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 – 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:
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 – 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: ΔU = ΔQ + ΔW we have ΔU = -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 = C
VdT where C
V 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 C
P
– C
V = R we have
C
VdT/T + (C
P- C
V)dV/V = 0
and writing γ = C
P/C
V and integrating
gives:
TV
(γ-1) = constant and since the gas still obeys PV
= RT we have: PV
γ =
constant
Adiabatic change : PVγ = constant TV(γ - 1) = constant
