Imperfect Gases Detail

Properties of Gases If a gas is sufficiently dilute it obeys the ideal gas law The ideal gas law can also be written

PV = nRT A molar quantity is indicated by the bar across the top. The ideal gas law is an equation of state.

PV = RT where V = V/n An equation of state relates the pressure, volume, and temperature of the gas given a quantity of n moles of gas. The properties of a gas are of two type.Extensive variables are proportional to the size of the system. Intensive variables do not depend on the size of the system. Extensive variables: volume, mass, energy Intensive variables: pressure, temperature, density If we divide an extensive quantity by the number of moles (or number of particles), we obtain an intensive quantity. For example, volume V (L) is an extensive quantity, but molar volume V/n (L/mole) is an intensive quantity. Pressure in the ideal gas law has units of N/m2 which is corresponds to force per unit area. When thinking about the unit of atmospheres which corresponds to the pressure at sea level, the force is that due to the weight of the atmosphere above the surface of the earth. We look at some problems below. ---------------------------------------------------

Problem: what is the mass of the atmosphere above 1 m2 of the earth surface? Solution: The pressure is a force per unit area. The force in this case is the weight of the atmosphere (F = mg). If we assume that g is a constant for the entire column of the atmosphere above 1 m2 we have m = PA/g = (1.0325 x 105 N/m2)(1 m2)/(9.8 m/s2) m = 1.05 x 104 kg. SI units of pressure refer to the N/m2 as the Pascal (Pa). There are 1.0325 x 105 Pa per atm. A new standard unit of pressure is the bar, where 1 bar = 105 Pa. If m is the mass of the liquid and g is the gravitational acceleration, the force is F = mg. The pressure is where ρ is the density, ρ = m/V. Note that the area of the object cancels exactly as in our example

mg ρhAg F P= = = = ρhg A A A

above for the mass of the atmosphere. The application of P = ρhg is most common in liquids. Problem: What is the pressure due to water at a depth of 100 m? (You can solve it). We can also consider the pressure of a gas in gravitational field using the fluid formula above and the ideal gas law. For example, we can solve the following problem. Problem: What is the pressure of the atmosphere at an elevation of 6000 m (the top of Mt. Kilimanjaro)? You may assume that T = 273 K. We must recognize that the the pressure is a function of height P(h) = ρgh. The change in pressure dP is related to the change in height dh by dP = - ρg dh. The minus arises from the fact that the pressure decreases as the height above the earth's surface increases. We need to know how the

density changes with height. We obtain this from the ideal gas law. The density is ρ = nM/V where n is the number of moles, M is the molar mass and V is the volume. We can obtain n/V from the ideal gas law: n/V = P/RT, thus ρ = MP/RT. We can now substitute the density into the pressure vs. height formula. MPg dP = – dh RT P

1

dP = – Mg P RT

ln P = –

h

dh 0

Mgh RT

Mgh RT At this point you can plug in numbers. We will use M = 29 grams/mole as an average molar mass for the atmosphere. P = exp –

0.029 kg/ mol 9.8 m/s 2 6000 m P = 1 atmospere exp – 8.31 J / mol – K 273 K P = exp – 0.751 = 0.471 atmospheres The big mistake you can make in a problem like this one is to get the units wrong. For example, if you forget to convert the molar mass to kilograms and plug in the number 29 instead of 0.029 then you get an answer of P = exp{ - 751}, which is essentially zero pressure. --------------------------------------------------Temperature is perhaps the most difficult quantity to conceptualize. We shall see that temperature depends on microscopic motions of molecules. However, the temperature can be defined macroscopically based on the ideal gas law. Since the PV product cannot be less than zero, T cannot be less than zero. This implies an absolute zero of temperature. We can consider a definition of a thermometer based on the following theorem. We consider object A that is in thermal equilibrium with object B. Further we can consider object B in equilibrium with C. The zeroth law of thermodynamics states that A is in equilibrium with C. This law implies that object C can act as a thermometer for other objects. The definition of a temperature scale based on the properties of water is known as the Kelvin scale. The triple point of water is defined to be at 273.16 K. K represents the degrees in the Kelvin temperature scale. The triple point is the unique point in the phase diagram where solid, liquid, and vapor coexist. The boiling point of water at 1bar of pressure is defined as the being 100 degrees higher (373.16 K).

Ideal vs. Real Gases All gases obey the ideal gas equation of state provided they are sufficiently dilute. The ratio Z = PV/nRT = 1 at all pressures for an ideal gas. For a real gas Z is deviates from one as the pressure increases. First Z < 1 at intermediate pressures of tens to several hundreds of bar due to the attractive forces between molecules. At higher pressure the repulsive forces begin to dominate resulting in Z > 1. Microscopic view of the deviation of gases from ideal behavior Real gases differ from ideal gases in two ways. First, they have finite size. Secondly, there are forces acting between the particles or molecules in a real gas. The microscopic description of these forces has a consequence for the macroscopic equation of state. Hard sphere gas We can begin with the finite size. If we think of the gas a collection of solid spheres of diameter d then the potential energy between them has the form shown in the Figure below.

The hard sphere diameter is often called σ. The volume of the sphere is 4π(σ/2)3/3 = πσ3/6. In the hard sphere model of a gas we call the molar volume of the hard spheres b. It appears in the gas law as follows: P(V-nb) = nRT. In this equation nb is the "excluded volume". This means that it is the volume occupied by the gas molecules themselves. Conceptual problem: what is the maximum possible density of a hard sphere gas? Answer: A hard sphere gas differs from an ideal gas in that it has a finite size. As you increase the pressure on a hard sphere gas you will eventually start packing the spheres into a solid. Since the spheres are "hard" you will reach a point where you cannot compact them any further. To find the formula for that point we solve for the density of a hard sphere gas.At infinite pressure RT/P Æ 0 and so ρ = M/b is the maximum density. P V – nb = nRT P V n – b = RT V = RT + b n P ρ = nM = M RT + b V P

Van der Waal's gas: microscopic considerations The van der Waal's gas includes both the finite extent of a hard sphere gas and intermolecular attractions due to induced dipole-induced dipole interactions. Next we consider the origin of attractive forces between molecules. These are the so-called London dispersion forces and they arise due to the fluctuating electron clouds in molecules. London dispersion interactions or induced ipole-induced dipole interactions are always attractive. μ 21α 2 μ 22α 1 u ind(r) = – – 4πε 0 r 6 4πε 0 r 6 where α is the molecular polarizability. Induced-dipole-induced-dipole interactions are dominant if the dipole moment is zero. where I is the ionization potential. This term is also known as the dispersion term (or the London dispersion attraction). Surprisingly, this is the dominant term in the attractive part of the potential I 1I 2 α 1α 2 u disp(r) = – 3 2 I 1 + I 2 4πε 2r 6 0 energy surface.

If the molecule has a permanent dipole moment as well there are dipoledipole interactions. where μ is the ground state dipole moment. Note that the factor of kT in the denominator signifies that

u d,d(r) = –

2μ 1μ 2 1 6 4πε 0 3kT r

thermal fluctuations tend to disrupt interactions between the dipole moments and to reduce the magnitude of this term.

Potential Energy Surface The attractive force is a mutual electrostatic interaction. The repulsive force is due to the finite dimension of molecules. One common model for the potential energy surface is the LennardJones potential.

c 12 c 6 u(r) = 12 – 6 r r

Repulsive Attractive Energy

u(r) Internuclear Distance , r

-12

The r term accounts for repulsion due to molecular radii. This term replaces the hard sphere model. The r-6 term accounts for attraction at larger distances. Contributions to the r-6 term include both dipole-dipole and induceddipole-induced-dipole interactions. Van der Waal's gas: macroscopic equation

d Finite Size

Electrostatic Attraction

There are non-ideal equations of state designed to account for the deviations of gases from ideal behavior. The best known of these is the van der Waals equation Where V bar designates the molar volume. The constants a and b are called van der Waals constants, and they depend on the gas being described. The parameter a depends on the strength of intermolecular interactions. The parameter b depends on the size of the molecules.

P + a2 V – b = RT V

The van der Waals equation can be rewritten a number of ways: (a) in terms of pressure

P=

RT – a 2 V V–b

(b)

in terms of the compressibility factor Z

Z=

V – a RTV V–b

(c)

in terms of the molar volume

3 2 V – b + RT V + a V – ab = 0 P P P

The last form shows clearly that the van der Waals equation of state is a cubic polynomial. This is important since it implies that below a critical temperature there will be a phase transition. In other words this simple equation also serves as an equation of state for the liquid! Equations of State: Virial and van der Waal's The ideal gas law (PV = nRT) assumes that atoms and molecules have no extent (no finite size) and there are no interactions between particles. The virial equation of state shows a natural connection between the microscopic and macroscopic view. The coefficients B(T) are called virial coefficients.

B (T) B3V(T) Z = PV = 1 + 2V + + ... 2 RT V V

Z is called the compressibility. The second virial coefficient is negative at low temperature. A negative B2V represents the dominance of intermolecular attractions. B2V can be related to the intermolecular potential through ∞

e – u(r) / kT – 1 r 2dr

B2V(T) = – 2πN A 0

NA is Avagadro’s number. kB is the Boltzmann constant. kB = R/NA. kB is a microscopic constant that corresponds to the macroscopic universal gas constant, R. If u(r) is the Lennard-Jones potential the equation is not analytic, but has been solved numerically. This is an important equation since it establishes a connection between a microscopic potential energy function and macroscopic term in the compressibility expansion. The second virial coefficient can be expressed in terms of molecular volume analytically if we use a hard sphere potential. The hard sphere potential is u(r) = ∞ for r < σ and u(r) = 0 for r > σ. ∞

e – u(r) / kT – 1 r 2dr

B2V(T) = – 2πN A 0

σ

∞

2

= 2πN A

e – 1 r 2dr

0 – 1 r dr 2πN A 0

σ

= 2πN A 0

0

0

3 NA 2πσ 2 r dr = 3

which is equal to four times the volume of NA hard spheres. The van der Waals equation has a natural interpretation in terms of finite size since b is the volume reduction due to the size of the gas molecules and a is a pressure reduction due to interactions between molecules.

P = RT – a2 V–b V The interpretation the van der Waals parameters can be obtained by writing a virial expansion 2 b PV 1 a Z= =1+ b– + + ... RT RT V V 2

Microscopic Interpretation of the van der Waals Equation Since

B2V(T) = b – a RT we can use a hard sphere 1/r6 potential,

u(r) = ∞ for r < σ and u(r) = -c6/r-6 for r > σ. ∞

e – u(r) / kT – 1 r 2dr

B2V(T) = – 2πN A 0

σ

∞

– 1 r 2dr – 2πN A

= –2πN A 0

σ

≈

∞

2

2πN A

r dr – 2πN A 0

σ

e – c 6 / kTr 6 – 1 r 2dr σ

r 2dr r6

2πσ 3 N A 2πN Ac 6 = – 3 3kTσ 3 2πσ 3 N A 2πN A2c 6 Thus b = and a = 3 3σ 3 Isotherms Isotherms are constant temperature curves in pressure-volume space (PV). Using the ideal gas law we can plot a number of isotherms. To evaluate simple expression and plot functions while at the

computer we will use the program Maple. To plot an ideal gas isotherm (P = RT/V) we go to the Maple prompt and type in the command ¾ with(plots):plot(0.082*300/v,v=0.1..1); where we have used R = 0.082 L-atm/mol-K and T = 300 K. To see three isotherms on a plot type ¾ plot({0.082*300/v,0.082*200/v,0.082*100/v},v=0.05..0.3); Your plot should look like this

where the red, yellow, green curves are isotherms. The {} bracket allows more than one function to be plotted on the same graph as {function1, function2,..}. We can make the same plots for a van der Waals gas. Let’s look at ammonia (a = 4.25 2 L atm/mol2 and b = 0.037 L/mol). At 298 K we have ¾ with(plots):plot(0.082*298/(v-0.037)-4.248/(v*v),v=0.05..1); You will note that this isotherm has a strange shape. The reason is that the cubic polynomial is giving rise to inflection. At high enough temperature this will no longer be important (and the van der Waals equation will begin to approach the ideal gas law).

¾

¾ ¾

plot({0.082*298/(v-0.037)-4.248/(v*v),0.082*400/(v-0.037)-4.248/(v*v),0.082*500/(v-0.037)4.248/(v*v)},v=0.05..0.3); Note that we have changed the range on the abcissa (the molar volume axis) to better see how the shape of the curves is changing. You can compare the ideal gas law to the van der Waals equation at 300 K with(plots):plot({0.082*300/v,0.082*300/(v-0.037)-4.248/(v*v)},v=0.05..0.3); and 500 K with(plots):plot({0.082*500/v,0.082*500/(v-0.037)-4.248/(v*v)},v=0.05..0.3);

The plot at 500 K shows more similarity than the plot at 300 K. Above a critical temperature there is no inflection point and the curves for real and ideal equations of state begin to have the same shape. This behavior corresponds to the known behavior of substances. They all have a critical temperature (and pressure) beyond which there is no distinction between liquid and gas. Below the critical temperature there will be a phase transition for a given temperature and pressure. Since the molar volume changes (think about liquid water turning to vapor) there should be an abrupt change in the molar volume as shown in Figure 16.7 in your text for isotherms below the critical point. The critical point in a van der Waal's gas

The critical point is defined mathematically as an inflection point. At the critical temperature the curve will no longer turn down. For ammonia the critical point must be between 300 K where the isotherm shows a change in the sign of curvature and 500 K where it does not. At the inflection point the slope of the curve will be zero and the curvature will be zero.

In pressure-volume space this means 2 ∂P = 0 and ∂ P = 0 2 ∂V ∂V

There is an elegant and simple way of solving this. They note that at the critical point there is only one root. Therefore, the cubic equation can be written 3

3

2

2

3

V – Vc = 0 or V – 3VcV + 3Vc V – Vc = 0 The coefficient Vc bar can be equated with the coefficients above for the van der Waals equation of state expressed as a cubic polynomial. A single point in P-V space is solved for, thus, there is a critical temperature, Tc, pressure, Pc, and volume, Vc. To find these we equate the terms in the van der Waal's expression to the above cubic polynomial. At the critical point: 3

V – b+

RTc 2 a V + V – ab = 0 Pc Pc Pc

Equating the coefficients we have the following equations:

3V c = b + 2 3V c = a Pc 3 V c = ab Pc

RTc Pc

Vc = 3b , Pc =

a , and T = 8a c 27bR 27b 2

The critical values are: in terms of the van der Waals parameters. For example, we can now calculate the critical temperature of ammonia, Tc = 8(4.25)/(27(0.039)(0.082)) = 393.7 K. I did not explicity write out the units. I used R = 0.082 L-atm/mol-K in this case since we are working in the L-atm units of P-V space. In practice, experimental critical data are used to obtain the parameters a and b. We will see further that the virial equation of state can be related to the van der Waals equation of state and to parameters that describe molecular interactions. The Law of Corresponding States We can define reduced quantities PR = P/Pc, VR = V/Vc, TR = T/Tc By substitution into the van der Waals equation we find

PR + 32 VR – 1 = 8 TR 3 3 VR

This equation is a universal equation for all gases. Although the actual pressures and volumes may differ, two gases are said to be in corresponding states if their reduced pressure, volume, and temperature are the same. The compressibility factor Z can also be cast into the form of corresponding states showing that Z also can be expressed as a universal function of VR and TR or any other two reduced quantities.

Z =

VR – 9 VR – 1/3 8V RTR