Chemistry 360
Spring 2015
Dr. Jean M. Standard
January 30, 2015
Exact Differentials
Exact Differentials
A general form of the differential df of a function of two variables, f ( x, y ) , is given by the expression
df = A( x, y ) dx + B( x, y ) dy .
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The differential df is sometimes referred to as the total differential of the function f ( x, y ) .
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A differential is said to be exact if the functions A( x, y ) and B( x, y ) take very specific forms. For the function
f ( x, y ) , an exact differential df is defined as
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#∂ f &
# ∂€ f &
df = %
( dx + %
( dy .
$ ∂x ' y
$ ∂y 'x
If the following is true for the functions A( x, y ) and B( x, y ) , then the differential is exact:
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# ∂A &
# ∂B &
% ( = % ( .
$ ∂x ' y
$ ∂y 'x
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This exactness test follows from the knowledge that the mixed second partial derivatives of a function are equal,
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Example 1
# ∂2 f &
# ∂2 f &
%%
(( = %%
(( .
$ ∂x∂y '
$ ∂y∂x '
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Consider the differential df given by
df = 3x 2 y dx + x 3 dy .
For this expression, A( x, y ) = 3x 2 y and B( x, y ) = x 3 . Determining the partial derivatives required for the test for
exactness, we have
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# &
€% ∂A ( = 3x 2
$ ∂y 'x
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and
# ∂B &
2
% ( = 3x .
$ ∂x 'y
# ∂A &
# ∂B &
Since we have shown that % ( = % ( , the differential df is exact in this case.
$ ∂x ' y
$ ∂y 'x
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2
Example 2
The volume of a cylinder is
V = π r 2h .
The exact differential dV is given by
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#∂ V &
#∂ V &
dV = %
( dr + %
( dh .
$ ∂r ' h
$ ∂h 'r
Evaluating the partial derivatives yields
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#∂ V &
%
( = 2π rh ,
$ ∂r ' h
#∂ V &
2
( = πr .
and %
$ ∂ h 'r
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Substituting the partial derivatives leads to a total differential given by
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dV = 2π rh dr + π r 2 dh
We can verify that the differential dV is exact. For this differential, A( r,h ) = 2πrh and B( r,h ) = π r 2 . Determining
the partial derivatives required for€the test for exactness, we have
# ∂A &
% ( = 2π r
$ ∂h 'r
and€
# ∂B &
% ( =€ 2π r .
$ ∂r 'h
# ∂A &
# ∂B &
Since we have shown that % ( = % ( , the differential dV is exact.
$
'
$ ∂r ' h
∂
h
r
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3
What do the parts of the total differential mean?
Since the volume of the cylinder V depends on two variables, the radius r and the height h, any variation of either of
these variables (dr or dh) leads directly to a change in the volume of the cylinder (dV). The partial derivatives
provide the proportionality between changes in the variables r and h and changes in the volume V.
For the cylinder, we have shown that the total differential is
dV = 2π rh dr + π r 2 dh .
Part 1
The first part of the total differential of the cylinder is 2π rh dr . This term corresponds to the volume of a hollow
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cylinder of thickness dr, as shown in Figure 1.
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Figure 1. Illustration of the term 2π rh dr .
Part 2
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The second part of the total differential of the cylinder is π r 2 dh . This term corresponds to the volume of a
cylindrical disk of thickness dh, as shown in Figure 2.
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Figure 2. Illustration of the term π r 2 dh .
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[Figures taken from Applied Mathematics for Physical Chemistry, 2nd ed., James R. Barrante, Prentice Hall, New Jersey, 1998, p. 39.]