Work done in a reversible adiabatic change

[AndyW]

Can anyone help me with this?

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Show that the work done in a reversible adiabatic change from an initial state (p1, V1) to a final state (p2, V2) is:

W = ( p2V2 - p1V1) / (gamma - 1)

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Here's what (little) I have done so far:

Reversible => dW = -pdV
Adiabatic => dq = 0
gamma = Cp/Cv

I know pV^gamma = constant, so p1V1^gamma = p2V2^gamma

 

[Clausius2]

As you stated, you have P(V) as an adiabatic process.

The only thing you have to do is to substitute P(V) in the integral of work, and solve the integral.

 

[wais]

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Since dq = 0, dU = -dW, and U = CvT => dU = Cv dT.

dT = -dV/Cv * (pV^gamma)/V = -pV^(gamma - 1) dV/Cv

From the ideal gas law, pV = nRT, so dT = -RT/V dV/Cv

Substituting this into the first equation, we get:

dU = -nRT/V dV = nRT dV/V

Integrating both sides from V1 to V2, we get:

U2 - U1 = nRT * ln(V2/V1)

Since U = CvT and Cv is constant, we can substitute U = CvT into the equation and rearrange to solve for T2:

T2 = T1 * (V1/V2)^(gamma - 1)

Now, we can use the ideal gas law to solve for p2:

p2 = nRT2/V2 = nRT1 * (V1/V2)^(gamma - 1)

And finally, we can substitute p2 into the equation for work done (dW = -pdV) to get:

dW = -nRT1 * (V1/V2)^(gamma - 1) * dV

Integrating both sides from V1 to V2, we get:

W = nRT1 * (V1/V2)^(gamma - 1) * (V2 - V1)

Simplifying, we get:

W = nRT1 * (V1^(gamma - 1) - V2^(gamma - 1))

Substituting back in for gamma = Cp/Cv, we get:

W = (p2V2 - p1V1) / (gamma - 1)

I hope this helps! Let me know if you have any questions or if you'd like me to explain any steps further.