How is Work Done in an Adiabatic Process of an Ideal Gas Calculated?

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SUMMARY

The work done in an adiabatic process of an ideal gas can be calculated using the formula W = (1/(\gamma - 1))(PfVf - PiVi), where Pf is the final pressure and Pi is the initial pressure. The derivation involves substituting P = constant/V^γ into the integral W = -∫Pdv and recognizing the relationship PV^γ = constant. A key insight is to express the exponential term as a product of V and V^(1-γ), simplifying the integration process.

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  • Understanding of ideal gas laws
  • Familiarity with adiabatic processes
  • Knowledge of calculus, specifically integration techniques
  • Concept of thermodynamic variables (pressure, volume, temperature)
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thenewbosco
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hello the problem is as stated:
a cylinder containing n moles of an ideal gas undergoes an adiabatic process. using W=-\int Pdv and using the condition PV^\gamma=constant, show that the work done is:
W=(\frac{1}{\gamma - 1}(PfVf - PiVi) where Pf is final pressure, Pi is initial pressure...
I tried substituting that P=\frac{constant}{V^\gamma} into the integral, and evaluating from Vi to Vf, but this still leaves the gamma as an exponent. how can i go about solving this one?
thanks
 
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Right, you have an exponential term that looks like V^{1- \gamma}. Write this as a product of two terms, one of which is V itself. What can you do with the other part ?
 
awesome thanks for your help. all i needed was to see it as 1-gamma, instead of how i had it as -gamma +1. it made a lot of difference.
 

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