Adiabatic Process: Work

  • Thread starter sisigsarap
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  • #1
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I have 5.00 mol of O2(Oxygen) at 20.0 Celsius and 1.00atm. I will compress this to 1/10 the original volume.

Find the work? The correct answer is 46.1 Kj

I am having a terrible time with this.

This is what I know: Q = 0 for an adiabatic process, and the change in internal energy is equal to work.

I think I use integrate Pdv, but Im getting very confused. I am really lost and need a push in the right direction!
 

Answers and Replies

  • #2
691
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Ideal gas law? You know pressure, temperature and the number of atoms and 1 atm is much less than the critical pressure.
 
  • #3
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PV = nRT
I just dont see how to plug it in. Pleaseee help its driving me crazy!
 
  • #4
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sisigsarap said:
PV = nRT
I just dont see how to plug it in. Pleaseee help its driving me crazy!

What variable do you not have a value for? There's only one. Solve for the unknown plug in your known values. This will yield your initial condition. Apply the condition in the question. Then use:

[tex]W=\int_{V1}^{V2}PdV[/tex]
 
  • #5
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In an adiabatic process [tex]Q=0[/tex] then [tex]\Delta U+L=0[/tex] and [tex]L=-\Delta U[/tex]. You can find now
[tex]L=-\nu C_V \Delta T[/tex].
(Because you deal with a diatomic gas, you'll have [tex]C_V=\frac{5}{2}R[/tex]).

The final temperature is taken from [tex]TV^{\gamma-1}=const[/tex] (here [tex]T[/tex] is in Kelvin, as you probably know).

OBS. In my opinion, the work made by the gas must be negative!
 
Last edited:

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