Download chapter 21 heat, work, and the first law of

CHAPTER 21
HEAT, WORK, AND THE FIRST LAW
OF THERMODYNAMICS
ActivPhysics can help with these problems:
Activities 8.5-8.13
Section 21-1: The First Law of
Thermodynamics
Problem
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Section 21-2:
Problem
Thermodynamic Processes
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8. An ideal gas expands from the state (PI, Vl) to the
state ( 9 , Vz), where P2= 2P1 and V2 = 2V1. The
expansion proceeds along the straight diagonal path
AB shown in Fig. 21-26. Find an expression for the
work done by the gas during this process.
1. In a perfectly insulated container, 1.0 kg of water
is stirred vigorously until its temperature rises by
7.0°C. How much work was done on the water?
Solution
Since the container is perfectly insulated thermally, no
heat enters or leaves the water in it. Thus, Q = 0 in
Equation 21-1. The change in the internal energy of
the water is determined from its temperature rise,
AU = rnc AT (see comments in Section 19-4 on
internal energy), so W = -AW = -(1 kg) x
(4.184 kJ/kg.K)(7 K) = -29.3 kJ. (The negative sign
signifies that work was done on the water.)
Problem
FIGURE
5. The most efficient large-scale electric power
generating systems use high-temperature gas
turbines and a so-called combined cycle system that
maximizes the conversion of thermal energy into
useful work. One such plant produces electrical
energy at the rate of 360 hlW, while extracting
energy-from its natural gas fuel at the rate of
670 MW. (a) At what rate does it reject waste heat
to the environment? (b) Find its efficiency, defined
as the percent of the total energy extracted from
the fuel that ends up as work.
21-26 Problems 8 and 9.
Solution
The work done by the gas equals the area under the
straight diagonal path AB in Fig. 21-26. The area of
this trapezoid is W = +(PI P2)(V2 - Vl ) = +(PI
2Pl) ( 2 4 - Vl) = PIv]. W can also be obtained from
Equation 21-3. On the path AB, P = PI (V - Vl)x
(P2 - PI)/(% - Vl). Then
2
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Solution
(a) If we assume that the generating system operates
in a cycle and choose it as "the system," then dli/dt =
0 and Equation 21-2 implies dQ/dt = dW/dt. Here,
dW/dt is the rate that the generator supplies energy to
its surroundings (360 h/lW in this problem) and dQ/dt
is the net rate of heat flow into the generator from the
surroundings. Since the system is just the generator,
the net heat flow is the difference between the heat
extracted from its fuel and the heat exhausted to the
environment, i.e., dQ/dt = (dQ/dt)i, - (dQ/dt)our = 670 MW - (dQ/dt),,,. Therefore, (dQ/dt),,, =
670 MW - 360 MW = 310 MW. (Note: If the system
is assumed to be the generator and its fuel, as in
Example 21-1, then dW/dt is still 360 MW, but the
system's internal energy decreases because energy is
extracted from the fuel, dU/dt = -670 MW, and there
is no heat input. Then dQ/dt = -670 MW
360 MW = -310 MW, representing the rate of heat
rejected to the environment.) (b) The efficiency is
(dW/dt)/(dQ/dt)i, = 360 MW/670 MW = 53.7% (see
Section 22-2).
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Problem
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11. A balloon contains 0.30 mot of helium. It rises,
while maintaining a constant 300 K temperature,
to an altitude where its volume has expanded
5 times. How much work is done by the gas in the
balloon during this isothermal expansion? Neglect
tension forces in the balloon.
Solution
During an isothermal expansion, the work done by a
given amount of ideal gas is W = nR'TIn(h/V1) =
(0.3 mo1)(8.314 J/mol.K)(300 K)ln(5) = 1.20 kJ (see
Equation 21-4).
Problem
19. A gas with y = 1.4 is at 100 kPa pressure and
occupies 5.00 L. (a) How much work does it take
to compress the gas adiabatically to 2.50 L?
(b) What is its final pressure?
Solution
The work done by an ideal gas undergoing an
adiabatic process is W12 = ( P I K - P2V2)/(y - 1) (see
Equation 21-14). Since the compression is specified by
given values of PI, Vl, and V2, we first find the final
pressure from the adiabatic gas law. (b) P2 = Pl (Vl+
V2l7 = (100 kPa)(5 L12.5 L)1.4 = 264 kPa. (a) Then
the work done on the gas (which is -W12) is -W12 =
(P2V2-PlVl)/(y-l) = [(264 kPa)(2.5 L)-(100 kPa)x
(5 L)]/0.4 = 399 J.
25:By how much must the volume of a gas with
7 = 1.4 be changed in an adiabatic process if the
kelvin temperature is to double?
Solution
V/G = (T~/T)'/(Y-') = ( 0 . 5 ) ' / ~ =
. ~ 0.177
(Equation 21-13b).
Problem
31. An ideal gas with y = 1.67 starts a t point A in
Fig. 21-29, where its volume and pressure are
1.00 m3 and 250 kPa, respectively. It then
undergoes an adiabatic expansion that triples its
volume, ending at point B. It's then heated a t
constant volume to point C, then compressed
isothermally back to A. Find (a) the pressure a t
B, (b) the pressure at C, and (c) the net work
done on the gas.
Problem
.
37. A bicycle pump consists of a cylinder 30 cm long
when the pump handle is all the way out. The
pump contains air (7 = 1.4) at 20°C. If the pump
outlet is blocked and the handle pushed until the
internal length of the pump cylinder is 17 cm, by
how much does the air temperature rise? Assume
that no heat is lost.
.
Solution
If no heat is lost (or gained) by the gas, the
compression is adiabatic and Equation 21-13b gives
TV7-' = ~ ~ ~ o 7Therefore,
-l.
the temperature rise is
T - To = A T = To[(Vo/V)7-I - 1). Since Vo/V =
(30 cm/17 cm), AT = [(30/17)O.~
- 1](293 K) =
74.7 CO.
Problem
43. A mixture of monatomic and diatomic gases has
specific heat ratio y = 1.52. What fraction of the
molecules are monatomic?
Solution
The internal energy of a mixture of two ideal gases is
U = flN E ~ fiNE2, where f i is the fraction of the
total number of molecules, N, of type 1, and El is
the average energy of a molecule of type 1, etc.
Classically, E = g(;kT), where g is the number of
degrees of freedom. The molar specific heat at constant
volume is Cv = ( i ) ( d ~ / d T =
) (N~/N)dldTx
(flNgl;kT + f2Ng24kT) = @(fig1 + f2g2). Suppose
that the temperature range is such that 91 = 3 for the
monatomic gas, and 92 = 5 for the diatomic gas, as
discussed in Section 21-3. Then Cv = R(1.5fl
2.5 f2) = R(2.5 - fl), where f2 = 1 - f l since the sum
of the fractions of the mixture is one. Now, Cv can
also be specified by the ratio 7 = Cp/Cv = 1 R/CV,
or Cv = R/(y - I), so in this problem, 2.5 - fl =
110.52, o r 5 = 57.7%.
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Problem
45. A gas mixture contains monatomic argon and
diatomic oxygen. An adiabatic expansion that
doubles its volume results in the pressure dropping
to one-third of its original value. What fraction of
the molecules are argon?
FIGURE 21-29 Problem 31.
Solution
(a) From the adiabatic law for an ideal gas
(Equation 21-13a), PB= PA(v~/VB)7= (250 kPa) x
= 39.9 kPa. (b) Point C lies on an isotherm
with A, so the ideal gas law (Equation 20-2) yields
PC= PAVA/VC= (250 kPa)(i) = 83.3 kPa.
(c) Wnet = WAB WBC WCA. WAB is for an
adiabatic process (Equation 21-14) and equals
(PAVA - P B V B ) / ( ~- 1) = [(250 kPa) (1 m3) (39.9 kPa)(3 m3)]/0.67 = 194 kJ; WBC is for an
isovolumic process and equals zero; WCA is for an
isothermal Process (Equation 21-4) and equals
~ WI nA( v ~ / v c )= PAVA l n ( v ~ / V =
~ )250 kJ In(?) =
-275 kJ. Thus, Wnet = -80.2 kJ. The work done on
the gas is the negative of this.
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Solution
From the pressures and volumes in the described
adiabatic expansion, PoV2 = (iPo)(2Vo)7,we can
calculate that y = In 3/ In 2 = 1.58. Then the result of
Problem 43 gives 2.5 - fA, = 110.58, Or f A r = 79.0%.