with both 1900J (total translational kinetic energy) and 3200J (thermal energy) for my E1. Using 3200 J ends up in 1,825 m/s and 1900 J ends with 1080 m/s. (both came back as incorrect)

Sorry for the confusion!!
 
  • #6
Alright, So the calculation itself appears to you to be fine, just make sure my #'s are correct?
 
  • #7
So I plugged in the new value:

Hydrogen atom (diatomic) 3.34 x 10^-27 kg (used 1.007g/mol / avagraddo's)

and my overall answer came out as 1,288 m/s and the assignment is still saying it is wrong.
I am obviously missing something, but I don't see what. Any help/hints would be appreciated!

Thanks
 
  • #8
collinsmark
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So I plugged in the new value:

Hydrogen atom (diatomic) 3.34 x 10^-27 kg (used 1.007g/mol / avagraddo's)
Don't you mean 3.34 x 10-27 kg per diatomic molecule?

Anyway, you might want to combine your formulas before plugging any numbers into them. It ends up that a lot of things might cancel out, making the overall calculations easier. It turns out that you might not even need Avogadro's number, molar masses, or any of that.
and my overall answer came out as 1,288 m/s and the assignment is still saying it is wrong.
I am obviously missing something, but I don't see what. Any help/hints would be appreciated!
I ended up with a different answer.
 
  • #9
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I just registered purely for this question here.
I'm also working on a similar problem but with different numbers.
Your work actually helped me since at first I ignored the initial thermal energy. So my calculation ended up being just the Ethermal from the work and heat given(ie the final). Which was wrong.

I managed to figure it out and I think I know what the problem is. Keep using the mass already given to you. 1.2g = 0.0012kg
And don't forget that Hydrogen is a diatomic gas with 3 translational 2 rotational degrees of freedom. I'm pretty sure I also saw you post the first two parts of this question. You had to use 5/3 in the first two parts. Same thing here.

E = (5/3) (1/2)mc^2

Hope that helps.
 
  • #10
collinsmark
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Oh, yeah. The rotational part too. I forgot about that. :redface:

[Edit: However, I'm not sure if the OP is supposed to consider that or not. I think it depends on the model being used.]
 
Last edited:
  • #11
Perfect,

I returned to this problem and got it correct. Thank you for all of your help.

@WonSol
Just out of curiosity, do you go to UofC?
It would be quite the coincidence if two institutions had the same question, during the same week haha.
 

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