Thermodynamics and Chemistry

CHAPTER 1 INTRODUCTION

1.3 DIMENSIONALANALYSIS 24

1.3 Dimensional Analysis

Sometimes you can catch an error in the form of an equation or expression, or in the dimen-
sions of a quantity used for a calculation, by checking for dimensional consistency. Here
are some rules that must be satisfied:

 both sides of an equation have the same dimensions
 all terms of a sum or difference have the same dimensions
 logarithms and exponentials, and arguments of logarithms and exponentials, are di-
mensionless
 a quantity used as a power is dimensionless
In this book thedifferentialof a function, such as df, refers to aninfinitesimalquantity.
If one side of an equation is an infinitesimal quantity, the other side must also be. Thus,
the equation df DadxCbdy(whereaxandbyhave the same dimensions asf) makes
mathematical sense, but df DaxCbdydoes not.
Derivatives, partial derivatives, and integrals have dimensions that we must take into
account when determining the overall dimensions of an expression that includes them. For
instance:

 the derivative dp=dTand the partial derivative.@p=@T /Vhave the same dimensions

asp=T

 the partial second derivative.@^2 p=@T^2 /Vhas the same dimensions asp=T^2

 the integral

R

TdThas the same dimensions asT^2
Some examples of applying these principles are given here using symbols described in
Sec.1.2.
Example 1.Since the gas constantRmay be expressed in units of J K^1 mol^1 , it has
dimensions of energy divided by thermodynamic temperature and amount. Thus,RThas
dimensions of energy divided by amount, andnRThas dimensions of energy. The products
RTandnRTappear frequently in thermodynamic expressions.
Example 2. What are the dimensions of the quantitynRTln.p=p/and ofpin
this expression? The quantity has the same dimensions asnRT (or energy) because the
logarithm is dimensionless. Furthermore,pin this expression has dimensions of pressure
in order to make the argument of the logarithm,p=p, dimensionless.
Example 3.Find the dimensions of the constantsaandbin the van der Waals equation

pD

nRT

Vnb

n^2 a

V^2

Dimensional analysis tells us that, becausenbis subtracted fromV,nbhas dimensions
of volume and thereforebhas dimensions of volume/amount. Furthermore, since the right
side of the equation is a difference of two terms, these terms have the same dimensions
as the left side, which is pressure. Therefore, the second termn^2 a=V^2 has dimensions of
pressure, andahas dimensions of pressurevolume^2 amount^2.
Example 4.Consider an equation of the form

@lnx
@T



p

D

y

R