1. Chapter 21: Molecules in motion Diffusion: the migration of matter down a concentration gradient. Thermal conduction: the migration of energy down a temperature gradient. Electric conduction: the migration of electric charge along an electrical potential gradient. Viscosity: the migration of linear momentum down a velocity gradient. Effusion: the emergence of a gas from a container through a small hole.

2. Molecular motion in gases PV = 1/3 nMc 2 (21.1) where M = mN A, the molar mass of the molecules, c is the root mean square speed of the molecules: c = 1/2 (21.2) c = (21.3) The root mean square speed of the gas molecules is proportional to the square root of the temperature and inversely proportional to the square root of the molar mass.

3. Important concepts for gases Mean speed The most probable speed Relative mean speed: (reduced mass) Collision frequency

4. 21.2 Collisions with walls and surfaces The collision flux, Zw, the number of collisions with the area in a given time interval divided by the area and the duration of the interval. Collision frequency can be obtained by multiplication of the collision flux by the area of interest.

5. 21.3 The rate of effusion Graham’s law of effusion: the rate of effusion is inversely proportional to the square root of the molar mass. Vapor pressures of liquids and solids can be measured based on the above equation.

6. Example 21.2 Cesium (m.p. 29 o C, b.p. 686 o C) was introduced into a container and heated to 500 o C. When a hole of diameter 0.500mm was opened in the container for 100s, a mass loss of 385 mg was measured. Calculate the vapor pressure of liquid cesium at 500K. Solution: Despite the effusion, the vapor pressure is constant inside the container because the hot liquid metal replenishes the vapor. Consequently, the effusion rate is constant! The mass loss Δm in an interval Δt is related to the collision flux by: Δm = ZwA 0 mΔt Where A 0 is the area of the hole and m is the mass of one atom. Zw = plug in numbers, one gets p = 11k Pa

7. Self-test 21.2: How long would it take 1.0 g of Cs atoms to effuse out of the over under the same conditions as listed in example 21.2? (260 s) Self-test: There is 1.0 g of Cs solid in the effusion oven, how long does it take to effuse out of the oven?

8. Step 1: Derive an expression that shows how the pressure of a gas inside an effusion oven varies with time if the oven is not replenished as the gas escapes. step 2: calculate the derive of P: step 3: the rate of change of the number of molecules is equal to the collision frequency with the hole multiplied by the area of the hole: so step 4: integrate the above equation P = P 0 e -t/τ ( Remark: how does the temperature (or the size of the hole) affect the decrease of pressure?)

9. 21.4 Transport properties of a perfect gas Experimental observations on transport properties shows that the flux of a property is proportional to the first derivative of other related properties. The flux of matter is proportional to the first derivative of the concentration (Fick’s first law of diffusion): J(matter)  The rate of thermal conduction is proportional to the temperature gradient: J(energy)  J(matter) = - D D is called the diffusion coefficient (m 2 s -1 ); J(energy) = - k dT/dz k is called the coefficient of thermal conductivity (J K -1 m -1 s -1 )

11. J(x-component of momentum) =, η is the coefficient of viscosity.

12. Table 21.3

13. Diffusion

14. As represented by the above Figure, on average the molecules passing through the area A at z = 0 have traveled about one free path. The average number of molecules travels through the imaginary window A from Left to Right during an interval Δt is ZwA Δt (L→R) Because Zw = So A Δt (L→R) The average number of molecules travels through the imaginary window A from Right to Left during an interval Δt is A Δt (R →L) The net number of molecules passing through the window A along the z direction is: A Δt - A Δt By definition the flux of molecules along z direction can be calculate as J(z) = ( A Δt - A Δt )/(A Δt ) J(z) = The number density N(-λ) and N(λ) can be represented by number density N(0) at z =0 N(-λ) = N(0) - λ N(λ) = N(0) + λ Therefore: J(z) = then we get D = (different from what we expected)

15. A factor of 2/3 needs to be introduced. So we get D =

16. Thermal conduction k = where C V, m is the molar heat capacity at constant volume. Because λ is inversely proportional to the molar concentration of the gas, the thermal conductivity is independent of the concentration of gas, and hence independent of the gas pressure. One exception: at very low pressure, where the mean free path is larger than the size of the container.

17. J(x-component of momentum) =, η is the coefficient of viscosity.

18. Viscosity The viscosity is independent of the pressure. Proportional to T 1/2

19. Measuring the viscosity Poiseuille’s formula:

20. Calculations with Poiseuille’s formula Example: In a poiseuille flow experiment to measure the viscosity of air at 298K, the sample was allowed to flow through a tube of length 100cm and internal diameter 1.00mm. The high-pressure end was at 765 Torr and the low-pressure end was at 760Torr. The volume was measured at the latter pressure. In 100s, 90.2cm3 of air passed through the tube. Solution: Reorganize Poseuille’s equation: