and so according to Equation 2.31,
ΔS system = nR ln
V f
V i
(2.33)
Equation 2.33 gives the entropy change in our system (ΔS system ) due to the
reversible expansion of a gas from some initial volume V i to final volume
V f . Since ΔS is a state function, its value does not depend on the path
taken. This means that ΔS has the same value if the process occurred as a
single-step irreversible expansion.
When we talk about entropy we must consider entropy changes in both
the system and the surroundings, because if heat enters the system then it
must leave the surroundings and vice versa. Therefore, the total change in
entropy must be given by the changes that occur in both the system and
the surroundings (Equation 2.34):
ΔS total = ΔS system + ΔS surroundings
(2.34)
Example 2.8 Entropy Change of Expanding Gas
Determine ΔS system , ΔS surroundings , and ΔS total when an ideal gas
expands isothermally and reversibly from 1 L to 10 L.
Solution Let’s consider the system and the surroundings separately.
System: ΔS system = (1 mol)(8:314 Jmol
−1 K
−1 ) ln
10 L
1 L
= 19:14 JK
−1
Surroundings: The surroundings have lost a quantity of heat equal to
exactly −q rev . Therefore,
ΔS surroundings = −
q rev
T
= −nR ln
V f
V i
= −(1 mol)(8:314 Jmol
−1 K
−1 ) ln
10 L
1 L
= −19:14 JK
−1
Total: ΔS total = ΔS system + ΔS surroundings = 0
What is the physical meaning of ΔS? We can examine ΔS to see how it
changes for various processes. In the example above, we’ve already seen
that it’s zero for a reversible process (for the total system and surroundings). Let’s examine some other processes. We will not go through
the details here, but rather focus on the findings. It turns out that the
CHAPTER 2: Thermodynamics and Nanoscience
40
ΔS system = nR ln
V f
V i
(2.33)
Equation 2.33 gives the entropy change in our system (ΔS system ) due to the
reversible expansion of a gas from some initial volume V i to final volume
V f . Since ΔS is a state function, its value does not depend on the path
taken. This means that ΔS has the same value if the process occurred as a
single-step irreversible expansion.
When we talk about entropy we must consider entropy changes in both
the system and the surroundings, because if heat enters the system then it
must leave the surroundings and vice versa. Therefore, the total change in
entropy must be given by the changes that occur in both the system and
the surroundings (Equation 2.34):
ΔS total = ΔS system + ΔS surroundings
(2.34)
Example 2.8 Entropy Change of Expanding Gas
Determine ΔS system , ΔS surroundings , and ΔS total when an ideal gas
expands isothermally and reversibly from 1 L to 10 L.
Solution Let’s consider the system and the surroundings separately.
System: ΔS system = (1 mol)(8:314 Jmol
−1 K
−1 ) ln
10 L
1 L
= 19:14 JK
−1
Surroundings: The surroundings have lost a quantity of heat equal to
exactly −q rev . Therefore,
ΔS surroundings = −
q rev
T
= −nR ln
V f
V i
= −(1 mol)(8:314 Jmol
−1 K
−1 ) ln
10 L
1 L
= −19:14 JK
−1
Total: ΔS total = ΔS system + ΔS surroundings = 0
What is the physical meaning of ΔS? We can examine ΔS to see how it
changes for various processes. In the example above, we’ve already seen
that it’s zero for a reversible process (for the total system and surroundings). Let’s examine some other processes. We will not go through
the details here, but rather focus on the findings. It turns out that the
CHAPTER 2: Thermodynamics and Nanoscience
40