Lecture 18 Molecular Motion and Kinetic Energy

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1 Physical Principles in Biology Biology 3550 Fall 2017 Lecture 18 Molecular Motion and Kinetic Energy Monday, 2 October c David P. Goldenberg University of Utah goldenberg@biology.utah.edu

2 Fick s First and Second Laws of Diffusion C(x) First law: J = D dc dx Flux, J, at position x is proportional to the concentration gradient at that position. x x Second law: dc dt = D d 2 C dx 2 Rate of change in concentration at position x is proportional to the derivative of the concentration gradient.

3 C(x) and its Derivatives, for Diffusion from a Sharp Boundary

Estimating D from Diffusion from a Sharp Boundary C(x, t) = 1 2 + x

7 10 5 s D 2 10 10 m 2 /s D = δ2 x 2τ 0 1 4πDt e x2 /(4Dt) dx

4 Estimating D from Diffusion from a Sharp Boundary C(x, t) = x t = 48 hr = s D m 2 /s D = δ2 x 2τ 0 1 4πDt e x2 /(4Dt) dx Relative Concentration x (cm) What are the average length (δ x ) and duration (τ) of the random walk steps?

5 Warning! Direction Change

6 Molecular Motion and Kinetic Energy What is energy? Capacity to do work. What is work? Mechanical work: The application of force over distance: w = b a Fdx The units of work and energy. Force: Units defined by Newton s second law: F = mass acceleration SI unit of mass: Kg Acceleration: change in velocity (m/s) with time. SI units: m/s 2 SI units of Force: Kg m/s 2 1 N = 1 Kg m/s 2 Work or energy: Kg m 2 /s 2 1 J = 1 N m = 1 Kg m 2 /s 2

7 Kinetic Energy A object of mass, m, moving with velocity, v, in the x-direction has kinetic energy in that direction of: E k,x = mv 2 /2 Check the units: Kg (m/s) 2 = Kg m 2 /s 2 It s OK! What does this mean? The energy required to accelerate the mass, m, from rest to velocity, v. Also the energy released during the deceleration of the mass from velocity, v, to rest. Kinetic energy does not depend on the rate of acceleration, only the final velocity. But, amount of wasted energy likely does depend on rate of acceleration! E k,x is proportional to v 2. What are the implications?

8 Kinetic Energy of Molecules Temperature is the measure of kinetic energy of molecules. How do we measure temperature? Pressure of a gas is due to the collision of molecules against container walls. PV = nrt P = pressure, V = volume, n = number of moles, T = temperature, R = gas constant.

9 Clicker Question #1 What are the units of the gas constant? R = PV nt 1 pascal L K 1 mol 1 2 Kg m 2 s 2 K 1 mol 1 3 m 3 bar K 1 mol 1 4 JK 1 mol 1 5 L atm K 1 mol 1 6 N m K 1 mol 1 All of the above!

10 Units of the Gas Constant From the ideal gas law: In SI basic units: R = PV nt Pressure, force per unit area: Kg m s 2 m 2 = Kg m 1 s 2 = Pa Volume: m 3 Gas constant: Kg m 2 s 2 K 1 mol 1 Joule (unit of energy) = Nm = Kg m 2 s 2 The gas constant expressed in energy units: R JK 1 mol cal/k 1 mol 1 RT is proportional to the kinetic energy of one mole of molecules at a temperature T.

11 Kinetic Energy of Molecules Temperature is the measure of kinetic energy of molecules. How do we measure temperature? Pressure of a gas is due to the collision of molecules against container walls. PV = nrt P = pressure, V = volume, n = number of moles, T = temperature, R = gas constant.

12 Clicker Question #1 What are the units of the gas constant? R = PV nt 1 pascal L K 1 mol 1 2 Kg m 2 s 2 K 1 mol 1 3 m 3 bar K 1 mol 1 4 JK 1 mol 1 5 L atm K 1 mol 1 6 N m K 1 mol 1 All of the above!

13 Units of the Gas Constant From the ideal gas law: In SI basic units: R = PV nt Pressure, force per unit area: Kg m s 2 m 2 = Kg m 1 s 2 = Pa Volume: m 3 Gas constant: Kg m 2 s 2 K 1 mol 1 Joule (unit of energy) = Nm = Kg m 2 s 2 The gas constant expressed in energy units: R JK 1 mol cal/k 1 mol 1 RT is proportional to the kinetic energy of one mole of molecules at a temperature T.

14 Kinetic Energy of Molecules The Boltzmann constant, k, the gas constant per molecule : k = R/N A = JK 1 mol molecule/mol = JK 1 Molecules at a given temperature do not have unique velocities or kinetic energies. Molecules have a broad distribution of energies, with RMS average kinetic energy in each direction (x, y or z): RMS(E k,x ) = kt /2 Translational kinetic energy does not depend on mass or structure. In context of molecular motion, E k,x will imply RMS value.

15 Distribution of Molecular Kinetic Energies p(e k,x ) = 1 πekt e E/(kT )

16 Clicker Question #2 How fast does a small molecule move at room temperature (between collisions)? m/s m/s m/s 4 1 m/s m/s m/s All answers count for now.

17 Velocities of Molecules Kinetic energy of a molecule, in x-direction: RMS(E k,x ) = kt /2 Kinetic energy of any object: E k,x = mv 2 x /2 Solving for v: mv 2 x /2 = kt /2 vx 2 = kt /m v x = kt /m (also an RMS value) Velocity increases with T and decreases with m.

18 Velocity of an N 2 Molecule at Room Temperature Mass m = 28 g/mol molecules/mol = g = Kg Temperature = 25 C= 298 K Velocity: RMS(v x ) = = kt /m Kg m 2 s 2 K K/ Kg 300 m/s (1, 000 km/hr)

19 Distribution of N 2 Velocities in a Gas at 298 K The Maxwell-Boltzmann distribution m p(v x ) = (assumes ideal gas behavior) 2πkT e mv 2 x /(2kt)

20 Diffusion of Bromophenol Blue Revisited From diffusion experiment: D = m 2 /s Definition of the diffusion coefficient: D = δ2 x 2τ. δ x /τ = velocity, which we can calculate now! v = kt /m Molar mass = 670 gm/mol Molecular mass = Kg v = Kg m 2 s 2 K K/ Kg = 61 m/s

21 Distribution of Velocities at 298 K Bromophenol blue Average (RMS) kinetic energies of molecules are the same. Based on ideal gas behavior. Distribution is similar in liquids.

22 Diffusion of Bromophenol Blue Revisited D = δx/(2τ) 2 = m 2 /s Express D in terms of velocity: D = vδ x /2 v = δ x /τ = 61 m/s Solve for δ x δ x = 2D/v = m 2 /s 61 m/s = m

23 Diffusion of Bromophenol Blue Revisited From the previous slide δ x = v τ = m Solve for τ v = δ x /τ τ = δ x /v = m 61 m/s = s = 0.1 ps VERY short distances and times!

24 RMS Distance of Diffusion Random walk along one direction: x 2 = nδ 2 x For diffusion: D = δ2 x 2τ δ 2 x = 2Dτ n = t/τ x 2 = nδx 2 = t 2Dτ = 2Dt τ RMS(x) = 2Dt For bromophenol blue (and molecules of similar size): RMS(x) = t/s m

25 RMS Distance of Diffusion

26 RMS Distance of Diffusion

27 RMS Distance of Diffusion

Biology 3550 Physical Principles in Biology Fall Semester 2017 Mid-Term Exam 6 October 2017

100 points total Please write your name on each page. In your answers you should: Biology 3550 Physical Principles in Biology Fall Semester 2017 Mid-Term Exam 6 October 2017 Show your work or provide an