Question

Derive the relationship between the temperature and volume for a reversible adiabatic process.

Answer 1

From the First Law of thermodynamics;

$\Delta$$E$ $= q - w$

In adiabatic change q =0

Therefore; $\Delta$$E$ $= -w$

When we consider small changes;

$dE = -dw$

And in terms of heat capacity, we know that;

$Cv=\dfrac{dE}{dT},$

hence;

$CvdT = dE$

and therefore;

$CvdT = -dw$

Since $dw = PdV$ , then;

$CvdT = -PdV$ .....................................................(i)

Now, if we express Pressure (P) in terms of volume (V) and Temperature (T);

We know the Ideal gas equation for one mole is;

$PV = RT$

$P$ $=\dfrac{RT}{V}$

Therefore;

$CvdT = \dfrac{-RTdV}{V}$

dividing both sides by T will give;

$\dfrac{CvdT}{T} = \dfrac{-RdV}{V}$ ...............................................(ii)

Integrating the equation will give;

$Cv\int \dfrac{dT}{T} = -R\int dV\dfrac{1}{V}$

hence;

$Cvln(\dfrac{T2}{T1}) = -Rln(\dfrac{V2}{V1})$