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Unformatted text preview: HW5_Chapter 20:
205:
Q. In this problem, we will prove that Equation 20.5 is valid for an arbitrary system. To do this,
consider an isolated system made up of two equilibrium subsystems, A and B, which are in
thermal contact with each other; in other words, they can exchange energy as heat between
themselves. Let subsystem A be an ideal gas and let subsystem B be arbitrary. Suppose now
that an infinitesimal reversible process occurs in A accompanied by an exchange of energy as
heat
(ideal). Simutaneously, another infinitesimal reversible process takes place in B
accompanied by an exchange of energy as heat
(arbitrary). Because the composite
system is isolated, the First Law requires that Now use Equation 20.4 to prove that A. W e use the First Law as suggested in the problem and substitute Equation 20.1 for δqrev
(ideal) to write Then
is the derivative of a state function. We know that the cyclic integral of
a state function is equal to 0. Therefore, we can write and Equation 20.4 holds for any system. 206:
Q. Calculate
and
for a reversible cooling of one mole of an ideal gas at a constant
volume V1 from P1, V1, T1 to P2, V1, T4 followed by a reversible expansion at constant pressure
P2 from P2, V1, T4 to P2, V2, T1 (the final state for all the processes showed in Figure 20.3).
Compare your result for
with those for paths A, B+B, and D+E in Figure 20.3. A. Step 1. P1, V1, T1 P2, V1, T4
Because there is no change in the volume of the ideal gas, , and we can write Step 2. P2, V1, T4 P2, V2, T1
In this case, we write (by the First Law) For the entire process, P1, V1, T1 P2, V2, T1, we have The value of
differs from those found for paths A, B+C, and D+E (section 203), but the
value of
is the same (because entropy is a pathindependent function). 209:
Q. Calculate the value of
if one mole of an ideal gas is expanded reversibly and isothermally
from 1.00 bar to 0.100 bar. Explain the sign of .
A. Because the reaction is isothermal, . For an ideal gas, So we write The value of as is positive because the gas expands. 2014:
Q. Show that If one mole of an ideal gas is taken from T1, V1 to T2, V2, assuming that
is independent of
temperature. Calculate the value of
if one mole of N2(g) is expanded from 20.0 dm3 at 273 K
to 300 dm3 at 400 K. Take
= 29.4 J∙K1∙mol1.
A. For the path (T1, V1) (T2, V2),
write
as Because For N2, and for an ideal gas, we can write this as . W e can then 2016:
Q. In this problem, we will illustrate the condition
with a concrete example. Consider
the twocomponent system in Figure 20.8. Each compartment is in equilibrium with a heat
reservoir at different temperatures T1 and T2, and the two compartments are separated by a
rigid heatconducting wall. The total change of energy as heat of compartment 1 is where
is the energy as heat exchanged with the reservoir and
exchanged with compartment 2. Similarly, is the energy as heat Clearly, Show that the entropy change for the twocompartment system is given by where
is the entropy exchanged with the reservoirs (surroundings) and is the entropy produced within the twocompartment system. Now show that the condition
implies that energy as heat flow spontaneously from a higher temperature to a lower
temperature. The value of
, however, has no restriction and can be positive, negative
or zero. Figure 20.8. A twocompartment system with each compartment in contact with an (essentially
infinite) heat reservoir, one at temperature T1 and the other at temperature T2. The two
compartments are separated by a rigid heatconducting wall.
A. As stated in the text of the problem, we can write The energy as heat exchanged between compartments 1 and 2 is involved in the entropy
transferred between the two compartments. We can therefore express Similarly, the energy as heat exchanged between the compartments and reservoirs is involved
in the entropy produced within the twocompartment system, so we can write
as These are the only two means of changing the entropy of the system, so we can find
be Now take the condition . This is the same as saying that to Arbitrarily, let , then , so and heat is flowing from compartment 1 to compartment 2. If
, by the same reasoning
compartment 2 to compartment 1. and heat is flowing from 2018:
Q. Vaporization at the normal boiling point (
) of a substance (the boiling point at one atm)
can be regarded as a reversible process because if the temperature is decreased infinitesimal
below
, all the vapor will condensed to liquid, whereas if it is increased infinitesimal above
, all the liquid will vaporize. Calculate the entropy change when two moles of water vaporize
at 100.0 . The value of
is 40.65 kJ∙mol1. Comment on the sign of
.
A. At constant pressure and temperature,
isothermal process, (Equation 19.37). And for a reversible As the water becomes more disordered, changing from liquid to vapor, the entropy increases. 2025:
Q. Calculate the change in entropy of the system and of the surroundings and the total change
in entropy if one mole of an ideal gas is expanded isothermally and reversibly from a pressure of
10.0 bar to 2.00 bar at 300 K.
A. Because this is an isothermal reversible expansion,
gas equation to write For an isothermal expansion of an ideal gas,
entropy of the gas as For a reversible expansion, , so . W e then use the ideal . We can then write the change of . ...
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This note was uploaded on 11/08/2011 for the course CH 353 taught by Professor Bersuker during the Fall '08 term at University of Texas.
 Fall '08
 Bersuker
 Physical chemistry, Equilibrium, pH

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