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Monatomic Ideal Gas volume expansion

  • Thread starter TFM
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  • #1
TFM
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[SOLVED] Monatomic Ideal Gas volume expansion

Homework Statement



A monatomic ideal gas expands slowly to twice its original volume, doing 260 Joules of work in the process.

(a)

Find the heat added to the gas if the process is isobaric.

(b)

Find the change in internal energy of the gas if the process is isobaric.


Homework Equations



[tex] U = \frac{1}{2}nRT [/tex] per degree of freedom

pV = nRT

pV = const.
p/T = Constant
V/T = const.

The Attempt at a Solution



I am not quite sure what to do for this equation. I now that the pressure is constant, and the the volue has doubled. and it has done 260 Joul;es of work in the process.

Any ideas where to start?

TFM
 

Answers and Replies

  • #2
Doc Al
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Try to figure out the change in internal energy. You'll have to play around a bit. Assemble all the equations that apply to this problem and see what you can do.
 
  • #3
TFM
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Have I started this the right way.

The gas is monatomic, so

[tex] U = \frac{3}{2} nRT [/tex]

R is a constant, 8.31

[tex] U = \frac{3}{2} 8.31*nT [/tex]

However, we don't know thenumber of moles.

[tex] \frac{v_1}{T_1} = \frac{v_2}{T_2} [/tex]

We know the volume of v2 is twice that of V1

[tex] \frac{1}{T_1} = \frac{2}{T_2} [/tex]

This gives the ratio of temperatures as:

[tex] \frac{T_1}{T_2} = \frac{1}{2} [/tex]

Does this look right so far? If so, what could be a possible next step?

TFM
 
  • #4
Doc Al
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Have I started this the right way.

The gas is monatomic, so

[tex] U = \frac{3}{2} nRT [/tex]
Good.

Combine this with the ideal gas law and an expression for the work done.

R is a constant, 8.31

[tex] U = \frac{3}{2} 8.31*nT [/tex]
Don't plug in numbers until the last step.
 
  • #5
TFM
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Following on then,

[tex] pV = nRT [/tex]

[tex] n = \frac{pV}{RT} [/tex]

Insert into U ewquation,

[tex] U = \frac{3}{2} \frac{pV}{RT} RT [/tex]

Cancels to:


[tex] U = \frac{3}{2} pV [/tex]

The Work Equation:

[tex] W = p \Delta V [/tex]

thus,

[tex] p = \frac{W}{\Delta V} [/tex]

putting into the equation gives:

[tex] U = \frac{3}{2} \frac{W}{\Delta V}V [/tex]

Sincve the V is changing anyway:

tex] U = \frac{3}{2} W [/tex]

and we have the work.

Does this look right?

TFM
 
  • #6
TFM
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Sorry, that final equation hould have been:

[tex] U = \frac{3}{2} W [/tex][/B]

Is this correct?

TFM
 
  • #7
Doc Al
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I assume you mean:
Sorry, that final equation hould have been:

[tex] U = \frac{3}{2} W [/tex]

Is this correct?
Good! Now use it to find the heat added.
 
  • #8
TFM
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Would you use:

[tex] \Delta U = Q - W [/tex]

[tex] \Q = U + W [/tex]

?

TFM
 
  • #9
Doc Al
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Looks good to me.
 
  • #10
TFM
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I put the values ion and the right answer did come out!!!

Thanks for your assistance, Doc Al, :smile:

TFM
 

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