Page 122 - Physical chemistry eng

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5.7 THE CHANGE OF ENTROPY IN THE SURROUNDINGS AND ¢S total =¢S +¢S surroundings
between the system and the surroundings. Therefore, the surroundings always remain
in internal equilibrium during heat transfer.
Next consider the entropy change of the surroundings, whereby the surroundings
are at either constant V or constant P. We assume that the system and surroundings are
at the same temperature. If this were not the case, heat would flow across the boundary
until T is the same for system and surroundings, unless the system is surrounded by adi-
abatic walls, in which case q surroundings = 0. The amount of heat absorbed by the sur-
roundings, q surroundings , depends on the process occurring in the system. If the
surroundings are at constant V, q surroundings =¢U surroundings , and if the surroundings
. Because H and U are state functions,
are at constant P, q surroundings =¢H surroundings
the amount of heat entering the surroundings is independent of the path; q is the same
whether the transfer occurs reversibly or irreversibly. Therefore,

dq surroundings
dS surroundings = or for a macroscopic change,
T
q surroundings
¢S surroundings = (5.37)
T
Note that the heat that appears in Equation (5.37) is the actual heat transferred because
the heat transferred to the surroundings is independent of the path as discussed earlier.
By contrast, in calculating ¢S for the system, dq reversible for a reversible process that
connects the initial and final states of the system must be used, not the actual dq for the
process. It is essential to understand this reasoning in order to carry out calculations
for ¢S and ¢S surroundings .
This important difference is discussed in calculating the entropy change of the system
as opposed to the surroundings with the aid of Figure 5.7. A gas (the system) is enclosed
in a piston and cylinder assembly with diathermal walls. The gas is reversibly compressed
by the external pressure generated by droplets of water slowly filling the beaker on top of
the piston. The piston and cylinder assembly is in contact with a water bath thermal reser-
voir that keeps the temperature of the gas fixed at the value T. In Example Problem 5.7, w
¢S and ¢S surroundings are calculated for this reversible compression.
Piston
T q
EXAMPLE PROBLEM 5.7 P,V
One mole of an ideal gas at 300. K is reversibly and isothermally compressed from a
volume of 25.0 L to a volume of 10.0 L. Because the water bath thermal reservoir in
the surroundings is very large, T remains essentially constant at 300. K during the FIGURE 5.7
process. Calculate ¢S , ¢S surroundings , and ¢S total . A sample of an ideal gas (the sys-
tem) is confined in a piston and
Solution cylinder assembly with diathermal
walls. The assembly is in contact
Because this is an isothermal process, ¢U = 0 , and q reversible = –w. From Section 2.7,
with a thermal reservoir that holds
the temperature at a value of 300 K.
V f
dV V f Water dripping into the beaker on the
q reversible =-w = nRT = nRT ln piston increases the external pressure
V V i
3 slowly enough to ensure a reversible
V i
compression. The value of the pres-
10.0 L 3 sure is determined by the relative
-1 -1
= 1.00 mol * 8.314 Jmol K * 300. K * ln =- 2.285 * 10 J
25.0 L rates of water filling and evaporation
from the beaker. The directions of
The entropy change of the system is given by work and heat flow are indicated.
3
dq reversible q reversible -2.285 * 10 J
¢S = = = = - 7.62 J K -1
T T 300. K
L
The entropy change of the surroundings is given by
3
q surroundings q system 2.285 * 10 J
¢S surroundings = =- = = 7.62 J K -1
T T 300. K