Physical Chemistry Third Edition

10.4 Transport Processes in Liquids 467

10.25Show that for fixed pressure, the self-diffusion coefficient
of a hard-sphere gas is proportional toT^3 /^2. Explain in
words why the temperature dependence is different for
constant pressure than for constant number density.

10.26For diffusion in a two-substance hard-sphere gas in which
the substances have different masses and sizes and in
which neither substance is dilute, one uses the linear
equation
u 1 −u 2 
1
x 1 x 2

D 12 ∇x 1

instead of Fick’s law. In this equation,u 1 is the mean drift

velocity of molecules of component 1 andu 2 is the same

quantity for component 2. The quantitiesx 1 andx 2 are

the two mole fractions andD 12 is called the mutual

diffusion coefficient. For a mixture of hard spheres^8

D 12 

3

8 π^1 /^2

1

d^212 Ntot

(

kBT

2 μ

) 1 / 2

whereμis the reduced mass,Ntotis the total number

density, andd 12 is the mean collision diameter

d 12 

d 1 +d 2

2

Find the mutual diffusion coefficient for helium and

argon at 273.15 K and 1.000 atm pressure. Compare with

the self-diffusion coefficient of each substance from

Problem 10.22.

10.27Gaseous argon has a viscosity of 2. 099 × 10 −^5 Pa s
 2. 099 × 10 −^5 kg m−^1 s−^1 at 0◦C and 1.000 atm.
a.Estimate its viscosity at 100◦C and 1.000 atm.
b.Estimate its viscosity at 0◦C and 0.500 atm.
c.Estimate its viscosity at 100◦C and 0.500 atm.

10.28Liquid uranium hexafluoride has a vapor pressure at 56◦C
equal to 765 torr, so UF 6 is a gas at 60◦C and 1.000 atm.

Various diffusion and thermal diffusion processes were

used in World War II to separate gaseous^235 UF 6

molecules from^238 UF 6 molecules. Calculate the mutual

diffusion coefficient of these substances at 60◦C and

1 .000 atm. Calculate the self-diffusion coefficient of

(^238) U at this temperature and pressure. Make a reasonable
estimate of the effective hard-sphere diameter of UF 6 and
see Problem 10.26 for the necessary equation.
10.29Calculate the mutual diffusion coefficient for helium and
carbon dioxide at 298.15 K and 1.000 bar. (See Problem
10.26 for equations.)
10.30At 20.0◦C, the viscosity of ammonia gas is equal to
9. 82 × 10 −^6 kg m−^1 s−^1. Find the effective hard-sphere
diameter of ammonia molecules at this temperature.
10.31From values of the viscosity in Table A.18, calculate the
effective hard-sphere diameter of gaseous water at 100◦C
and at 200◦C. Explain any temperature dependence.
10.32For a temperature of 293 K and a pressure of 1.00 atm,
calculate the viscosities of helium gas and of carbon
dioxide gas from the hard-sphere diameters. Explain why
the values compare as they do.
10.33Find the value of each of the ratios:
a.The self-diffusion coefficient of helium divided by the
self-diffusion coefficient of argon at the same pressure
and temperature.
b.The viscosity of helium divided by the viscosity of
argon at the same pressure and temperature.
c.The viscosity of helium at 1.00 atm and 298.15 K
divided by the viscosity of helium at 0.100 atm and
298.15 K.
d.The thermal conductivity of argon divided by the
thermal conductivity of xenon at the same pressure
and temperature.

10.4 Transport Processes in Liquids

In Section 9.9 the molecular environment of a molecule in a liquid was described as a

cage made up of neighboring molecules, in which the molecule is confined by repulsive

intermolecular forces. If a molecule were absolutely confined to such a cage there could

be no diffusion or viscous flow, and this is almost the case in solids. In a liquid there

are voids among the neighbors. There is a chance that, after colliding many times with

the neighboring molecules in a given cage, a molecule can push past some neighbors

(^8) J. O. Hirschfelder, C. F. Curtiss, and R. B. Bird,Molecular Theory of Gases and Liquids, Wiley, New York, 1954, pp. 14, 518.