# Thermodynamics basics

1. Dec 18, 2005

### 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.

2. Dec 18, 2005

### 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.?

Last edited: Dec 18, 2005
3. Dec 19, 2005

### Andrew Mason

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.
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

Last edited: Dec 19, 2005
4. Dec 19, 2005

### siddharth

Thanks for clearing the second part Andrew.