Page 108 - Physical Chemistry

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Section 3.4
produces a smaller increase in entropy for the higher-temperature process, and Calculation of Entropy Changes
S is smaller for the 50°C to 75°C heating. The higher the temperature, the
smaller the entropy change produced by a given amount of reversible heat.
Exercise
Find S when 100 g of water is reversibly heated from 50°C to 75°C at 1 atm.
(Answer: 31.2 J/K.)

Now suppose we heat the water irreversibly from 25°C to 50°C at 1 atm (say, by
using a bunsen-burner flame). The initial and final states are the same as for the re-
versible heating. Hence the integral on the right side of (3.28) gives S for the irre-
versible heating. Note that S in (3.28) depends only on T , T , and the value of P
1
2
(since C will depend somewhat on P); that is, S depends only on the initial and final
P
states. Thus S for heating 100 g of water from 25°C to 50°C at 1 atm is 33.7 J/K,
whether the heating is done reversibly or irreversibly. For irreversible heating with a
bunsen burner, portions of the system nearer the burner will be at higher temperatures
than portions farther from the burner, and no single value of T can be assigned to the
system during the heating. Despite this, we can imagine doing the heating reversibly
and apply (3.28) to find S, provided the initial and final states are equilibrium states.
Likewise, if we carry out the change of state by stirring at constant pressure as Joule
did, rather than by heating, we can still use (3.28).
To heat a system reversibly, we surround it with a large constant-temperature bath
that is at the same temperature as the system, and we heat the bath infinitely slowly.
Since the temperature of the system and the temperature of its surroundings differ only
infinitesimally during the process, the process is reversible.

6. Reversible change of state of a perfect gas. From the first law and Sec. 2.8, we
have for a reversible process in a perfect gas

dq rev dU dw rev C dT P dV C dT nRT dV>V (3.29)
V
V
dS dq >T C dT>T nR dV>V
V
rev
2 dT 2 dV
¢S C 1T2 nR
V
1 T 1 V
T 2 C 1T2 V 2
V
¢S dT nR ln perf. gas (3.30)
T 1 T V 1
If T T , the first integral is positive, so increasing the temperature of a perfect
2 1
gas increases its entropy. If V V , the second term is positive, so increasing the
2 1
volume of a perfect gas increases its entropy. If the temperature change is not
large, it may be a good approximation to take C constant, in which case S
V
C ln (T /T ) nR ln (V /V ). A mistake students sometimes make in using (3.30)
V 2 1 2 1
is to write ln (V /V ) ln (P /P ), forgetting that T is changing. The correct
2 1 1 2
expression is ln (V /V ) ln (P T /P T ).
2 1 1 2 2 1
7. Irreversible change of state of a perfect gas. Let n moles of a perfect gas at P ,
1
V , T irreversibly change its state to P , V , T . We can readily conceive of a
1 1 2 2 2
reversible process to carry out this same change in state. For example, we might
(a) put the gas (enclosed in a cylinder fitted with a frictionless piston) in a large
constant-temperature bath at temperature T and infinitely slowly change the pres-
1
sure on the piston until the gas reaches volume V ; (b) then remove the gas from
2
contact with the bath, hold the volume fixed at V , and reversibly heat or cool the
2
gas until its temperature reaches T . Since S is a state function, S for this
2