Ideal Gas Behavior. NC State University

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1 Chemistry 331 Lecture 6 Ideal Gas Behavior NC State University

2 Macroscopic variables P, T Pressure is a force per unit area (P= F/A) The force arises from the change in momentum as particles hit an object and change direction. Temperature derives from molecular motion (3/2RT = 1/2M) Mi l Greater average velocity results in a higher temperature. M is molar mass u is the velocity

3 Mass and molar mass We can multiply the equation: 3 2 RT = 1 2 M <u2 > by the number of moles, n, to obtain: 3 2 nrt = 1 2 nm <u2 > If m is the mass and M is the molar of a particle then we can also write: nm = Nm (N is the number of particles)

4 Mass and molar mass In other words nn A = N where N A is Avagadro s number nrt = 1 2 Nm <u2 >

5 Kinetic Model of Gases Assumptions: 1. A gas consists of molecules that move randomly. 2. The size of the molecules l is negligible. ibl 3. There are no interactions between the gas molecules. Because there are such large numbers of gas molecules in any system we will interested in average quantities. We have written average with an angle bracket. For example, the average speed is: =c 2 = s s s s N 2 N c = s 1 + s 2 + s s N N We use s for speed and c for mean speed.

6 Velocity and Speed When we considered the derivation of pressure using a kinetic model we used the fact that the gas exchanges momentum with the wall of the container. Therefore, the vector (directional) quantity velocity was appropriate. However, in the energy expression the velocity enters as the square and so the sign of the velocity does not matter. In essence it is the average speed that is relevant for the energy. Another way to say this is the energy is a scalar E = 1 2 m<u2 >= 1 2 mv2 = 1 2 mc2 p = mu = mv All of these notations mean the same thing.

7 The root-mean-square speed The ideal gas equation of state is consistent with an interpretation of temperature as proportional p to the kinetic energy of a gas M u = RT If we solve for we have the mean-square speed. u 2 = 3RT M If we take the square root of both sides we have the r.m.s. speed. 2 1/2 3RT u 2 = 3RT M

8 Example The r.m.s. speed of oxygen at 25 o C (298 K) is 482 m/s. Note: M is converted to kg/mol! u 2 1/2 = 38.31J/mol K 298 K kg/mol = m / s

9 The Maxwell Distribution Not all molecules have the same speed. Maxwell assumed that the distribution of speeds was Gaussian. F(s) = 4 M 3/2 s 2 exp Ms 2 2 RT RT As temperature increases the r.m.s. speed increases and the width of the distribution increases. Moreover, the fucntions is a normalized distribution. This just means that the integral over the distribution function is equal to 1. 0 F(s) ds = 1 See the MAPLE worksheets for examples.

10 Units of Pressure Force has units of Newtons F = ma (kg m/s 2 ) Pressure has units of Newtons/meter 2 P= F/A = (kg m/s 2 /m 2 = kg/s 2 /m) These units are also called Pascals (Pa). 1 bar = 10 5 Pa = 10 5 N/m 2. 1 atm = x 10 5 Pa

11 Units of Energy Energy has units of Joules 1 J = 1 Nm Work and energy have the same units. Work is defined as the result of a force acting through a distance. We can also define chemical energy in terms of the energy per mole. 1 kj/mol 1 kcal/mol l = kj/mol

12 Thermal Energy Thermal energy can be defined as RT. Its magnitude depends d on temperature. R = 8.31 J/mol-K or 1.98 cal/mol-k At 298 K, RT = 2476 J/mol (2.476 kj/mol) Thi i th l il bl t This is the molar energy available to a system at 298 K.

13 Thermal Energy Thermal energy can also be expressed on a per molecule basis. The molecular equivalent of R is the Boltzmann constant, k. R = N A k N A = x molecules/mol At 298 K, RT = 2476 J/mol (2.476 kj/mol) RT = 4.11 x J

14 Extensive and Intensive Variables 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

15 Equation of state relates P, V and T The ideal gas equation of state is PV = nrt An equation of state relates macroscopic properties which result from the average behavior of a large number of particles. P Macroscopic Microscopic

16 Microsopic view of momentum c u x b area = bc a A particle with velocity u x strikes a wall. The momentum of the particle changes from mu The momentum of the particle changes from mu x to mu x. The momentum change is p = 2mu x.

17 Transit time c u x b area = bc The time between collision is t = 2a/u x. a

18 Transit time c Round trip distance is 2a u x b area = bc a The time between collision is t = 2a/u x. velocity = distance/time. time = distance/velocity.

19 The pressure on the wall force = rate of change of momentum F = p t = 2mu x = mu 2 x 2a/u x a The pressure is the force per unit area. The area is A = bc and the volume of the box is V = abc P = F = mu x 2 = mu 2 x bc abc V

20 Average propertiesp Pressure does not result from a single particle striking the wall but from many particles. Thus, the velocity is the average velocity times the number of particles. 2 P = Nm u x V PV = Nm u2 x

21 Average propertiesp There are three dimensions so the velocity along the x-direction is 1/3 the total. 2 u x 3 = 1 3 u2 2 PV = Nm u 3 From the kinetic theory of gases 1 u Nm u2 = nrt 2

22 Putting the results together When we combine of microscopic view of pressure with the kinetic theory of gases result we find the ideal gas law. PV = nrt This approach applies to a monatomic gas like neon or argon. What about internal motions of molecules?

23 RT is a natural energy scale We can rewrite the ideal gas law in terms of fthe molar volume V = V/n The ideal gas law has the form PV = RT The molar volume at standard T and P V = RT P 8.31 J/mol K = 298 K = m 3 =24.4L N/m 2

24 Microscopic variables Monatomic gases: translation Pressure and temperature can be described solely in terms of the ballistic motion of the gas. Diatomic gases: translation, vibration, rotation Center of mass

25 Quantized energy levels The constant h, known as Planck s constant gives the scale for quantized energy levels. h = x J Translation particle in a box Vibration harmonic oscillator Rotation rigid rotator The energy levels for each of these is obtained by solution of the Schrödinger equation.

26 The energy level spacing The constant h, known as Planck s constant gives the scale for quantized energy levels. h = x J Motion Vibration Rotation Translation # v J n Formula (v + 1/2)h [h 2 /8 2 I]J(J+1) [h 2 /8ma 2 ]n 2 kj/mol We will see how to obtain these in the second half of the course.

27 Levels are thermally ypopulatedp E RT Vibration Rotation ti Translation

28 Key points regarding the microscopic view Translational energy levels are so densely spaced that these can be treated using classical methods. We can treat particles as ideal even though they have vibrations and rotations. The dynamics of the gas are not affected. We will see that the heat capacity of the gas is affected by the internal degrees of freedom.

29 Key points regarding the microscopic view The kinetic energy of a large number of individual particles is proportional to the temperature of the system. As the system heats up we can picture the molecules moving more rapidly. Pressure results from the net momentum transfer between the particles and wall of the container.

30 Pressure of a dense fluid For a dense fluid (or a liquid) such as water we can think of the pressure arising i from the weight of the column of fluid above the point where the measurement is made. The force is due to the mass of water m (kg) accelerated by gravity (g = 9.8 m/s 2 ). P = F = mg = mgh = mgh = gh A A Ah V where is the density = m/v.

31 The dependence of atomspheric pressure on altitude We can think of the atmosphere is a fluid, but it is not dense. Moreover, unlike water the density of the atmosphere decreases with altitude. Thus, at high elevations both the pressure and the density are decreased. To obtain the dependence of pressure on height h above the earth s surface we use the ideal gas law to define the density of an ideal gas.

32 The dependence of atomspheric pressure on altitude The density of an ideal gas is: = m/v = nm/v = MP/RT The dependence of pressure on elevation is: dp = g dh = MPg RT dh We need to collect variables of integration on the same side of the equation. dp P = Mg RT dh

33 The barometric pressure formula Then we integrate (assuming P 0 =1 at h=0): P dp P = Mg RT P 0 =1 RT 0 ln P P P 0 = Mgh RT P = P Mgh 0 exp RT h dh or P = exp Mgh RT atm

34 Isotherms We can plot the pressure as a function of the volume as shown below. Each of the curves on the plot has a constant temperature.

35 Partial pressure For any gas in a mixture of gases the partial pressure is defined as: P j = x j P where x j is the mole fraction of component j and P is the total pressure. The mole fraction is defined d as: x j = n j n i i

36 The molar volume We can rewrite the ideal gas law in terms of fthe molar volume V = V m n The ideal gas law has the form PV m = RT The molar volume at standard d T and P V 8.31 J/mol K 298 K 0244 m = RT 4 P = = m 3 = 24.4 L N/m 2

37 The Compression Factor One way to represent the relationship between ideal and real gases is to plot the deviation from ideality as the gas is compressed (i.e. as the pressure is increased). The compression factor is defined as: Compression Factor = Written in symbols this becomes: Molar volume of gas Molar volume of perfect gas V m PV m Z = = perfect Vm RT Note that perfect gases are also called ideal gases. Imperfect gases are sometimes called real gases.

38 The Compression Factor A plot of the compression factor reveals that many gases exhibit Z < 1 for low pressure. This indicates that attractive forces dominate under these conditions. As the pressure increases Z crosses 1 and eventually becomes positive for all gases. This indicates that the finite molecular volume leads to repulsions between closely packed gas molecules. These repulsions are not including the ideal gas model. Attractive Region Repulsive Region

39 The Virial Expansion One way to represent the deviation of a gas from ideal (or perfect) behavior is to expand the compression factor in powers of the inverse molar volume. Such an expansion is known as a virial expansion. Z =1+ B V m + C V m The coefficients B, C etc. are known as virial coefficients. For example, B is the second virial coefficient. Virial coefficients depend on temperature. From the preceding considerations we see the B < 0 for ammonia, ethene, methane and B > 0 for hydrogen.

40 The Virial Equation of State We write Z in complete form as: PV m RT =1+ V B + C m V An then solve for the pressure: m V m P = RT 1+ V B + C m V m V m This expression is known as the virial equation of state. Note that if B, C etc. are all equal to zero this is just the ideal gas law. However, if these are not zero then this equation contains corrections to ideal behavior.

41 Relating the microscopic to the macroscopic Real gases differ from ideal gases in two ways. First they have a real size (extent). The excluded volume results in a repulsion between particles and larger pressure than the corresponding ideal gas (positive contribution to compressibility). Secondly, they have attractive ti forces between molecules. l These are dispersive forces that arise from a potential energy due to induced-dipole induced-dipole interactions. We can relate the potential energy of a particle to the terms in the virial expansion or other equation of state. While we will not do this using math in this course we will consider the graphical form of the potential energy functions.

42 Hard Sphere Potential A hard sphere potential is the easiest potential to parameterize. The hard sphere diameter corresponds to the interatomic spacing in a closest packed geometry such as that t shown for the noble gas argon. The diameter can be estimated Ar Ar Ar Ar from the density of argon in Ar Ar Ar Ar the solid state. The hard sphere Ar Ar Ar Ar potential is widely used because Ar Ar Ar Ar of its simplicity. u(r) = r < u(r) =0 r > u(r) r

43 The Hard Sphere Equation of State As a first correction to the ideal gas law we can consider the fact that a gas has finite extent. Thus, as we begin to decrease the volume available to the gas the pressure increases more than we would expect due to the repulsions between the spheres of finite molar volume, b, of the spheres. P = nrt V nb Gas molecule of volume B

44 The Hard Sphere Model Low density: These are ideal gas conditions

45 The Hard Sphere Model Increasing density: the volume is V b is the molar volume of the spheres.

46 The Hard Sphere Model Increasing density

47 The Hard Sphere Model Increasing density

48 The Hard Sphere Model High density: At sufficiently high density the gas becomes a high density fluid or a liquid.

49 The Hard Sphere Model Limiting density: at this density the hard spheres have condensed into an ordered lattice. They are a solid. The gas cannot be compressed further. If we think about the density in each of these cases we can see that it increases to a maximum value.

50 Lennard-Jones Potential Function The Lennard-Jones potential is a most commonly used potential function for non-bonding interactions in atomistic computer simulations. V LJ (R) = R R The potential has a long-range attractive tail 1/r 6, and negative well depth, and a steeply rising repulsive wall at R =. Typically the parameter is related to the hard sphere diameter of the molecule. For a monoatomic condensed phase is determined either from the solid state or from an estimate of the packing in dense liquids. The well depth e is related to the heat of vaporization of a monatomic fluid. For example, liquid argon boils at ~120K at 1 atm. Thus, ~ kt or 1.38x10-23 J/K(120 K) = 1.65x10-21 J. This also corresponds to 1.03 kj/mol.

51 Graphical Representation L-J Potential The L-J potential function has a steep rise when r <. This is the repulsive term in the potential that arises from close contacts t between molecules. l The minimum i is found for Rmin = 2 1/6. The well depth is in units of energy. R min

52 The van der Waal s Equation of State The microscopic terms and in the L-J potential can be related to the a and b parameters in the van der Waal s equation of state below. P = nrt n2 a V nb V 2 The significance of b is the same as for the hard sphere potential. The parameter a is related to the attractive force between molecules. It tends to reduce the pressure compared to an ideal gas.

53 The van der Waal s Equation of State in terms of molar volume Recall that V m = V/n so that the vdw equation of state becomes: P = V RT m b a 2 V m We can plot this function for a variety of different temperatures. As we saw for the ideal gas these are isotherms. At sufficiently high temperature the isotherms of the vdw equation of state resemble those of the ideal gas.

54 The argon phase diagram For argon T c = K P c = 48.7 bar = Pa V c = 74.9 cm 3 /mol Critical Point

55 Significance of the critical point Note that the vdw isotherms look very different from those of the ideal gas below the critical point. Below the critical point there are two possible phases, liquid and gas. The liquid phase is found at small molar volumes. The gas phase is observed at larger molar volumes. The shape of the isotherms is not physically reasonable in the transition region between the phases. Note that the implication is that there is a sudden change in volume for the phase transition from liquid to gas.

56 View of the liquid region of fthe argon phase diagram Liquid Phase Equilibrium Region

57 Critical Parameters The critical parameters can be derived in terms of the vdw a and b parameters as well as the gas constant R. The derivation can use calculus since the derivative of the vdw equation of state is zero at the critical point. Given that this is also an inflection point the second derivative is also zero. P c = a 27b 2 T c = 27Rb 8a V c =3b

Imperfect Gases. NC State University

Imperfect Gases. NC State University Chemistry 431 Lecture 3 Imperfect Gases NC State University The Compression Factor One way to represent the relationship between ideal and real gases is to plot the deviation from ideality as the gas is