Physical Chemistry , 1st ed.

To evaluate the total heat, we integrate both sides of this infinitesimal equation

to get

qV

Ti

Tf

CVdTU (2.25)

where the final equality is taken from the fact that Uqfor a constant-

volume change. Equation 2.25 is the most general form for a constant-volume

change. However, if the heat capacity is constant over the temperature range

(for small temperature ranges not involving changes in phase, it is), it can be

taken out of the integral to yield

UCV

Ti

Tf

dT CV(TfTi) CVT (2.26)

For nmoles of gas, this is rewritten simply as

UnCVT (2.27)

where CVis the molar heat capacity.If the heat capacity does vary substantially

with temperature, some expression for CVin terms of temperature will have to

be substituted in equation 2.25 and the integral evaluated explicitly. If this is

the case, the temperatures for the integral limits must be expressed in kelvins.

If the heat capacity is divided by the mass of the system, it will have units

of J/gK or J/kgK and is referred to as the specific heat capacityor, commonly,

the specific heat.Care should be taken to note the units of a given heat capac-

ity to determine if it is really a specific heat.

Example 2.10

Evaluate Ufor 1.00 mole of oxygen, O 2 , going from 20.0°C to 37.0°C at

constant volume, in the following cases. (Uwill have units of J.)

a.It is an ideal gas with CV20.78 J/molK.

b.It is a real gas with an experimentally determined CV21.6 + 4.18 

10 ^3 T(1.67  105 )/T^2.

Solution

a.Because we are assuming that the heat capacity is constant, we can use

equation 2.27, where the change in temperature is 57°:

UnCVT(1.00 mol)20.78 

mo

J

lK

(57.0°) 1184 J

Here, we are using the unit for CVthat includes the mole unit in the de-

nominator.

b.Since the heat capacity varies with temperature, we have to integrate the

expression in equation 2.25. We must also convert our temperatures to kelvins:

U

Ti

Tf

CVdT

310 K

253 K

 21.6 + 4.18  10 ^3 T

167

T

,0

2

00

dT

Integrating term by term:

U21.6 T+ 

1

2

4.18  10 ^3 T^2 + 

167

T

,000



310

253

and evaluating at the limits:

U6696.0 + 200.0 + 538.7 5464.8 133.8 660.1 1176.8 J

40 CHAPTER 2 The First Law of Thermodynamics