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Chemical thermodynamics 25
dsuniverse = dssystem + dssurroundings (2.20a)
dsuniverse = 0 for reversible (equilibrium) transformations (2. 2 0b)
dsuniverse>O for irreversible (spontaneous) transformations (2.20c)
The second law of thermodynamics cannot be proved. It is believed
to be valid because it leads to deductions that are in accord with
observations and experience. The following exercise provides an ex
ample of such a deduction.
Exercise 2 .4. Assuming the truth of the second law of thermodynam
ics, prove that an isolated ideal gas can spontaneously expand but not
spontaneously contract.
Solution. We will consider a unit mass of the gas. If the gas is
isolated, it has no contact with its surroundings; therefore, dssurroundings
= 0 , and
d su ni verse = dssystem + dssurroundings = ds gas (2.2 1 )
i
Also, f the gas s i s olated , dq = dw = O ; therefore, from Eqs. (2.5) and
i
(2. 6 ) , du = O. If du = O , it follows from Joule' s law for an ideal gas (see
Note I in this chapter) that dT= O . Hence, the gas must pass from its
initial state I ) to its final state (2) isothermally.
(
s
To obtain an expression for d s ga • we can follow any reversible and
isothermal path from state I to state 2, and evaluate the integral
f dqrev
ga
ds s = 2
I T
For an ideal gas (see Exercise . 2 8)
2
dqrev dT dp
-- = c - - R -
T P T p
where R is the gas constant for a unit mass of the gas and c the
P
specific heat at constant pressure of the gas. Therefore,
or,