Calculating Work in an Adiabatic Process for a Diatomic Gas

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In the discussion about calculating work in an adiabatic process for a diatomic gas, the user is working with 5.00 mol of O2 at 20.0 Celsius and 1.00 atm, compressing it to 1/10 of its original volume. They correctly note that in an adiabatic process, heat transfer (Q) is zero, and the change in internal energy equals the work done. To find the work, they are advised to integrate the pressure with respect to volume and apply the ideal gas law. The conversation emphasizes using the relationship between temperature and volume in adiabatic conditions, specifically that the final temperature can be derived from the equation TV^(γ-1) = constant. The user is reminded that the work done by the gas should be considered negative due to the nature of compression.
sisigsarap
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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 I am getting very confused. I am really lost and need a push in the right direction!
 
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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.
 
PV = nRT
I just don't see how to plug it in. Pleaseee help its driving me crazy!
 
sisigsarap said:
PV = nRT
I just don't 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:

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

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

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