Page 125 - Physical Chemistry

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3.17 Calculate S for the mixing of 10.0 g of He at 120°C per cycle to a reservoir at t . Let Carnot engine B absorb heat
2
and 1.50 bar with 10.0 g of O at 120°C and 1.50 bar. q 2B per cycle from the reservoir at t and discard heat q per
1
2
2
cycle to a reservoir at t . Further, let q q , so that engine
3.18 A system consists of 1.00 mg of ClF gas. A mass spec- 1 2A 2B
37
35
2
trometer separates the gas into the species ClF and ClF. B absorbs an amount of heat from the t reservoir equal to the
35
19
Calculate S. Isotopic abundances: F 100%; Cl 75.8%; heat deposited in this reservoir by engine A. Show that
37 Cl 24.2%. g1t 2 , t 3 2g1t 1 , t 2 2 q 1 >q 3
3.19 Let an isolated system be composed of one part at T and where the function g is defined as 1 e . The heat reservoir
1
rev
a second part at T , with T T ; let the parts be separated by at t can be omitted, and the combination of engines A and B
1
2
2
2
a wall that allows heat flow at only an infinitesimal rate. Show can be viewed as a single Carnot engine operating between t 3
that, when heat dq flows irreversibly from T to T , we have and t ; hence g(t , t ) q /q . Therefore
1
2
1
3
3
1
1
dS dq/T dq/T (which is positive). (Hint: Use two heat
1
2
reservoirs to carry out the change of state reversibly.) g1t 1 , t 3 2
g1t 1 , t 2 2 (3.59)
g1t 2 , t 3 2
Section 3.5
3.20 True or false? (a) For a closed system, S can never be Since t does not appear on the left side of (3.59), it must can-
3
negative. (b) For a reversible process in a closed system, S cel out of the numerator and denominator on the right side.
must be zero. (c) For a reversible process in a closed system, After t is canceled, the numerator takes the form f(t ) and the
3
1
S univ must be zero. (d) For an adiabatic process in a closed denominator takes the form f(t ), where f is some function;
2
system, S cannot be negative. (e) For a process in an isolated we then have
system, S cannot be negative. ( f ) For an adiabatic process in
a closed system, S must be zero. (g) An adiabatic process can- g1t 1 , t 2 2 f1t 1 2 (3.60)
not decrease the entropy of a closed system. (h) For a closed sys- f1t 2 2
tem, equilibrium has been reached when S has been maximized.
which is the desired result, Eq. (3.42). [A more rigorous de-
3.21 For each of the following processes deduce whether each rivation of (3.60) from (3.59) is given in Denbigh, p. 30.]
of the quantities S and S is positive, zero, or negative.
univ
(a)Reversible melting of solid benzene at 1 atm and the normal 3.26 For the gaussian probability distribution, the probability
melting point. (b) Reversible melting of ice at 1 atm and 0°C. of observing a value that deviates from the mean value by at
(c)Reversible adiabatic expansion of a perfect gas. (d)Revers- least x standard deviations is given by the following infinite
ible isothermal expansion of a perfect gas. (e) Adiabatic expan- series (M. L. Abramowitz and I. A. Stegun, Handbook of
sion of a perfect gas into a vacuum (Joule experiment). Mathematical Functions, Natl. Bur. Stand. Appl. Math. Ser. 55,
(f) Joule–Thomson adiabatic throttling of a perfect gas. 1964, pp. 931–932):
(g)Reversible heating of a perfect gas at constant P. (h) Revers-
ible cooling of a perfect gas at constant V. (i) Combustion of 2 x >2 1 1 3 # # #
2
e a b
benzene in a sealed container with rigid, adiabatic walls. ( j) Adi- 22p x x 3 x 5
abatic expansion of a nonideal gas into vacuum.
where the series is useful for reasonably large values of x.
3.22 (a) What is S for each step of a Carnot cycle? (b) What (a) Show that 99.7% of observations lie within 3 standard de-
is S univ for each step of a Carnot cycle? viations from the mean. (b) Calculate the probability of a devi-
6
3.23 Prove the equivalence of the Kelvin–Planck statement ation 10 standard deviations.
and the entropy statement [the set-off statement after Eq. (3.40)]
3.27 If the probability of observing a certain event in a single
of the second law. [Hint: Since the entropy statement was de-
trial is p, then clearly the probability of not observing it in one
rived from the Kelvin–Planck statement, all we need do to show
trial is 1 p. The probability of not observing it in n indepen-
the equivalence is to assume the truth of the entropy statement n
dent trials is then (1 p) ; the probability of observing it at
and derive the Kelvin–Planck statement (or the Clausius state- n
least once in n independent trials is 1 (1 p) . (a) Use these
ment, which is equivalent to the Kelvin–Planck statement) from
ideas to verify the calculation of Eq. (3.58). (b) How many
the entropy statement.]
times must a coin be tossed to reach a 99% probability of
observing at least one head?
Section 3.6
3.24 Willard Rumpson (in later life Baron Melvin, K.C.B.) General
defined a temperature scale with the function f in (3.43) as 3.28 For each of the following sets of quantities, all the quan-
“take the square root” and with the water triple-point tempera- tities except one have something in common. State what they
ture defined as 200.00°M. (a) What is the temperature of the have in common and state which quantity does not belong with
steam point on the Melvin scale? (b) What is the temperature of the others. (In some cases, more than one answer for the prop-
the ice point on the Melvin scale?
erty in common might be possible.) (a) H, U, q, S, T; (b) T, S,
3.25 Let the Carnot-cycle reversible heat engine A absorb q, w, H; (c) q, w, U, U, V, H; (d) r, S , M, V; (e) H, S,
m
heat q per cycle from a reservoir at t and discard heat q 2A dV, P; ( f) U, V, H, S, T.
3
3