Understanding Adiabatic Expansion in Thermodynamics

[vaishakh]

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]

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

 

[Andrew Mason]

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:

nC_VdT = dU = PdV and

VdP + PdV = nRdT = n(C_P - C_V)dT

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:

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}
Work that out to get the expression for Work.

AM

 

[siddharth]

Thanks for clearing the second part Andrew.