Adiabatic and Isothermal Changes

Adiabatic and Isothermal changes and the First law of thermodynamics

(a) an isothermal change

This is a change where the temperetaure of the gas remains constant and to achieve this we will have to either add or remove heat from the gas during the change - adding heat energy if is expanded and removing if it is compressed. Think about how hot a bicycle tyre valve gets as you pump up the tyre. Since the temperature of the gas remains constant dU = 0.

The first law of thermodynamics for such a change becomes:

This equation backs up the idea of addition or removal of heat energy – work has been put into the gas and this means that heat must be taken out to keep the temperature constant!

Compression

If the gas is compressed dW is positive and so dQ must be negative – heat energy must be removed from the gas to keep its temperature constant.

Expansion

If the gas is expanded dW is negative and so dQ must be positive – heat energy must be supplied to the gas to keep its temperature constant.

(b) a reversible adiabatic change

In such a change there is no change in the heat content of the gas and therefore dQ = 0.

Such a change must therefore take place in an insulated chamber.

The first law of thermodynamics for an adiabatic change is therefore:

Remembering that dW is the work done on the gas this shows that for an ideal gas a decrease in volume results in an increase in the internal energy of the gas – in other words the temperature of the gas goes up.

In this case dV = 0 and the first law becomes

This means that for an ideal; gas an input of energy goes purely to raise the temperature of the gas as long as the volume of the gas remains constant.

Proof of the formula for a reversible adiabatic change

We will assume that we are dealing with an ideal gas and therefore the first law of thermodynamics is obeyed and so: dU = dQ + dW

Let the gas expand from volume V to volume V+dV at constant pressure P and let the temperature fall from T to T-dT. (Remember here that all temperatures are measured on the Kelvin scale).

Consider one mole of gas.

For a gas at constant volume the increase in internal energy of the gas is the mass of the gas times its specific heat capacity we have:

dU = CVdT where CV is the specific heat of the gas at constant volume

so for an adiabatic change CVdT +PdV = 0

But PV = nRT and for one mole n= 1 and so PV = RT P = RT/V so CVdT + RdV/V = 0

But for an ideal gas CP – CV = R we have

CVdT/T + (CP- CV)dV/V = 0 and writing γ = CP/CV and integrating gives:

TV(γ- 1) = constant and since the gas still obeys PV = RT we have: PVg = constant