From WolfWikis

LECTURE 5: Isotherms and adiabats

Isothermal expansion of an ideal gas

For a monatomic ideal gas we have seen that energy <E> observed as U= 3/2 nRT. This means that energy is only dependent on temperature. That means that if we compress a gas isothermally it does not change:

ΔU_{isothermal-ideal gas} = 0

This means that the reversible work must cancel the reversible heat:

ΔU_rev = w_rev + q_rev = 0 => w_rev = - q_rev

If V₂>V₁ (expansion) this means that you (or the environment) must put heat into the system because this is a positive number.

Adiabatic expansion of an ideal gas

Now suppose you make sure that no heat can enter the cylinder. (Put it in styrofoam or so). Then the path can still be reversible (slow pulling) but the process is then adiabatic.

This bat- part comes from a Greek verb βαινω (baino) that means walking, compare acrobat, someone who goes high places (acro-). The δια (dia) part means 'through' (cf. diagram, diorama, diagonal etc.) and the prefix α- (a-) denies it all (compare atypical versus typical).

So the styrofoam prevents the heat from walking through the wall.

When expanding the gas from V₁ to V₂ it still does reversible work but where does that come from? It can only come from the internal energy itself. So in this case any energy change should consist of work (adiabatic means: δq=0!).

dU = δw_rev.

This implies that the temperature must drop, because if U changes, T must change.

The change of energy with temperature at constant volume is known as the heat capacity (at constant volume) C_v

For an ideal gas U only changes with temperature, so that

or:

We can now compare two paths to go from state 1 to state 2

1. isothermal reversible expansion A
2. adiabatic reversible expansion B followed by isochoric (V=constant) heating C

Notice that the temperature remains T₁ for path A (isotherm!), but that it drops to T₂ on the adiabat (B), so that the cylinder has to be warmed up (C) to regain the same temperature.

ΔU_tot should be the same for both A and (B+C), because the end points are the same (U is a state function!). As the and points are at the same temperature and U only depends on T ΔU_tot = 0

Along the adiabat B q_rev=0.

Along the heating path C, there is no volume work because the volume is kept constant, so that.

This is the only reversible heat involved in path B+C.

However, we know that ΔU_tot for path A is zero (isothermal!). This means that the volume work along B must cancel the heat along C:

The book keeping looks as follows, all paths are reversible:

ΔU_B+C = ΔU_A = 0 = q_B + w_B + q_C + w_C (A is isothermal path: T=const, U=constant)

ΔU_B+C = ΔU_A = 0 = 0 + w_B + q_C + w_C (B is adiabatic q=0)

ΔU_B+C = ΔU_A = 0 = 0 + w_B + q_C + 0 (C is isochoric, V=constant => w=0)

So: w_B = - q_C

We had already seen before that along the isotherm A

w_A = - q_A = - nRTln V₁/V₂

As you can see w_A and w_B are not the same thing. (WORK IS A PATH FUNCTION even if reversible).

As we are working with an ideal gas we can be more precise about wB and qc as well.

The term wB along the adiabat is reversible volume work. Since there is no heat along B we can write a straight d instead of ∂ for the work contributions (It is the only contribution and must be identical to the state function dU):

(See 19.5)

We can make the same argument for the heat along C.

If we do the three processes A and B+C only to a tiny extent we can write:

And now we can integrate from V₁ to V₂ over the reversible adiabatic work along B and from T1 to T2 for the reversible isochoric heat along C. To separate the variables we do need to bring the temperature to the right side of the equation.:

The latter expression is valid for a reversible adiabatic expansion of a monatomic ideal gas (say Argon) because we used the C_v expression for such a system.

We can use the gas law PV=nRT to translate this expression in one that relates pressure and volume see Eq 19.23

Read section 19-6

Statistical interpretation

We can use what we know about the statistical side of thermodynamics to give a simple interpretation to a change dU:

See 19.26 and 19.29 /19.30

we see that because δw_rev = -PdV

See also section 17-5 :

In this section it is shown that we can manipulate the partition function to find the pressure of a system by calculating the above moment of the distribution. Again we take the derivative of the logarithm of the partition function Q, this time versus V and show that the result resembles the last equation pretty closely (apart from a factor ). Thus we get:

Once again we can find an important quantity of our system by manipulating the partition function Q.

Section 19-7