Physical Chemistry I
Physical Chemistry I CHM 3410
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This 22 page Class Notes was uploaded by Dr. Jamil Donnelly on Monday October 12, 2015. The Class Notes belongs to CHM 3410 at Florida Atlantic University taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/221666/chm-3410-florida-atlantic-university in Chemistry at Florida Atlantic University.
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Date Created: 10/12/15
Distribution Over Quantum States Using the kineticmolecular derivation of PV nRT we saw AVg translational energies per molecule kT 2 and AVg translational energies per mole RT 2 In order to understand how the molecules are distributed in each allowed energy states we must first look at their probability of distribution Let us consider an example of throwing marble randomly in to a box with equal sized compartment Let us consider the fourmarble two compartment distribution as shown below Refer Figure 213 The number of ways W of distribution of these marbles is given as W 4 l234 The same analogy can be extended to the distribution of molecules through out the set of quantum states W This gives an expression with which we can calculate the number of ways a particular distribution can be achieved This number is proportional to the probability of distribution Qualitative Consideration Let us find out the values of W for various distributions subject to following constraints 1 The total number of particles is fixed 2 The system of the particles has some fixed total energy eg The most probable distribution of 14 particles with an average energy as 2 units Since N and Ni values are small we can calculate the value of W The largest value for W is obtained when the particles are spread out as much as possible Refer figure 214 e g The most probable distribution of 14 particles with varying average energy of the system All distributions have same general shape Lower the average energy the more particles are crowded in lower energy state Higher the average energy more particles are spread out into higher energy level Refer Figure 215 Boltzmann Distribution This a mathematical expression for a large number of molecules which gives the most probable distribution The Boltzmann distribution gives the number if particles per state at one energy 8i compared to that of another energy 81 as 818 kT Ni e J More conveniently we compare the population of the states at one energy level with the population of the sates of lowest energy level Then we get 1i N0 6SiSokT 1 Thus according to Boltzmann distribution The population depends upon the ratio of 8i 80 to kT The population of ith state is always less than that of the lowestenergy state At low temperature The eXponential factor makes population decrease rapidly with increasing energy and particles are crowded into the states of lower energy At high temperature The eXponential factor falloff is less rapid and the particles can spread out into the higherenergy states Problem 1 Use the Boltzmann distribution expression to calculate the ratio of the number of molecules per quantum state at an energy equal to l2kT the average energy for onedimensional motion to the number per quantum state at the lowest allowed energy N No 26SiSo kT 1 8 12 kT 80 o N N0 e4121ltT01ltT 1 Ni N0 e412 Ni N0 O606 In most cases we are given the total number of particle ie N The Boltzmann distribution relates the number of particles Ni in the quantum states at any energy to the number N0 in the states at the lowest energy Ni N0 e 48180 Ni N0 e 48180 The total number of particles is given as N 2 Ni N Z N0 6 481 80kT N No 2 e ver 1 N No q 2 Where q is the partition function and its given as q Z 6 481 80kT 3 2 e 81 80kT 2 e 82 80kT 2 e 783 80kT N0Nq 4 This relation gives us how many molecules are in the lowestenergy state if we know the total number of molecules in a given sample Onedimensional Translational Motion The Boltzmann distrubution can be used to find the distribution of the molecules of a gas over the allowed energies and speeds Let us consider the molecule of a gas with mass m moves along in X direction and is confined to a linesegment of length a Then its translational energy is given as Snx nX2 h2 8 m a2 where nX 123 The energy difference is given as Snx Sno nX2 h2 8 m a2 n02 h2 8 m a2 nX2 hz8 m a2 12 h2 8 m a2 nX2 1h28ma2 Since most of the nX values are very much larger than 1 and thus le2 1 E nxz 8nX Sno E nX2 h2 8 m a2 5 The partition function q from equation 3 can be written as 8 8 15ka no1ltT Summation can be replaced by integration since summation is over so many closely spaced energy levels Integration can be evaluated to give l n 12 qx 2 112 8ma2 6 Thus the partition function q for a particle of a gas with mass m moving in Xdirection confined to a linesegment of length a at any temperature can be calculated From equation 4 we can write NoNqx lt7gt Thus now we can calculate the number of molecules present in the lowest energy state Problem 2 Calculate the value for qx and the value for N0 for 1mol sample of nitrogen gas in 1L container at 25 C Mass of one N2 molecule m 0028 kg 602 x 1023 465 x 103926 kg Assuming the container to be cubic the length of container is a 10 cm 01 m k 138066 x 103923 Jk391 T 25 27315 29815 K h 6626 x 103934 J s l n 12 qX 523 X 109 2 h2 8ma2 and N0 N qX With a total 602 x 1023 molecuels in this 1mol sample N0 602 X1023523 X 109 115 gtlt1014 Distribution of Quantum States The number of molecules per unit I lX interval can be expressed as dN dI lX Then The fraction of the molecules per unit I lX interval as dN an N Then The Boltzmann distribution gives the number of molecules in a unit quantumnumber interval as Nn dN 111 N0 e SnxSOVkT The fraction of the molecules per unit quantum state can be expressed as dN an N N0 N e SnxkaT 8 For onedimensional translational motion from equation 7 N N0 qx And from equation 5 8nX Sno 11X2 h2 8 m a2 Then Fraction per unit nX interval dN an N N auxSQkT 1 qx e SmSo kT 2 h2 8ma2 2 ln kT lt9gt Problem 3 Use the equation for fraction per unit nx interval to calculate he number of molecules per unit quantumnumber interval at nx 0 for he system of 1mol sample of nitrogen in a 1L container at 25 C Mass of one N2 molecule m 0028 kg 602 x 1023 465 x 103926 kg Assuming the container to be cubic the length of container is a 10 cm 01 m k 138066 x 103923 Jk391 T 25 27315 29815 K h 6626 x 103934 J s Thus the fraction per unit interval at nx0 is l dN 2 hZ8ma2 2 N an HFO l 7 kT 191 gtlt103910 This is the fraction of the molecules that are in the first quantumnumber unit interval The number of molecules in a lmol sample that are in this interval is obtained by multiplying this fraction by Avogadro s number Thenweget l dN NAgtlt NAxl9lgtltlO3910 N dnx nx0 602 X 1023 191 X 1010 115 X 1014 Distribution Over Energies To obtain the distribution molecules over the allowed energies we convert the derivative dNan which gives the number of molecules per unit interval nx to dNdSX which gives the number of molecules per unit energy interval 8x We can write dNdsx dNdnx andex We have ex 11X2 h2 8 m a2 nX2 8X 8 m a2 h2 nX 5 2J2Jm ah nX 2ah 2m8X andsX ah 2 mSX 10 Now the fraction per unit energy interval 8x on be written as Fraction per unit energy interval lN dNdsx lN dNdnx andsx From equation 9 and 10 1 1 esX kn m Distribution over Speeds We know 8X 12 m UX2 Differentiate with respect ox We get dSX d1X 12 In duxz doX m1X Wax 1 1 Using equation 11 and 12 we can write Fraction per unit speed interval lN dNdex e12mux2 kT An operator is a symbol that designates a process that will transform one function into another function For example An operator DX given as follows DX d dX Means a first derivative of function fX with respect to X should be taken If two or more operators are applied simultaneously to a function the operator immediately adjacent to the function will operate on the function first giving a new function For example If DX d dX and Dy ddX to the function fXy X3y2 DX Dy X3y2 DX 2 X3y 6 Xzy Or Dy DX X3y2 Dy 3 Xzyz 6 Xzy The result of the two operations is independent of the order in which the operators are applied then these operators are said to commute That is DX Dy fXy Dy DX fXy Or DX Dy DX Dy fXy 0 Here the brackets are called as commuter brackets and the expression in the bracket itself an operator called the commutator of the operators In general we can say that two operators A and B commute if and only if AB BA fX 0 QuantumMechanical Operators The Schrodinger equation of a particle of mass m confined to a thre dimensional region and with a potential energy Uxyz is given as follow h2 dzw dzw dzw UXYZl I 8W 872m dx2 dy2 dz2 The mathematical expression for w that satisfies this equation generally exist only for certain values of 8 and these values are the energies of the states of the system We can write the above equation as h2 d2 d2 d2 UXYZl I 8W 872m dx2 dy2 dz2 The expression in the curly bracket is an example of an quantum mechanical operator and is known as Hamiltonian operator and its symbol is H The A hat signifies it as a operator Thus the equation can be written as Hw 8w The function which satisfies such equations are called as eigenfunctions and the values of constants calculated from such equations are called as eigenvalues Eigen is a German word and it means characteristic Postulates of Quantum Mechanics The wave functions from the Schrodinger equation or eigen functions from the operator equation can be used to deduce the information about the systems in each of allowed states Thus we introduce a set of postulates that are expressed in terms of quantummechanical operators 1 The value of each physical property can be deduced by operating on eigenfunctions with the operator corresponding to that physical property or by solving Schrodinger equation 2 There are two basic operator which can be used to deduce physical properties a Position Operator is for one dimensional system and it is the position coordinate of the system Position Operator X X b The momemtum operator in X direction is Momentum operator p h 2 7 i d dX 3 There are two different situations arises when we a value of any physical property is obtained by working on eigen functions a When quantized values are obtained For a particular physical property the operator A is such that Aw aw Where a is the number of allowed values of physical property For example when the Hamiltonian operator is used Hw 2 8w yields the values for 8 Any measurement of the system would show that it had an energy equal to one of the 8 values b When average values are obtained For a another physical property the operator B is such that But 2 bw Where ltbgt is average value of physical property obtained as ltbgt le dT M1sz For eg in the particleonline problem the particle is not restricted to certain positions along the line but we can find out the average position of the particle using above expression A derived operator for kinetic energy Operators for the other physical properties can be deduced from the position and momentum operators Kinetic energy is given as KE 111022 mo2 2m momentum2 2m Thus the kinetic energy operator T is given as Kineticenergy operator T 12mh27iddxh2niddx h28n2md2dX2 For threedimensional system Kineticenergy operator T 112 8 72 m dzdxz dZdy2 dZdzz Rotation in a plane Angular Features Angular momentum is an important physical property of atoms and molecules Angular momentum values are used to characteristics the electronics states of atoms and some molecules and for the rotational states of the molecules of gas Angular Momentum Operators Similar to kinetic energy operators angular momentum operators can be deduced from the basic position operator and linear momentum operator The direction and the magnitude of the angular momentum can be described by a vector This vector is perpendicular to the plane that contains he radius vector and the linear momentum vector Refer figure 97 Refer 96 The magnitude of this vector since the radius and linear momentum vector are perpendicular to each other is equal to the product of the magnitudes of these two vectors This can be eXpressed in term of coordinates X and y of particle and the values of angular momentum components PK and Py Thus the contribution to the angular momentum along the 2 directions are XPy and ny Thus The angular momentum LZ X Py y Px Thus in order to get the value of angular momentum operator LZ We use position operator X X and y y an d linear momentum operators Px h 2 7 i d dX and Pyh21iddy Thus we get Angular momentum operator 2 L2 h 2 7 i X d dy y ddX The polar coordinates are more convenient than Cartesian coordinates to calculate the angular momentums Refer figure 98 Thus X r COSQ and 1X do r sino y Also y r sine and dyd0 rcoso x In addition the derivative df d9 of any function f with respect to a can be written in terms of X and y as follows df d0 df dX dX do df dy dy do but dXdQ y and dydo X Therefore df d0 df dX y df dy X dfdo X df dy y df dX Thus in general we can write dde Xddyyddx Therefore the angular momentum operator LZ can be written in terms of its polar coordinates Angular momentum operator 2 L2 h 2 Tl l 1 d0 Rotational Kinetic Energy Operator The kinetic energy can be expressed in terms of angular velocity 0 oa2nv 2nxu2nr Ur The moment of Inertia I mr2 Therefore we can write KE 12 mo2 12 m r2 1 r 2 12 1 032 10221 Thus the we can express KE operator TZ as Tz h28n21d2d20
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