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Unformatted text preview: ly, the temperature remains constant and
so does the internal energy of the gas. Therefore, ∆U = 0 J. To evaluate W we use
V W = nRT ln f (Equation 15.3), where Vf and Vi are the final and initial volumes of the
V i
gas, respectively.
SOLUTION According to the first law of thermodynamics, as given in Equation 15.1, the
heat added to the gas is
Q = ∆U + W Using the fact that ∆U = 0 J for an ideal gas undergoing an isothermal process and the fact
V that W = nRT ln f (Equation 15.3), we can rewrite the expression for the heat as
V i
follows:
V
Q = ∆U + W = W = nRT ln f
V
i Since the volume of the gas doubles, we know that Vf = 2 Vi. Thus, it follows that
V
Q = nRT ln f
V
i 2 Vi = ( 2.5 mol ) 8.31 J/ ( mol ⋅ K ) ( 430 K ) ln Vi = 6200 J 778 THERMODYNAMICS 27. REASONING During an adiabatic process, no heat flows into or out of the gas (Q = 0 J).
For an ideas gas, the final pressure and volume (Pf and Vf) are related to the initial pressure
and volume (Pi and Vi) by PViγ = Pf Vfγ
i ( Equation 15.5 ) , where γ is the ratio of the specific heat capacities at constant pressure and constant volume (γ = 7
5 in this problem). The initial and final pressures are not given. However, the initial and final temperatures are known, so
we can use the ideal gas law, PV = nRT (Equation 14.1) to relate the temperatures to the
pressures. We will then be able to find Vi/Vf in terms of the initial and final temperatures.
SOLUTION Substituting the ideal gas law, PV = nRT, into PViγ = Pf Vfγ gives
i nRTi V
i γ nRTf Vi = V f γ Vf Ti Viγ −1 = Tf Vfγ −1 or Solving this expression for the ratio of the initial volume to the final volume yields
1 Vi Tf
=
Vf Ti γ −1 The initial and final Kelvin temperatures are Ti = (21 °C + 273) = 294 K and Tf = (688 °C + 273) = 961 K. The ratio of the volumes is
1 T
= f
Vf Ti Vi 1 γ −1 961 K ( 7 − 1)
5
= 19.3 = 294 K Chapter 15 Problems 779 28. REASONING The first law of thermodynamics states that due to heat Q and work W, the
internal energy of the system changes by an amount ∆U according to ∆U = Q − W
(Equation 15.1). This law will enable us to find the various quantities for the three
processes.
B C Pressure Process A→B There is no work done
for the process A→B. The reason is that
the volume is constant (see the
drawing), which means that the change
∆V in the volume is zero. Thus, the area
under the plot of pressure versus volume
is zero, and WA→B = 0 J. Since Q is
known, the first law can then be used to
find ∆U. A
Volume Process B→C Since the change ∆UB→C in the internal energy of the gas and the work
WB→C are known for the process B→C, the heat QB→C can be determined directly by using the first law of thermodynamics: QB→C = ∆UB→C + WB→C (Equation 15.1). Process C→A For the process C→A it is possible to find the change in the internal energy
of the gas once the changes in the internal energy for the processes A→B and B→C are
known. The total change ∆Utotal in the internal energy for the three processes is ∆Utotal = ∆UA→B + ∆UB→C + UC→A, and we can use this equation to find ∆UC→A, with the aid of values for ∆UA→B and ∆UB→C. This is possible because ∆Utotal is the change in the
internal energy for the total process A→B→C→A. This process begins and ends at the same
place on the pressureversusvolume plot. Therefore, the value of the internal energy U is
the same at the start and the end, with the result that ∆Utotal = 0 J.
SOLUTION
3
a. Since the change in volume is zero (∆V = 0 m ), the area under the plot of pressure versus
volume is zero, with the result that the work is WA→B = 0 J .
b. The change in the internal energy of the gas for the process A→B is found from the first
law of thermodynamics:
∆UA→B = QA→B − WA→B = +561 J − 0 J = +561 J
c. According to the first law of thermodynamics, the change in the internal energy of the gas
for the process B→C is ∆UB→C = QB→C − WB→C. Thus, the heat for this process is QB→C = ∆UB→C + WB→C = +4303 J + 3740 J = +8043 J 780 THERMODYNAMICS Since this heat is positive, it is added to the gas.
d. The change ∆Utotal in the total internal energy for the three processes is ∆Utotal =
∆UA→B + ∆UB→C + ∆UC→A. Solving this equation for ∆UC→A gives ∆UC→A = ∆Utotal − ∆UA→B − ∆UB→C. As discussed in the REASONING (Process C→A), ∆Utotal = 0 J, since
the third process ends at point A, which is the start of the first process. Therefore,
∆UC→A = 0 J − 561 J − 4303 J = −4864 J e. According to the first law of thermodynamics, the change in the internal energy of the gas
for the process C→A is ∆UC→A = QC→A − WC→A. The heat for this process is QC→A = ∆UC→A + WC→A = −4864 J + (−2867 J) = −7731 J
Since this heat is negative, it is removed from the gas. 29. SSM REASONING AND SOLUTION
Step A → B
The internal energy of a monatomic ideal gas is U = (3/2)nRT. Thus, the change is ∆U = 3 nR ∆T =
2 3
2 (1.00 mol ) 8.31 J/ ( mol ⋅ K ) ( 800.0 K – 400.0 K ) = 4990 J The work for this constant...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
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