School of Physics and Astronomy

Junior Honours Thermodynamics GJA 2017-2018

Lecture TOPIC 4 (Finn: 3.6, 3.7, 3.8)

Synopsis: The ideal gas and isothermal and adiabatic expansions; the van der Waals equation of state.Free and Joule-Kelvin expansion coefficients introduced.

IDEAL GAS PROCESSES

Reversible Isothermal Expansion of an Ideal gas (No temperature change)

The equation of state PV = nRT gives P = nRT/V = constant/V with different values of the constantfor different temperatures (ie a family of hyperbolae). The work done on the surroundings by a systemcomprising n moles of ideal gas expanding reversibly and isothermally is equal to nRT ln(V2/V1).

Reversible Adiabatic Expansion of an Ideal gas (No heat transfer)

The First Law gives d̄Q = dU + PdV = CV dT + PdV .Put d̄Q = 0 (adiabatic condition) to get −PdV = CV dT .Then substitute for P from the ideal gas equation, replace

nR by CP − CV and integrate to get (eventually)

TV γ−1 = a constant, where γ = CP /CV

The equation describes the variation of T with V in a reversibleadiabatic process. Alternative equations (use the ideal gas equa-tion to derive them) are:

PV γ = a different constant

TP1− 1

γ = yet another constant

For a reversible adiabatic expansion, from (Pi, Vi) to (Pf , Vf ), the work done on an ideal gas, by itssurroundings is d̄W = −

∫PdV =

∫CV dT = CV (Tf − Ti).

For an irreversible process, this expression for work is not true because pressure of the gas is not welldefined, so nor is the integral

∫PdV on which the result depends. BUT the change in internal energy

∆U = CV (Tf − Ti) IS correct because U is a state variable.

Irreversible Adiabatic Expansion and Free Expansion Coefficient (or Joule Coefficient)

Consider a container with rigid adiabatic outside walls. An innerrigid partition divides the container into two equal volumes V withgas initially occupying only one of the volumes and a vacuum in theother. The system is the combination of gas plus vacuum inside theouter wall. The partition is then broken and the gas moves irre-versibly, into the vacuum, until the two halves each contain thesame amount of gas. The system does no work on its surroundings,and no heat enters the system from its surroundings. Therefore thefinal internal energy of the gas, has to be equal to the initial value.If the gas is an ideal gas, internal energy depends only on temper-ature (U=U(T)) and vice versa (T=T(U)), so its final temperaturewill be the same as its initial temperature.

gas everywhere

2V

gas vacuum

V V

breakpartition

partition

�

? ?

For a real gas, intermolecular forces depend on how far apart the molecules are, so U is a function ofboth T and V . Hence, since we are interested in changes in T , write T = T (U, V ). Then

dT =

(∂T

∂V

)U

dV +

(∂T

∂U

)V

dU =

(∂T

∂V

)U

dV = µJdV

1

µJ is the Joule coefficient, whose value depends on the gas being used, and can be predicted from the gas’sequation of state. For a free expansion of a gas from Vi to Vf , the temperature change can be calculated

from ∆T =∫ VfViµJdV .

Continuous Flow Process: The Joule-Kelvin Expansion and Coefficient

In contrast with the Free (or Joule) Expansion, the Joule-Kelvin Expansion is a more practical expansionprocess, referred to as a throttling process. The gas is forced steadily through a porous plug (various designspossible) held in a cylinder with adiabatic walls, see upper sketch.

To analyse the process, consider (lower sketch) a sample of the gas initially in an equilibrium state withthermodynamic coordinates Pi, Vi and Ti. The gas behind it, pushing it forward, behaves like a piston whichmaintains the gas at its original equilibrium pressure (Pi) as it is forced slowly but irreversibly through theplug. Beyond the plug, another imaginary piston maintains the expanded gas at the equilibrium pressureof the gas sample as it emerges to form its final equilibrium state (Pf , Vf , Tf ).

�������

�

�� �������

�

��

�������

�

��

gas gas

gassample

gassample

Pi, Ti Pf , Tf

Pi, Vi, Ti Pf , Vf , Tf

Pi + dP Pf − dP Pi Pf

porous plug

before after

- -

- -

�

For calculation purposes, we can ignore irreversibility of the motion through the plug provided weconsider initial and final equilibrium state variables. Any redistribution of energy between gas and plug canbe regarded as temporary, leaving the plug in its initial state after all the gas sample has passed through it.The gas is assumed to emerge in its final equilibrium state (Pf , Tf ).

The process is adiabatic (∆Q = 0). For the work done, it is easiest to consider the pistons: the first“piston” does work

∫ 0ViPidV (the second is similar). So the first law gives

Uf − Ui = Pi(Vi − 0)− Pf (Vf − 0) = PiVi − PfVf .

In terms of enthalpy (H = U +PV ), this becomes Hi = Hf : Enthalpy in conserved and so the Joule-Kelvinprocess is traditionally called isenthalpic.

Although the flow process is irreversible, calculations of the change in state variables such as temperaturewith changing pressure are made using an equivalent, unspecified reversible isenthalpic process.

We can always write T as a function of any two other state variable. To calculate the change in T here,it is convenient to use the conserved quantity H and the applied quantity P , and write T = T (H,P ). Then

dT =

(∂T

∂P

)H

dP +

(∂T

∂H

)P

dH =

(∂T

∂P

)H

dP = µJKdP

µJK is the Joule-Kelvin coefficient, a state variable whose value depends on the gas being used and theconditions, and can be predicted from the gas’s equation of state. For a Joule-Kelvin expansion of a gas

from Pi to Pf , the temperature change can be calculated from ∆T =∫ PfPi

µJKdP .The temperature change for an ideal gas is always zero, but the temperature change for real gases can

be either positive or negative (or zero), depending on the sign of µJK . If ∆T is negative and large, theprocess can be used to liquefy gases. The sequence of equations used to describe the process is similar tothat for the free expansion process, but with the enthalpy rather than internal energy held constant.

2

NON-IDEAL GASES: THE VAN DER WAALS GAS

The van der Waals equation of state has the same low density limit as the ideal gas, but allows for:

1. The finite size of gas molecules – the ‘free’ volume available to the molecules is less than the volumeV of the containing vessel and given by V − nb for n moles, where b is a (gas dependent) constant.Physically b represents the ‘hard core’ volume taken up by 1 mole of the gas molecules.

2. Intermolecular forces acting at large distances between the molecules. They are generally attractiveand thus tend to hold the gas together exerting an additional retaining force that must be added tothat due to the pressure P exerted by the walls. P is then replaced by P+ a

(V/n)2, where a is a constant

that gives a measure of the combined strength and range of these interactions. Note the use of Vn in

the ‘pressure correction’ term ensures that, like P , the correction term is an intensive quantity, andthat (V/n)−2 implies the same r−6 dependence as the van der Waals interaction.

The van der Waals equation of state for n moles of molecules is:(P +

a

(V/n)2

)(V − nb) = nRT

from which, dividing through by n and writing V/n = v, the van der Waals equation for 1 mole of gas is(P +

a

v2

)(v − b) = RT ;

Gas equation of state: Virial Expansion

Another way to correct the ideal gas equation uses a power law expansion in a small parameter, 1/V or P .

PV

nRT=(

1 +B2(T )n

V+B3(T )

( nV

)2+ . . .

)A similar expansion also often referred to as a Virial expansion is:

PV = n(RT +B(T )P + C(T )P 2 + . . . )

All these approximations recover the ideal gas law in the low density limit. The van der Waals equation hassome physical insight, and gives a reasonably good description up to quite high densities.

Solid equation of state: Constant Bulk modulus and Murnaghan

A solid has a well defined equilibrium volume, which suggests an equation of state expanding about theequilibrium point using bulk modulus. If the bulk modulus is constant then from the definition K0 =−V

(∂P∂V

)T

we can integrate getP = K0 ln(V/V0)

and from P =(∂U∂V

)the energy is U = U0 + V K0 (ln(V/V0)− (V − V0/V ))

The Murnaghan equation allows the bulk modulus to increase with pressure K = K0 +K ′0P , leading to:

P (V ) =K0

K ′0

[(V0V

)K′0

− 1

]

Note that T enters only implicitly through temperature dependence of equilibrium volume V0, bulkmodulus K0 and the exponent K ′0.

3

Note on intensive/extensive variables

A sample of gas in equilibrium has the same pressure throughout its volume. A region of smaller volumewill contain less mass but have the same pressure. Quantities such as pressure which are unchanged whenthe extent of a system is changed are called intensive quantities or intensive variables. Quantities such asvolume which depend on the extent of a system, eg the amount of gas present, are extensive quantities(extensive variables). In any expression in which terms are summed the summed terms must all be of thesame type, either all intensive or all extensive. A product of such as PV is extensive. An extensive quantitydefined per mole becomes intensive, and is often written with a small letter, e.g. M is total mass, m is massper mole. This is not a convention to rely on - sometimes intensive quantities are per kg (e.g. density).When solving problems make sure you understand your own notation and keep it consistent.

Appendix: Comparison of Internal Energy and Enthalpy

Enthalpy is useful for analysing processes where the boundary conditions specify the pressure, whereas theinternal energy is more useful for processes that occur at constant volume.

Internal Energy, U Enthalpy H

In general dU = d̄Q− PdV dH = d̄Q+ V dP(∂U∂T

)V

= Cv

(∂H∂T

)P

= Cp

Useful for analysing ISOCHORIC processes ISOBARIC processes(constant volume) (constant pressure)Uf − Ui = Q Hf −Hi = Q

Uf − Ui =∫ TfTiCvdT Hf −Hi =

∫ TfTiCpdT

Associated Flow process Free Expansion (Joule) Throttling (Joule-Kelvin)one-off event constant flow processUf = Ui Hf = Hi

Adiabatic reversible processes

(any gas) Uf − Ui = −∫ VfViPdV = W Hf −Hi =

∫ PfPi

V dP 6= W

Ideal gas (needs Joule’s law: U=U(T)) (needs Joule’s law & PV=nRT)

any process Uf − Ui =∫ TfTiCvdT Hf −Hi =

∫ TfTiCpdT

Equilibrium Minimised for isochoric, Minimised for isobaric,adiabatic boundary adiabatic boundary

4