Page 35 - Physical Chemistry

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3
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Chapter 1 Since 1 atm 101325 N/m [Eq. (1.10)], we have 1 cm atm (10 2 m) 101325 N/m 2
Thermodynamics 3 2
0.101325 m N/m 0.101325 J. [One newton-meter one joule (J); see Sec. 2.1.]
Hence R 82.06 0.101325 J/(mol K), or
3
R 8.314 J>1mol K2 8.314 1m Pa2>1mol K2 (1.20)*
5
5
Using 1 atm 760 torr and 1 bar 750 torr, we find from (1.19) that R 83.14
5
3
(cm bar)/(mol K). Using 1 calorie (cal) 4.184 J [Eq. (2.44)], we find
R 1.987 cal>1mol K2 (1.21)*
Accurate values of physical constants are listed inside the back cover.
Ideal Gas Mixtures
So far, we have considered only a pure ideal gas. In 1810 Dalton found that the pres-
sure of a mixture of gases equals the sum of the pressures each gas would exert if
placed alone in the container. (This law is exact only in the limit of zero pressure.) If
n moles of gas 1 is placed alone in the container, it would exert a pressure n RT/V
1
1
(where we assume the pressure low enough for the gas to behave essentially ideally).
Dalton’s law asserts that the pressure in the gas mixture is P n RT/V n RT/V
2
1
(n n )RT/V n RT/V, so
2
1
tot
PV n RT ideal gas mixture (1.22)*
tot
Dalton’s law makes sense from the molecular picture of gases. Ideal-gas molecules do
not interact with one another, so the presence of gases 2, 3, . . . has no effect on gas 1,
and its contribution to the pressure is the same as if it alone were present. Each gas
acts independently, and the pressure is the sum of the individual contributions. For real
gases, the intermolecular interactions in a mixture differ from those in a pure gas, and
Dalton’s law does not hold accurately.
The partial pressure P of gas i in a gas mixture (ideal or nonideal) is defined as
i
P x P any gas mixture (1.23)*
i
i
where x n /n is the mole fraction of i in the mixture and P is the mixture’s
i i tot
pressure. For an ideal gas mixture, P x P (n /n ) (n RT/V) and
i
i
tot
i
tot
P n RT>V ideal gas mixture (1.24)*
i
i
The quantity n RT/V is the pressure that gas i of the mixture would exert if it alone
i
were present in the container. However, for a nonideal gas mixture, the partial pres-
sure P as defined by (1.23) is not necessarily equal to the pressure that gas i would
i
exert if it alone were present.
EXAMPLE 1.1 Density of an ideal gas
Find the density of F gas at 20.0°C and 188 torr.
2
The unknown is the density r, and it is often a good idea to start by writ-
ing the definition of what we want to find: r m/V. Neither m nor V is given,
so we seek to relate these quantities to the given information. The system is a
gas at a relatively low pressure, and it is a good approximation to treat it as an
ideal gas. For an ideal gas, we know that V nRT/P. Substitution of V
nRT/P into r m/V gives r mP/nRT. In this expression for r,weknow P
and T but not m or n. However, we recognize that the ratio m/n is the mass per
mole, that is, the molar mass M. Thus r MP/RT. This expression contains only
known quantities, so we are ready to substitute in numbers. The molecular