Page 44 - Physical Chemistry

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Section 1.8
Exercise Integral Calculus
Fora gas obeying the equation of state V RT/P B(T), where B(T)isa certain
m
function of T,(a)find a and k;(b)find 10P>0T2 in two different ways. [Answer:
V m
2
2
a (R/P dB/dT)/V ; k RT/V P ; 10P>0T2 V m P/T P (dB/dT)/RT.]
m
m
1
For solids, a is typically 10 5 to 10 4 K . For liquids, a is typically 10 3.5 to
1
10 3 K . For gases, a can be estimated from the ideal-gas a, which is 1/T; for tem-
1
peratures of 100 to 1000 K, a for gases thus lies in the range 10 2 to 10 3 K .
1
For solids, k is typically 10 6 to 10 5 atm . For liquids, k is typically 10 4
1
atm . Equation (1.47) for ideal gases gives k as 1 and 0.1 atm 1 at P equal to 1 and
10 atm, respectively. Solids and liquids are far less compressible than gases because
there isn’t much space between molecules in liquids and solids.
The quantities a and k can be used to find the volume change produced by a
change in T or P.

EXAMPLE 1.4 Expansion due to a temperature increase

Estimate the percentage increase in volume produced by a 10°C temperature in-
1
crease in a liquid with the typical a value 0.001 K , approximately independent
of temperature.
Equation (1.43) gives dV aVdT . Since we require only an approximate
P
P
answer and since the changes in T and V are small (a is small), we can approx-
imate the ratio dV /dT by the ratio V / T of finite changes to get V /V
P
P
P
P
P
1
a T (0.001 K ) (10 K) 0.01 1%.
P
Exercise
3
For water at 80°C and 1 atm, a 6.412 10 4 K 1 and r 0.971792 g/cm .
7
Using the approximation dV /dT V / T for T small, find the density of
P
P
P
P
P
3
water at 81°C and 1 atm and compare with the true value 0.971166 g/cm .
3
(Answer: 0.971169 g/cm .)
1.8 INTEGRAL CALCULUS
Differential calculus was reviewed in Sec. 1.6. Before reviewing integral calculus, we
recall some facts about sums.
Sums
The definition of the summation notation is
n
a a a a . . . a n (1.49)*
2
i
1
i 1
2
2
For example, 3 i 1 2 3 14. When the limits of a sum are clear, they
2
2
i 1
are often omitted. Some identities that follow from (1.49) are (Prob. 1.59)
n n n n n
a ca c a a , a 1a b 2 a a a b i (1.50)*
i
i
i
i
i
i 1 i 1 i 1 i 1 i 1
n m n m
a a a b a a i a b j (1.51)
i j
i 1 j 1 i 1 j 1