Unit-IV: Real Gases

Behavior of Real Gases. Critical Constants. Van der Waal’s Equation of State for Real Gases. Values of Critical Constants. Law of Corresponding States. Joule-Thomson Porous Plug Experiment. Joule-Thomson Effect for Real and Van der Waal Gases. Joule-Thomson Cooling.

Behaviour of real gases

In case of an ideal gas, the size of molecule and intermolecular forces are negligible. The behaviour of real gases usually agrees with the predictions of the ideal gas equation to within 5% at normal temperatures and pressures. At low temperatures or high pressures, real gases deviate significantly from ideal gas behaviour.

1. Van der Waal’s Equation of State

At high pressure, the molecules are close enough therefore as such size of molecules cannot be neglected in comparison to the volume of gas. At high pressure, molecule are close, so intermolecular forces also cannot be neglected.

Correction to the volume term

Because of finite size of molecules, in place of the volume of the container, volume available to the molecule should be taken.

Let the unavailable volume per mole=b, where 𝑏 ≅ , 𝑟 being the radius of the molecule.

If n moles are there, unavailable volume=nb. The effective volume, 𝑉 = (𝑉 − 𝑛𝑏) and the EOS becomes; 𝑃𝑉 = 𝑛𝑅𝑇 1. 1

𝑃(𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇 1.2 (P is pressure, V is total Volume, T is Temp, R is gas constant)

𝑃(𝑣 − 𝑏) = 𝑅𝑇 1.3 (𝑣 = 𝑉/𝑛 is the volume per mole)

Correction to the Pressure term

Let a given molecule be well inside the container as shown in Figure 1. Attractive forces will act from all the sides. When it reaches the surface or strikes the wall, it is not attracted from all the sides. Hence a particle at the wall experiences a resulting force directed towards the container’s interior, which reduces the pressure on the container’s wall. Therefore the velocity or momentum of molecule gets reduced. Reduction in momentum results in reduction in pressure. Since attraction occurs between pairs of molecules, it must increase as the square of the concentration(N/V) . In a container of volume V containing N gas molecules,

Reduction in pressure ∝ No. of molecules/volume, in the outer layer (n=N/V)

Or, Reduction in pressure ∝No. of molecules/volume, in the next layer (n=N/V)

Or, Reduction in pressure ∝ n

Or, Reduction in pressure = 𝛽n where 𝛽 is a specific constant for each gas.

(Caution: n is the gas concentration, 𝑛 is the no. of moles, N is the no. of gas molecules)

Reduction in pressure = 𝛽 = 𝛽 = 𝛽 = ,

Where, 𝑛 = , and 𝑎 = 𝛽𝑁 1. 4

The effective pressure, 𝑃 > 𝑃 ( the wall pressure).

Or, 𝑃 = 𝑃 − = 𝑃 − 1.5

Or, 𝑃 = 𝑃 + 1. 6

So, taking both the corrections into consideration the Vander Waal’s EOS of state can now be written as,

𝑃 𝑉 = 𝑛𝑅𝑇 1.7

𝑃 + (𝑉 − 𝑛𝑏) = 𝑛𝑅𝑇 1. 8

𝑃 + (𝑣 − 𝑏) = 𝑅𝑇 1.9 EOS of Van der Waal’s gas

1. Critical Constants of Van der Waal’s gas

In the graph in the figure below, the pressure, volume and temperature corresponding to the point P are called critical temperature𝑇 , critical pressure𝑃 , and critical volume𝑉 .

The critical temperature 𝑇 of a gas is defined as that temperature below which the gas can be liquefied by the application of pressure alone. Above the 𝑇 the gas cannot be liquefied howsoever large the applied pressure may be.

The critical pressure 𝑃 of a gas is the applied pressure at its critical temperature so that the gas is liquefied.

The critical volume 𝑉 of a gas is the volume at its critical temperature and critical pressure.

Andrew’s experiment to determine the critical constants

In his experiment on CO2, Andrews came to a conclusion that at high temperatures, despite high pressure, the gases cannot be liquefied. In the case of carbon dioxide, at 30.98° C, the gas started changing into a liquid. Andrews in his experiment observed that above a specific temperature, the gas sample couldn’t be liquefied, howsoever high the pressure becomes. The graph between the Pressure and Volume at a given constant temperature is the isotherm.

On both the right side of the critical point and on the left side of the critical point the curve is concave. It is also called as the point of inflexion. At the critical point slope is zero, we have

= 0 and = 0 2.1

We have from the Van der Waal’s EOS,

𝑃 + =( )

2.2

Or, 𝑃 =( )

− 2.3

=( )

+ 2.4

=( )

− 2.5

At the critical point, i.e., 𝑇 = 𝑇 , 𝑣 = 𝑣 and ,

= 0 and ,

= 0 , we get

,

=( )

+ =0 2.6

Or, ( )

= 2.7

And, ,

=( )

− = 0 2.8

Or, ( )

= 2.9

Dividing eq. 2.7 by eq. 2.9

( )

( )

= 2.10

Or, ( )

= 2.11

Or, 3(𝑣 − 𝑏)=2𝑣 2.12

Or, critical volume, 𝑣 = 3𝑏 2.13

Plugging the value of 𝑣 from eq. 2.13in eq.2.7, one gets

( )=

( ) 2.14

Or, = 2.15

Or, 𝑇 = 2.16

Feeding this value in eq.2.3, we get

𝑃 =( )

− 2.17

Or, 𝑃 =( )

−( )

2.18

Or, 𝑃 = − 2.19

Or, 𝑃 = = 2.20

Or, 𝑃 = 2.21

Now we have the set of critical constants

𝑣 = 3𝑏 2.13

𝑇 = 2.16

𝑃 = 2.21

From 2.16, we get ,

= 2.22

From 2.21, we get ,

= 27𝑃 2.23

Dividing eq. 2.22 by eq. 2.23, one obtains

= 2.24

𝑏 = 2.25

Substituting eq. 2.25 in eq.2.22, gives

= 2.26

Or, 𝑎 = 2.27

The eqs. 2.25 and 2.27 give the values of a and b in terms of critical constants.

2. Van der Waal’s reduced EOS and Law of Corresponding states

The Van der Waal’s EOS can be written in a form which is applicable to any substance. This is done by introducing the concept of reduced pressure, volume and temperature, i.e,

𝑃 = ; 𝑣 = ; 𝑇 = 3.1

𝑃 = 𝑃 𝑃 ; v= 𝑣 𝑣 ; T= 𝑇 𝑇 3.2

𝑃 = 𝑃 ; v= 3𝑏𝑣 ; T= 𝑇 3.3

Therefore, the Van der Waal equation 𝑃 + (𝑣 − 𝑏) = 𝑅𝑇 becomes

𝑃 + (3𝑏𝑣 − 𝑏) = 𝑅 𝑇 3.4

𝑃 + 3𝑏 𝑣 − = 𝑇 3.5

𝑃 + 𝑣 − = 𝑇 3.6

𝑃 + 𝑣 − = 𝑇 3.7

𝑃 + (3𝑣 − 1) = 8𝑇 3.8

This equation is called the van der Waal’s reduced EOS. This equation does not contain a and b, which are the characteristics of particular gas. The same equation applies to any Van der Waal’s gas. This equation is also called law of corresponding states. According to this law, two different gases are said to be in the corresponding states if their pressure, volume and temp are the same fraction (or multiple) of the critical pressure, volume and temperature of two substances.

4. Joule’s law

= 0, in word the internal energy of an ideal gas depends only on temperature and is

independent of volume. Real gases do not obey this law because they possess intermolecular interactions.

Joule-Thomson Effect (Joule–Kelvin effect or Kelvin–Joule effect)

When a gas or liquid at a constant high pressure is forced through a valve or porous plug thermally insulated from outside to a region of constant low pressure, it suffers a temperature change. This is called Joule–Thomson effect and the process as throttling, which is an irreversible process. At room temperature, all gases except hydrogen, helium, and neon cool upon expansion by throttling. These three gases experience the same effect but only at lower temperatures.

Most liquids such as hydraulic oils will be warmed by the Joule–Thomson throttling process. The gas-cooling throttling process is commonly exploited in refrigeration processes such as air conditioners, heat pumps, and liquefiers. In hydraulics, the warming effect from Joule–Thomson throttling can be used to find internally leaking valves as these will produce heat which can be detected by thermocouple or thermal-imaging camera.

Porous-Plug Experiment

The porous plug experiment established the existence of intermolecular interactions in the case of real gases. In the figure the gas is first allowed to pass through a water bath by compressing the piston P and then through a porous plug across which temperatures are measured.

The main results are of the experiment

1. All gases, when passed through the porous plug, suffer a change in temperature. 2. The change in temperature is directly proportional to the pressure-difference on the two

sides of the plug.

3. At ordinary temperatures, all gases, except hydrogen, helium and neon, suffer a fall in temperature (cooling effect), while hydrogen, helium and neon suffer a slight rise in temperature (heating effect). For example, at 00C, air shows a cooling of 0.2750C, while hydrogen shows a heating of 0.040C per atmosphere pressure difference.

4. The cooling effect diminishes as the initial temperature of the gas rises, and changes to a heating effect above a certain temperature, called the inversion temperature. The inversion temperature is different for different gases.

Theory of Joule-Thomson Expansion

(a) Constancy of Enthalpy

External work done on the gas by left piston in Fig.a,

𝑊 = 𝑃 𝐴𝑥 = 𝑃 𝑉 4.1

External work done by the escaping gas on right piston in Fig.b,

𝑊 = 𝑃 𝐴𝑥 = 𝑃 𝑉 4.2

Therefore, net external work done by thee gas in passing through the plug

∆𝑊 = 𝑊 − 𝑊 = 𝑃 𝑉 − 𝑃 𝑉 4.3

As the cylinder is thermally insulated no heat is excahged between the gas and its surroundings, i.e., ∆𝑄 = 0 4.4

Change in internal energy, ∆𝑈 = 𝑈 − 𝑈 4.5

First law of thermodynamics, ∆𝑄 = ∆𝑈 + ∆𝑊 4.6

0 = ∆𝑈 + ∆𝑊 = 𝑈 − 𝑈 + 𝑃 𝑉 -𝑃 𝑉 4.7

𝑈 + 𝑃 𝑉 = 𝑈 + 𝑃 𝑉 4.8

Or, 𝐻 = 𝐻 = 𝑎 constant, as 𝐻 = 𝑈 + 𝑃 𝑉 4.9

If, 𝑃 𝑉 >𝑃 𝑉 ; 𝑈 < 𝑈 , the internal energy will decrease and hence cooling will result.

If, 𝑃 𝑉 <𝑃 𝑉 ; 𝑈 > 𝑈 , the internal energy will increase and hence heating will result.

If, 𝑃 𝑉 =𝑃 𝑉 ; 𝑈 = 𝑈 , the gas is perfect and obeys Boyle’s law.

(b) Joule Thomson coefficient

It is seen from Joule Thomson expansion that although there is a pressure difference on the two sides of the porous plug, the enthalpy of the gas remains constant,

𝐻 = 𝑈 + 𝑃𝑉= A constant 4.10

𝑑𝐻 = 𝑑 𝑈 + 𝑃𝑑𝑉 + 𝑉𝑑𝑃 = 0 4.11

But, 𝑑 𝑈 + 𝑃𝑑𝑉 = 𝑑𝑄 = 𝑇𝑑𝑆 4.12 (1st & 2nd law)

Therefore, 𝑑𝐻 = 𝑑𝑄 + 𝑉𝑑𝑃 = 0 4.13

Or, 𝑇𝑑𝑆 + 𝑉𝑑𝑃 = 0 4.14

If we take entropy S as a function of two independent variables P and T

𝑆 = 𝑆(𝑃, 𝑇) 4.15

The change in entropy 𝑑𝑆 can be expressed as

𝑑𝑆 = 𝑑𝑃 + 𝑑𝑇 4.16

Feeding this in eq.(4.14),

𝑇 𝑑𝑃 + 𝑑𝑇 + 𝑉𝑑𝑃 = 0 4.17

Or, 𝑇 𝑑𝑃 + 𝑇 𝑑𝑇 + 𝑉𝑑𝑃 = 0 4.18

Now, 𝑇 = = 𝑐 4.19

And by Maxwell’s 4th relation, = − 4.20

Using eq.4.20, eq.4.18 becomes,

−𝑇 𝑑𝑃 + 𝑐 𝑑𝑇 + 𝑉𝑑𝑃 = 0 4.21

Or, 𝑐 𝑑𝑇 = 𝑇 𝑑𝑃 − 𝑉𝑑𝑃 4.22

Or, 𝑐 𝑑𝑇 = 𝑇 − 𝑉 𝑑𝑃 4.23

The Joule Thomson cooling, 𝛿𝑇 = 𝑇 − 𝑉 𝛿𝑃 4.24

Therefore, the Joule Thomson coefficient, 𝜇 = = 𝑇 − 𝑉 4.25

This expression gives the change in temperature 𝛿𝑇 of a gas due to Joule-Thomson effect. The enthalpy H remains constant in the process and 𝛿𝑃 is the pressure difference on the two sides of porous plug which is necessarily negative as pressure on the emergent side of the plug is lower.

© Joule Thomson coefficient for Real (Van der Waal) gases

For real (Van der Waal) gases the EOS is

𝑃 + (𝑣 − 𝑏) = 𝑅𝑇 4.26

Differentiating w.r.t T, taking P constant, we have

𝑃 +𝜕𝑉

𝜕𝑇 𝑃−

2𝑎

𝑣3

𝜕𝑉

𝜕𝑇 𝑃(𝑣 − 𝑏) = 𝑅 4.27

𝑃 + −2𝑎

𝑣3(𝑣 − 𝑏)

𝜕𝑉

𝜕𝑇 𝑃= 𝑅 4.28

Or, =𝑅

𝑃+𝑎

𝑣2(𝑣−𝑏)

4.29

Or, =( )

−2𝑎

𝑣3( )

4.30

Or, =( )

−2𝑎

𝑣3( )

4.31

Or, 𝑇 =( )

−2𝑎

𝑣3( )

4.32

Since, 𝑏 ≪ 𝑣, we can neglect it and write 𝑣 in place of (𝑣 − 𝑏) . Thus

𝑇 =( )

−2𝑎

𝑣3

=( )

−2𝑎

𝑣

=( )

−2𝑎

𝑣𝑅𝑇

4.33

Or, 𝑇 =( )

−2𝑎

𝑣

=( )

−2𝑎

𝑣𝑅𝑇

4.34

Or, 𝑇 =( )

−2𝑎

𝑣𝑅𝑇

4.35

Or, 𝑇 =(𝑣 − 𝑏) 1 −2𝑎

𝑣𝑅𝑇 4.36

Expanding binomially and neglecting smaller terms, we have

𝑇 =(𝑣 − 𝑏) 1 + ≅ 𝑣 − 𝑏 + , 4.37 neglecting

Or, 𝑇 − 𝑣 = −𝑏 + 4.38

Putting this in the expression for Joule Thomson coefficient eq. 4.25, yields

𝜇 = = 𝑇 − 𝑣 = − 𝑏 4.39

This gives the Joule-Thomson coefficient for a real Van der Waal’s gas. It is clear from this expression that,

If > 𝑏 or 𝑇 < , then is positive.

But since 𝛿𝑃 is the pressure difference on the two sides of porous plug is necessarily negative quantity as pressure on the emergent side of the plug is lower, 𝛿𝑇 is also negative and the gas will be cooled on passing through the porous plug.

If < 𝑏 or 𝑇 > , then is negative and hence 𝛿𝑇 will be positive and the gas will be

heated up on passing through the porous plug.

At,𝑇 = , then = 0 or 𝛿𝑇=0, the gas shows no change in temperature on passing

through the porous plug. This is why the temperature 𝑇 = is called the temperature of

inversion and is denoted by 𝑇 and is characteristic of a particular gas.

(d) Joule–Thomson Cooling (Internal Work)

If > 𝑏 or 𝑇 < , then is positive.

But since 𝛿𝑃 is the pressure difference on the two sides of porous plug is necessarily negative quantity as pressure on the emergent side of the plug is lower, 𝛿𝑇 is also negative and the gas will be cooled on passing through the porous plug.

The physical reason for this cooling is, as the gas expands from volume V1 to Volume V2, an amount of internal work is needed to draw the molecules further apart against their mutual attractions. This work is drawn from the internal energy of the gas which therefore cools. Thus the effect due to deviation from Joule’ law or of the internal work is always a cooling effect.

Joule-Thomson effect is the resultant due to external and internal works

Below the Boyle temperature (Boyle temperature is defined as the temperature for which the second viral coefficient becomes zero. It is at this temperature that the attractive forces and repulsive forces acting on the gas particle balance out.), there is cooling due to external work. This together with the cooling due to internal work produces a greater cooling effect. Above the Boyle temperature, there is heating due to external work which lessens the cooling effect due to internal work. Hence the resultant effect is a weaker cooling effect. As the initial temperature of the gas is further raised, the external work effect (heating) increases and at a certain temperature it neutralises the internal work cooling effect. Hence the resultant Joule-Kelvin effect becomes zero. This temperature is the temperature of inversion. If the initial temperature of the gas is still raised, the heating effect become greater than the cooing effect so that the Joule-Thomson effect is a heating effect. For example the inversion temperatures for hydrogen and helium are -800C and -2400C respectively. Hence the gases show a heating effect at ordinary temperatures. If the initial temperatures of hydrogen and helium be brought below -800C and -2400C respectively, these gases would show a cooling effect.

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Q1. Show that for a perfect gas 𝝁 = 𝟎.

A1. The EOS of perfect gas is 𝑃𝑣 = 𝑅𝑇

Differentiating w.r.t T, taking P constant, we get,

𝑃𝜕𝑣

𝜕𝑇= 𝑅

⟹𝜕𝑣

𝜕𝑇 𝑃

=𝑅

𝑃

Or, 𝑇 = = 𝑣

Now, 𝜇 = 𝑇 − 𝑣 = (𝑣 − 𝑣) = 0.

Hence the result.

Q2. What is the difference between the Joule-kelvin and adiabatic expansion?

A2. In Joule-Kelvin expansion the cooling is produced mainly due to the internal work done by the gas against the molecular attractions. The contribution of the net external work may be a cooling or heating effect depending upon the initial temperature of the gas. In adiabatic expansion the compressed gas is suddenly released to atmosphere when it performs external work against the pressure of the atmosphere. Since the expansion is sudden, no appreciable heat flows into it from the surroundings. Hence the external work is drawn from the internal energy of the gas itself which therefore cools. Thus in adiabatic expansion the cooling is mainly due to internal work.