Dr.Salwa Al Saleh Salwams@ksu.edu.sa Thermal Physics 2009 Dr.Salwa Al Saleh Salwams@ksu.edu.sa

Lecture 7 Deviations from Ideality Van Der Waals Equation Virial equations of state Compressibility factor

Derivation of some laws from kinetic theory of gases

KINETIC THEORY

For a single molecule, the average force is: For N molecules, the average force is: root-mean-square speed volume

Example : The Speed of Molecules in Air Air is primarily a mixture of nitrogen N2 molecules (molecular mass 28.0u) and oxygen O2 molecules (molecular mass 32.0u). Assume that each behaves as an ideal gas and determine the rms speeds of the nitrogen and oxygen molecules when the temperature of the air is 293K.

For nitrogen…

The internal energy of monatomic ideal gas

High temp. % of Molecules Low temp. Kinetic Energy

High temp. % of Molecules Low temp. Kinetic Energy Few molecules have very high kinetic energy Kinetic Energy

High temp. % of Molecules Low temp. Kinetic Energy Average kinetic energies are temperatures Kinetic Energy

Kinetic Theory – Summary Using Newtonian mechanics we have established: the relationship between p, N/V, T; the universality of the gas constant; the relationship between temperature and K.E. the internal energy of a monatomic ideal gas

Real Gases J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910. 29

Real Gases General Observations Deviations from ideal gas law are particularly important at high pressures and low temperatures Real gases differ from ideal gases in that there can be interactions between molecules in the gas state Repulsive forces important only when molecules are nearly in contact, i.e. very high pressures Gases at high pressures , gases less compressible Attractive forces operate at relatively long range (several molecular diameters) Gases at moderate pressures are more compressible since attractive forces dominate At low pressures, neither repulsive or attractive forces dominate → ideal behavior →

Real Gases Deviations from Ideality . The Ideal Gas Law ignores both the volume occupied by the molecules of a gas and all interactions between molecules, whether attractive or repulsive In reality, all gases have a volume and the molecules of real gases interact with one another. For an ideal gas, a plot of PV/nRT versus P gives a horizontal line with an intercept of 1 on the PV/nRT axis. J. van der Waals, 1837-1923, Professor of Physics, Amsterdam. Nobel Prize 1910.

Real Gases Deviations from Ideality For an ideal gas, a plot of PV/nRT versus P gives a horizontal line with an intercept of 1 on the PV/nRT axis. Real gases behave ideally at ordinary temperatures and pressures. At low temperatures and high pressures real gases do not behave ideally. The reasons for the deviations from ideality are: The molecules are very close to one another, thus their volume is important. The molecular interactions also become important.

Real Gases Van Der Waals Equation Real gases do not follow PV = nRT perfectly. The van der Waals equation corrects for the nonideal nature of real gases. a corrects for interaction between atoms. b corrects for volume occupied by atoms. 29

Real Gases Van Der Waals Equation A non-zero volume of molecules = “nb” (b is a constant depending on the type of gas, the 'excluded volume'). The molecules have less free space to move around in, so replace V in the ideal gas equation by V - nb Very roughly, b  4/3 r3 where r is the molecular radius.

In the van der Waals equation, where “nb” represents the volume occupied by “n” moles of molecules. 29

Real Gases Van Der Waals Equation The attractive forces between real molecules, which reduce the pressure: p  wall collision frequency and p  change in momentum at each collision. Both factors are proportional to concentration, n/V, and p is reduced by an amount a(n/V)2, where a depends on the type of gas. [Note: a/V2 is called the internal pressure of the gas].

Also, in the van der Waals equation, where “n2a/V2” represents the effect on pressure to intermolecular attractions or repulsions. 29

So… " p = nRT/V " becomes p = nRT/(V - nb) - a(n/V)2 i.e. [ p + an2/V2 ][ V - nb ] = nRT This is what already could ″Van der Waals equation of state″ When V or T are very large i.e. at low p or high T, then this van der Waals equation of state becomes equivalent to the ideal equation. a and b are empirical Van der Waals constants. It is easy to solve for p given V. To find V given p you need to solve a cubic equation (with Vm=V/n)

V.D.W Cubic equations of state Simple equation capable of representing both liquid and vapor behavior. The van del Waals equation of state: a and b are positive constants unrealistic behavior in the two-phase region. In reality, two, within the two-phase region, saturated liquid and saturated vapor coexist in varying proportions at the saturation or vapor pressure. Three volume roots, of which two may be complex. Physically meaningful values of V are always real, positive, and greater than constant b.

Real Gases Deviations from Ideality Problem Calculate the pressure exerted by 84.0 g of ammonia, NH3, in a 5.00 L container at 200. oC using the van der Waal’s equation. The van der Waal's constants for ammonia are: a = 4.17 atm L2 mol-2 b =3.71x10-2 L mol-1 n = 84.0g x 1mol/17 g T = 200 + 273 P = (4.94mol)(0.08206 L atm mol-1 K-1)(473K) (4.94 mol)2*4.17 atm L2 mol-2 5 L – (4.94 mol x3.71E-2 L mol-1) (5 L)2 P = 39.81 atm – 4.07 atm = 35.74 P = 38.3 atm 7% error

Real Gases Nonideal Conditions when gas gets close to conditions where it will liquify Lower Temperature Higher Pressure

Real Gases Compressibility factor, Z Deviations from ideality can be described by the COMPRESSION FACTOR, Z (sometimes called the compressibility). For ideal gases Z = 1 always.

Compressibility factor, Z Compression factor, Z, is ratio of the actual molar volume of a gas to the molar volume of an ideal gas at the same T & P Z = Vm/ (Vm)id , where Vm = V/n Using ideal gas law, p Vm = RTZ The compression factor of a gas is a measure of its deviation from ideality Depends on pressure (influence of repulsive or attractive forces) z = 1, ideal behavior z < 1 attractive forces dominate, moderate pressures z > 1 repulsive forces dominate, high pressures

… Compressibility factor, Z EXAMPLE What is the molar volume Vm for an ideal gas at SATP? V = nRT/p = 0.0248 m3 = 24.8 dm3 = 24.8 L = 24800 cm3 T=300K Z = PVm/RT =Vm/ (Vm)id “compressibility factor”: Ideal gas: z = 1 z < 1: attractive intermolecular forces dominate z > 1: repulsive intermolecular forces dominate

Microscopic interpretation: When p is very high, r is small so short-range repulsions are important. The gas is more difficult to compress than an ideal gas, so Z > 1. When p is very low, r is large and intermolecular forces are negligible, so the gas acts close to ideally and Z  1. At intermediate pressures attractive forces are important and often Z < 1.

Critical point of Van Der Waals Equation At fixed P and T, V is the solution of a cubic equation. There may be one or three real-valued solutions. The set of parameters Pc, Vc, Tc for which the number of solutions changes from one to three, is called the critical point. The van der Waals equation has an inflection point at Tc.

Critical point of Van Der Waals Equation Different gases have different values of p, V and T at their critical point Comparing Different Gases

Comparing different gases different gases fall on the same curves.

Critical point of Van Der Waals Equation p The isotherm at Tc has a horizontal inflection at the critical point → dp/dV = 0 and d2p/dV2 = 0. At the critical temperature the densities of the liquid and gas become equal - the boundary disappears. The material will fill the container so it is like a gas, but may be much denser than a typical gas, and is called a 'supercritical fluid'.} The general Van der Waals pVT surface V }

Consider 1 mol of gas, with molar volume V, at the critical point (Tc, pc, Vc) 0 = dp/dV = -RTc(Vc-b)-2 + 2aVc-3 0 = d2p/dV2 = 2RTc(Vc-b)-3 - 6aVc-4 The solution is Vc = 3b, pc = a/(27b2), Tc = 8a/(27Rb). THE PRINCIPLE OF CORRESPONDING STATES Define reduced variables pr = p/pc Tr = T/Tc Vr = Vm/Vm,c Van der Waals hoped that different gases confined to the same Vr at the same Tr would have the same pr.

Proof: rewrite Van der Waals equation for 1 mol of gas, p = RT/(V-b)-a/V2, in terms of reduced variables: Substitute for the critical values: Thus Thus all gases have the same reduced equation of state (within the Van der Waals approximation).

Problem Calculate the pressure exerted by 1 mol H2S behaving as A perfect gas A van der Waals gas a = 4.484 L2 atm mol-2 b = 4.34  10-2 L mol-1 At 500 K in 150 cm3 VAN DER WAALS LOOPS An artifact of the equation of state below Tc. V p The general Van der Waals pVT surface

a = 6.865 L2.atm/mol2 b = 0.05679 L/mol Problem If sulfur dioxide were an “ideal” gas, the pressure at 0°C exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the “real” pressure. a = 6.865 L2.atm/mol2 b = 0.05679 L/mol 29

R= 0.0821 L. atm/mol. K T = 273.2 K V = 22.41 L a = 6.865 L2.atm/mol2 Solution First, let’s rearrange the van der Waals equation to solve for pressure. R= 0.0821 L. atm/mol. K T = 273.2 K V = 22.41 L a = 6.865 L2.atm/mol2 b = 0.05679 L/mol 29

Problem The “real” pressure exerted by 1.00 mol of SO2 at STP is slightly less than the “ideal” pressure. 29

Virial equations of state PV along an isotherm: The limiting value of PV as P →0 for all the gases: , with R as the proportionally constant. Assign the value of 273.16 K to the temperature of the triple point of water: Ideal gas: the pressure ~ 0; the molecules are separated by infinite distance; the intermolecular forces approaches zero.

VIRIAL COEFFICIENTS and the Boyle temperature We can consider the perfect gas law as the first term of a general expression p Vm = RT (1 + B’p + C’p2 + ...) i.e. p Vm = RT (1 + B/Vm + C/Vm2 + ...) This is the virial equation of state and B and C are the second and third virial coefficients. The first is 1. B and C are themselves functions of temperature, B(T) and C(T). Usually B/Vm >> C/Vm2

Consider dZ/dp. This is zero for a perfect gas since Z is constant. dZ/dp = B’ + 2pC’ + .... In the limit as p tends to zero, dZ/dp = B’ which is zero only when B = 0. The temperature at which this occurs is the Boyle temperature, TB, and then the gas behaves ideally over a wider range of p than at other temperatures. Each gas has a characteristic TB, e.g. 23 K for He, 347 K for air, 715 K for CO2.

Real Gases - Other Equations of State Virial equation is phenomenolgical, i.e., constants depend on the particular gas and must be determined experimentally Other equations of state based on models for real gases as well as cumulative data on gases Berthelot (1898) Better than van der Waals at pressures not much above 1 atm a is a constant Dieterici (1899)

SUMMARY The STATE of a system is specified by a few variables, that may be linked by an EQUATION OF STATE e.g. for IDEAL GASES pV = nRT. ISOTHERMS and ISOBARS. Dalton's Law of partial pressures. The VIRIAL EQUATION and the BOYLE TEMPERATURE. REAL GASES: the COMPRESSION FACTOR and INTERMOLECULAR FORCES. pV diagrams: the CRITICAL POINT.

Question time! Question1 Consider a fixed volume of gas. When N or T is doubled the pressure doubles since pV=NkT T is doubled: what happens to the rate at which a molecule hits a wall? (a)  1 (b) 2 (c) 2 N is doubled: what happens to the rate at which a molecule hits a wall?

Question 2 Container A contains 1 l of helium at 10 °C, container B contains 1 l of argon at 10 °C. a) A and B have the same internal energy b) A has more internal energy than B c) A has less internal energy than B

Question 3 Container A contains 1 l of helium at 10 °C, container B contains 1 l of argon at 10 °C. a) The argon and helium atoms have the same average velocity b) The argon atoms are on average faster than the helium atoms c) The argon atoms are on average slower than the helium atoms

Question 4 Container A contains 1 l of helium at 10 °C, container B contains 1 l of helium at 20 °C. a) The average speeds are the same b) The average speed in A is only a little higher c) The average speed in A is about 2 higher d) The average speed in A is about twice as high