Thermodynamics: Proofs of work done on/by gas during adiabatic process

AI Thread Summary
The discussion focuses on deriving the work done on and by a gas during a quasistatic adiabatic compression. The user is confused about the signs in their equations and seeks clarification on the correct expressions for work. They present their attempts, showing that the work done on the gas is W = (P_f V_f - P_i V_i) / (γ - 1) and the work done by the gas is w = (P_i V_i - P_f V_f) / (γ - 1). A participant confirms the user's derivations and clarifies the sign changes involved. The user expresses gratitude for the assistance and acknowledges their struggle with the concepts.
Matty R
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Hello

I'm really confused with this and would appreciate any help.

Homework Statement



a) Show that the work done on a gas during a quasistatic adiabatic compression is given by:

W = \frac{P_f V_f - P_i V_i}{\gamma - 1}

b) Show that the work done by a gas during a quasistatic adiabatic compression is given by:

w = \frac{P_i V_i - P_f V_f}{\gamma - 1}


Homework Equations



w = -W
PV^{\gamma} = K


The Attempt at a Solution



a) dW = -PdV
P = KV^{- \gamma}
W = -K \left[\frac{V^{1- \gamma}}{1 - \gamma} \right]^{V_f} _{V_i}
W = - \frac{P_f V_f - P_i V_i}{1 - \gamma}
W = \frac{P_i V_i - P_f V_f}{\gamma - 1}

I don't understand where I've gone wrong.

b) w = -W
w = - \frac{P_i V_i - P_f V_f}{\gamma - 1}
w = \frac{P_f V_f - P_i V_i}{1 - \gamma}

I also get this answer if I derive w from:

dw = PdV

I've probably got a minus sign mixed up, or something like that, but I've been through the derivation so many times it's invading my dreams. Okay, maybe it isn't that bad, but I'm really struggling.
 
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Matty R said:
Hello

I'm really confused with this and would appreciate any help.

Homework Statement



a) Show that the work done on a gas during a quasistatic adiabatic compression is given by:

W = \frac{P_f V_f - P_i V_i}{\gamma - 1}


w = -W
PV^{\gamma} = K


The Attempt at a Solution



a) dW = -PdV
P = KV^{- \gamma}
W = -K \left[\frac{V^{1- \gamma}}{1 - \gamma} \right]^{V_f} _{V_i}
W = - \frac{P_f V_f - P_i V_i}{1 - \gamma}
W = \frac{P_i V_i - P_f V_f}{\gamma - 1}

I don't understand where I've gone wrong.


You have changed sign too many times:-p.
W = - \frac{P_f V_f - P_i V_i}{1 - \gamma}=\frac{P_f V_f - P_i V_i}{\gamma-1}
 
Thanks for the reply.

So, erm, is this true:

W = - \frac{P_f V_f - P_i V_i}{1 - \gamma}
= \frac{P_f V_f - P_i V_i}{\gamma-1}
= \frac{P_i V_i - P_f V_f}{1 - \gamma}
 
Matty R said:
So, erm, is this true:

W = - \frac{P_f V_f - P_i V_i}{1 - \gamma}
= \frac{P_f V_f - P_i V_i}{\gamma-1}
= \frac{P_i V_i - P_f V_f}{1 - \gamma}

Yes.

ehild
 
Thank you very much for your help.

I'm feeling quite embarrassed now, though most of the threads I've created here have been about really basic mistakes, so I should be used to it. :rolleyes:
 
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