Work done in a reversible adiabatic expansion

AI Thread Summary
To determine the work done during a reversible adiabatic expansion, it's important to recognize that the heat transfer (Q) is zero, leading to the relationship ΔU = W_rev. The discussion highlights that while both pressure and volume change, the formula W = p_exΔV cannot be applied directly. Instead, the change in internal energy can be expressed using ΔU = C_vΔT, which is valid for ideal gases regardless of whether volume or pressure is constant. The key is identifying the constant state variable that remains unchanged during the process. Understanding these principles is crucial for solving the problem effectively.
Trowa
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Hi there!

I have to determine the work done of a reversible adiabatic expansion. Becauce the system is adiabatic: Q = 0 so \DeltaU = Wrev

Becauce both the pressure and the volume changes I can't use W = pex\DeltaV.

Homework Statement


Cv, Cp, Ti, Tf, Pi, Pf, Vi, Vf are known

The Attempt at a Solution



I thought at first that I could use \DeltaU = CV.\DeltaT but the volume is not constant so I don't know if I could use it.

Who can help me find the right formula?

Thanx in Advance!
 
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Hi Trowa, welcome to PF.

(1) Because no heat goes in or out of the system, and because the process is reversible, one state variable remains constant. It's one that you haven't listed. What is it?

(2) \Delta U =c_V\Delta T=(c_P-R)\Delta T always holds for an ideal gas, and doesn't require constant volume, constant pressure, or any other condition.
 
Thanx for the fast response.
:smile:
 

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