Thermodynamics basics

Can anyone here help me to derive that during an adiabatic expansion, PV^gamma?is a constant, as well as other expressions similar to the above?
I have found the following equation using definite integration and the basic formulae, nRdT/gamma - 1 = work done.
 

siddharth

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vaishakh,
You can prove [tex] P V^{\gamma}[/tex] is constant for an ideal gas in an adiabatic process from the first law ( [itex] dU = dQ - pdV , dQ=0) [/itex] and the ideal gas law [itex] (pV=nRT) [/itex].
How did you get your formula nRdT/gamma - 1 = work done.?
 
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Andrew Mason

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vaishakh said:
Can anyone here help me to derive that during an adiabatic expansion, PV^gamma?is a constant, as well as other expressions similar to the above?
Since heat flow (Q) is zero, use:

[tex]nC_VdT = dU = PdV[/tex] and

[tex]VdP + PdV = nRdT = n(C_P - C_V)dT[/tex]

This will give two expressions for ndT. Integrate both expressions.
I have found the following equation using definite integration and the basic formulae, nRdT/gamma - 1 = work done.
This follows from the adiabatic condition. One can express the work as:

[tex]W = \int_{V_i}^{V_f} PdV = \int_{V_i}^{V_f} \frac{PV^\gamma}{V^\gamma}dV = K\int_{V_i}^{V_f} \frac{dV}{V^\gamma}[/tex]
Work that out to get the expression for Work.

AM
 
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siddharth

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Thanks for clearing the second part Andrew.
 

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