Kinetic Molecular Theory Glenn V. Lo Department of Physical Sciences Nicholls State University

Kinetic Molecular Theory Explains behavior of gases using basic principles of physics Newton’s second law: F = ma Newton’s third law: for every action, there is an equal and opposite reaction Kinetic energy, KE = (1/2) mv2

KMT assumptions to explain ideal behavior Gas particles are point masses in continuous random motion. Evidence of continuous random motion: Brownian motion No energy transfer occurs when gas particles collide with walls or with each other (collisions are elastic); although this is not valid for individual collisions, the total energy transfer for all the collisions does average out to zero. The gas particles exert no force on each other.

Interpreting Pressure Pressure = due to collisions of particles with walls of container Collision leads to acceleration. Wall exerts force on particle: F = ma Particle exerts force on wall (3rd law) P = F/A

Example Suppose one N2 molecule (mass per molecule = 4.65x10-26 kg) moving at an constant speed of 300.0 m s-1 collide with a wall (with an area of 10.0 cm2) and reverses direction during a 1.0 s time period. Assume the constant speed remains the same. Calculate the pressure on the wall due to this one particle.

Interpretating Temperature Pressure due to N particles, moving in 3-dimensions where Ktrans is the sum of the translational kinetic energies of the gas particles Compare with ideal gas equation: P = nRT/V

Interpreting Temperature Temperature is a measure of average translational kinetic energy. Note: atoms have translational kinetic energy only; molecules have translational AND vibrational and rotational kinetic energies. Animations of vibrational motion: http://www2.ess.ucla.edu/~schauble/molecular_vibrations.htm

Example Which has more kinetic energy? A. a molecule with mass of 32 u, moving at 300 m s-1 B. an atom with a mass of 32 u, moving at 300 m s-1

Example Calculate the translational kinetic energy per mole of gas particles at room temperature (298K).

Example Which sample of gas has more translational kinetic energy at 298K? A. 2.00 mol Ne, B. 1.00 mol H2, C. same

Example Which sample of gas has more translational kinetic energy per mole at 298K? A. 2.00 mol Ne, B. 1.00 mol H2, C. same

Example Calculate the average kinetic energy of He atoms at 300K.

Example Which gas sample has higher average kinetic energy? A. 2.0 mol He at 298K, B. 1.0 mol O2 at 298K C. same

Root mean square speed For a collection of particles with same mass (m) at temperature T RMS speed is directly proportional to square root of T and inversely proportional to square root of MM.

Average speed RMS speed is sort of like an average speed. The true average speed is slightly smaller but is also directly proportional to square root of T and inversely proportional to square root of MM.

Example Which gas sample has particles with the slowest average speed? A. 2.0 mol He at 298K, B. 1.0 mol He at 500K, C. 2.0 mol O2 at 298K

Example Calculate the average speed of O2 molecules at 298K.

Interpreting empirical gas laws Boyle: same T, smaller V  higher P Smaller V  less distance to travel in between collisions; more frequent collisions with walls; higher P Amontons: same V, higher T  higher P Higher T  faster speeds  more frequent collisions with walls; more force on wall Charles’ Law: same P, higher T  larger V see Amontons’; to maintain P with higher T, increase V to reduce collision frequency Avogadro’s Law: same T, P; higher n, larger V More particles – more collisions with wall --- need larger V (move walls farther apart) to maintain P.

Molecular speed distribution Particles do not all move at the same speed. At a given temperature Ex. O2 at 500K Most probable speed

Molecular speed distribution Total Area under the curve = 1 Area under curve from v1 to v2 is the fraction of molecules with speeds between v1 and v2 Ex. O2

Example Consider the molecular speed distributions for the same gas sample at two different temperatures, T1 and T2, as shown in the figure below. Which distribution corresponds to a higher temperature?

Example Consider the speed distribution for the He and O2 at the same temperature (1000K) as shown in the figure below. Which curve, A or B, represents He?

Diffusion and Effusion Diffusion: spreading out of a gas. Effusion: movement of gas through a pinhole into a vacuum. Rate of effusion depends on concentration (n/V), average speed (<v>) and area of pinhole (A): At same conditions (P, T, and A): rate of effusion is inversely proportional to square root of molar mass.

Example Cotton balls dipped in concentrated aqueous solutions of NH3 and HCl are simultaneously placed in opposite ends of a glass tube The NH3 and HCl gas from the cotton balls diffuse inside the tube and react to form NH4Cl(s) where they meet. The NH4Cl(s) will be expected to form A. closer to the NH3 end of the tube, B. closer to the HCl end of the tube, C. exactly halfway

Example CO2 effuses at a rate of 2.00 mmol min-1. At the same conditions, an unknown gas effuses at a rate of 3.32 mmol min-1. What is a possible identity of the unknown gas? A. CH4, B. O2, C. Cl2, D. CCl4

Example How much faster would He effuse compared to CH4 under the same conditions?

Example Consider a gas containing 0.7% 235UF6 and 99.3% 238UF6. If the gas is contained in chamber A, as illustrated below, and is allowed to escape through a pinhole into vacuum chamber B, the percentage of 235UF6 in the gas that initially emerges in chamber B will be... A. higher than 0.7%, B. lower than 0.7%, C. still 0.7%.