Physical Chemistry Third Edition

9.7 The Model System with Potential Energy 419

One-Body Forces

A one-body force is independent of the locations of other particles. Electrostatic forces

on a charged particle due to an external electric field are one-body forces. Near the

surface of the earth the force of gravity on a particle is also a one-body force. This

gravitational force on an object corresponds to a potential energy

Vgmgz (9.7-1)

wherezis the vertical coordinate of the particle andmis its mass. The acceleration

due to gravity is denoted byg. It is slightly dependent on latitude and is nearly equal to

9.80 m s−^2 at the latitude of Washington, DC, USA, and Seoul, South Korea. We will

use this value in our examples. The total energy of a particle subject to a gravitational

force near the surface of the earth is

ε

m

2

(v^2 x+v^2 y+v^2 z)+V(x,y,z)

m

2

v^2 +mgz (9.7-2)

We now assert without proof that the Boltzmann probability distribution holds for

this total energy:

(

probability of a

state of energyε

)

∝e−ε/kBTdvexp

(

−

mv^2 / 2 +V

kBT

)

(9.7-3)

where the symbol∝stands for “is proportional to.”

EXAMPLE9.14

Assume that the earth’s atmosphere has a constant temperature of 298 K (not an accurate

assumption). Estimate the pressure at 8900 m above sea level (roughly the altitude of Mount

Everest). Assume that air is a single ideal gas with a molar mass of 0.029 kg mol−^1.

Solution

Representing the high-altitude state as 2 and the sea-level state as 1, the probability ratio is

p 2

p 1



N 2

N 1



P 2

P 1



e−(K^2 +V^2 )/kBT

e−(K^1 +V^1 )/kBT

The average kinetic energy of a molecule depends only on the temperature, which we have

assumed to be independent of altitude. The kinetic energy terms will cancel after averaging:

P 2

P 1

exp

(

V 2 −V 1

kBT

)

Assuming that the ideal gas law is valid, the pressure is proportional to the number of

molecules per unit volume at constant temperature and is proportional to the population

given by the Boltzmann distribution.

V 2 −V 1 

(

0 .029 kg mol−^1

)(

9 .80ms^2

)

(8900 m)

6. 022 × 1023 mol−^1

 4. 2 × 10 −^21 J

IfP 1 .00 atm at sea level, and if we assume that air is an ideal gas:

P

1 .00 atm

exp

(

− 4. 2 × 10 −^21 J

(

1. 3807 × 10 −^23 JK−^1

)

(298 K)

)

 0. 36