Page 69 - Physical Chemistry

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Chapter 2 In contrast to U, the quantities q and w are not state functions. Given only the ini-
The First Law of Thermodynamics tial and final states of the system, we cannot find q or w. The heat q and the work w
depend on the path used to go from state 1 to state 2.
Suppose, for example, that we take 1.00 mole of liquid H O at 25.0°C and
2
1.00 atm and raise its temperature to 30.0°C, the final pressure being 1.00 atm.
What is q? The answer is that we cannot calculate q because the process is not speci-
fied. We could, if we like, increase the temperature by heating at 1 atm. In this case,
q mc T 18.0 g 1.00 cal/(g °C) 5.0°C 90 cal. However, we could instead
P
emulate James Joule and increase T solely by doing work on the water, stirring it with
a paddle (made of an adiabatic substance) until the water reached 30.0°C. In this case,
q 0. Or we could heat the water to some temperature between 25°C and 30°C and
then do enough stirring to bring it up to 30°C. In this case, q is between 0 and 90 cal.
Each of these processes also has a different value of w. However, no matter how we
bring the water from 25°C and 1.00 atm to 30.0°C and 1.00 atm, U is always the
same, since the final and initial states are the same in each process.

EXAMPLE 2.3 Calculation of U

Calculate U when 1.00 mol of H O goes from 25.0°C and 1.00 atm to 30.0°C
2
and 1.00 atm.
Since U is a state function, we can use any process we like to calculate U.
A convenient choice is a reversible heating from 25°C to 30°C at a fixed pres-
sure of 1 atm. For this process, q 90 cal, as calculated above. During the heat-
ing, the water expands slightly, doing work on the surrounding atmosphere. At
constant P, we have
2
2
w w rev P dV P dV P(V V )
1
1
2
1
where (2.27) was used. Because P is constant, it can be taken outside the inte-
gral. The volume change is V V V m/r m/r , where r and r are
2
1
1
2
1
2
the final and initial densities of the water and m 18.0 g. A handbook gives
3
3
3
r 0.9956 g/cm and r 0.9970 g/cm . We find V 0.025 cm and
1
2
1
1.987 cal mol K 1
3
3
w 0.025 cm atm 0.025 cm atm
1
3
82.06 cm atm mol K 1
0.0006 cal (2.43)
where two values of R were used to convert w to calories. Thus, w is com-
pletely negligible compared with q, and U q w 90 cal. Because vol-
ume changes of liquids and solids are small, usually P-V work is significant
only for gases.
Exercise
Calculate q, w, and U when 1.00 mol of water is heated from 0°C to 100°C
3
at a fixed pressure of 1 atm. Densities of water are 0.9998 g/cm at 0°C and
3
0.9854 g/cm at 100°C. (Answer: 1800 cal, 0.006 cal, 1800 cal.)
Although the values of q and w for a change from state 1 to state 2 depend on the
process used, the value of q w, which equals U, is the same for every process that
goes from state 1 to state 2. This is the experimental content of the first law.
Since q and w are not state functions, it is meaningless to ask how much heat a
system contains (or how much work it contains). Although one often says that “heat
and work are forms of energy,” this language, unless properly understood, can mislead