Page 50 - Physical chemistry eng

2.5 HEAT CAPACITY 27

the solid. The heat capacity increases discontinuously as the solid melts to form a liq-
uid. This is the case because the liquid retains all the local vibrational modes of the
solid, and more low-energy modes become available upon melting. Therefore, the 60
heat capacity of the liquid is greater than that of the solid. As the liquid vaporizes,
the local vibrational modes present in the liquid are converted to translations that 50
cannot take up as much energy as vibrations. Therefore, C P,m decreases discontinu-
ously at the vaporization temperature. The heat capacity in the gaseous state 40
increases slowly with temperature as the vibrational modes of the individual mole- C p,m /J K 1 mol 1
cules are activated as discussed previously. These changes in C P,m can be calculated 30
for a specific substance using a microscopic model and statistical thermodynamics,
as will be discussed in detail in Chapter 32.
20 Solid Liquid Gas
Once the heat capacity of a variety of different substances has been determined, we
have a convenient way to quantify heat flow. For example, at constant pressure, the heat
10
flow between the system and surroundings can be written as
T sys,f T surr,f
300
200
q = C system (T)dT =- C surroundings (T)dT (2.10) 100 Temperature/K 400
P
P
P
L L
T sts,i T surr,i
FIGURE 2.9
By measuring the temperature change of a thermal reservoir in the surroundings at con- The variation of C P,m with temperature is
stant pressure, q can be determined. In Equation (2.10), the heat flow at constant pres- shown for Cl 2 .
P
sure has been expressed both from the perspective of the system and from the
perspective of the surroundings. A similar equation can be written for a constant vol-
ume process. Water is a convenient choice of material for a heat bath in experiments
because C is nearly constant at the value 4.18 J g –1 K –1 or 75.3 J mol –1 K –1 over the
P
range from 0°C to 100.°C. Constant pressure heating
w
EXAMPLE PROBLEM 2.3 Mass
The volume of a system consisting of an ideal gas decreases at constant
pressure. As a result, the temperature of a 1.50 kg water bath in the sur- Mass Piston
roundings increases by 14.2°C. Calculate q for the system.
P
Piston
Solution
i
q P ,T i P ,T f
i
T surr,f
q = C surroundings (T)dT =-C surroundings ¢T
P
P
P
L
T surr,i Initial state Final state
-1
=-1.50 kg * 4.18 J g K -1 * 14.2 K =-89.1 kJ
Constant volume heating
and C related for a gas? Consider the processes shown in
How are C P V
Figure 2.10 in which a fixed amount of heat flows from the surroundings
into a gas. In the constant pressure process, the gas expands as its tem-
perature increases. Therefore, the system does work on the surround-
q
ings. As a consequence, not all the heat flow into the system can be used
to increase ¢U . No such work occurs for the corresponding constant V ,T i V ,T f
i
i
volume process, and all the heat flow into the system can be used to
increase ¢U . Therefore, dT 6 dT V for the same heat flow dq . For this
P
reason, C 7 C V for gases.
P
The same argument applies to liquids and solids as long as V increases Initial state Final state
with T. Nearly all substances follow this behavior, although notable excep-
FIGURE 2.10
tions occur, such as liquid water between 0°C and 4°C, for which the vol- Not all the heat flow into the system can be used to
ume increases as T decreases. However, because ¢V m upon heating is increase ¢U in a constant pressure process, because the
and C for a liq-
much smaller than for a gas, the difference between C P,m V,m system does work on the surroundings as it expands.
uid or solid is much smaller than for a gas. However, no work is done for constant volume heating.