Modern Engineering Thermodynamics

3.4 Some Exciting New Thermodynamic Properties 61
or
ν 2 = ν 1 fexp½βðT 2 − T 1 Þ − κðp 2 − p 1 ÞŠg (3.8)
Thus, v is seen to have an exponential dependence on p and T when β and κ are constant. Table 3.1 gives values of β
and κ for copper as a function of temperature, and Table 3.2 gives values of β and κ of various liquids at 20.°C (68°F).
Table 3.1 Values of β and κ for Copper as a Function of Temperature
β × 10 6 κ × 10 11
T (K) R −1 K −1 ft 2 /lbf m 2 /N
100. 17.5 31.5 34.51 0.721
150. 22.8 41.0 35.08 0.733
200. 25.3 45.6 35.80 0.748
250. 26.7 48.0 36.47 0.762
300. 27.3 49.2 37.14 0.776
500. 30.1 54.2 40.06 0.837
800. 33.7 60.7 44.13 0.922
1200. 38.7 69.7 49.30 1.030
Source: Material drawn from Zemansky, M. W., 1957. Heat and Thermodynamics, fourth ed. McGraw-Hill, New York. Reprinted by permission
of the publisher.
Table 3.2 Values of β and κ for Various Liquids at 20.°C (68°F)
β × 10 6 κ × 10 11
Substance R −1 K −1 ft 2 /lbf m 2 /N
Benzene 0.689 1.24 4550 95
Diethyl ether 0.922 1.66 8950 187
Ethyl alcohol 0.622 1.12 5310 111
Glycerin 0.281 0.505 1010 21
Heptane (n) 0.683 1.23 6890 144
Mercury 0.101 0.182 192 4.02
Water 0.115 0.207 2200 45.9
Source: Adapted by permission of the publisher from Zemansky, M. W., Abbott, M. M., Van Ness, H. C., 1975. Basic Engineering
Thermodynamics, second ed. McGraw Hill, New York.
EXAMPLE 3.2
A 1.00 cm 3 copper block at 250. K is heated in the atmosphere to 800. K (Figure 3.1). Find the volume of the block at 800. K.
∀ 1 = 1.00 cm 3
∀ 2 = ?
State 1
at 250. K
State 2
at 800. K
FIGURE 3.1
Example 3.2, problem.
Solution
Since the copper block changes state under atmospheric (constant) pressure, it undergoes an isobaric process. When p = constant,
Eq. (3.8) reduces to
v 2 = v 1 fexp½βðT 2 − T 1 ÞŠg
(Continued )