   # First law of thermodynamics and Adiabatic and Isothermal changes

### (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:

Isothermal change:    0 = dQ + dW    and so    dW = -dQ

This equation backs up the idea of addition or removal of heat energy ñ 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 ñ 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 ñ 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:

Adiabatic change:    dU = 0 + dW    dU = dW

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 ñ in other words the temperature of the gas goes up.

### (c) a change at constant volume

In this case dV = 0 and the first law becomes:

Change at constant volume:    dU = dQ

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.

Example problem
The air in a cylinder in the air suspension system of a car is compressed. Its volume decreases and its internal energy increases. (It acts rather like the energy stored in a compressed spring). If 200 J of work are done in compressing the air and 150 J of heat are conducted away through the walls of the cylinder what is the increase in the internal energy of the air?

Using the first law: DU = DQ + DW we have DU = -150J + 200J = 50J Notice that the change in heat energy is negative since heat leaves through the cylinder walls.

Increase in internal energy = 50J

### 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 ñ CV = R we have
CVdT/T + (CP- CV)dV/V = 0 and writing g = CP/CV and integrating gives:

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

Adiabatic change :    PVg = constant    TV(g - 1) = constant