EBK THERMODYNAMICS: AN ENGINEERING APPR
8th Edition
ISBN: 8220102809444
Author: CENGEL
Publisher: YUZU
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Textbook Question
Chapter 12.6, Problem 104FEP
For a gas whose equation of state is P(v − b) = RT, the specified heat difference cp − cv is equal to
- (a) R
- (b) R − b
- (c) R + b
- (d) 0
- (e) R(1 + v/b)
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p/atm
2.0
1.0
EXERCISE 3
(c)
3
1
thermal
(b)
Ⓡ
One mole of an ideal gas undergoes a
cyclic three-step process according to
Figure 2:
(a) 1->2: reversible and adiabatic
expansion starting at T₁, and pressure
= 2 atm.
pl
(b) 2->3: reversible isothermal
compression up to p3 = 1 atm and
(c) 3->1: heating under constant
volume, at final temperature Ti, and
pressure p1 = 2 atm.
Calculate the entropy changes of the
gas for the three steps and for the
cycle. The heat capacity under
constant volume is given: cx= 12.47 J
mol-¹ K-¹.
V
Enter your results (with units) in the Table below
AS₁1+2
AS2+3
AS3+1
AS1+2+3+1
The van der Wails equation of state is p ± (v-b = RT,
where p is pressure, v is specific volume, T is
temperature and R is characteristic gas constant The SI
unit of a is
(b)
A fixed mass of the ideal gas argon initially occupies a volume of 3 litres at a
pressure of 3 MPa. An isometric process, followed by an isentropic process, brings
the gas to an equilibrium state at 400°C, 4 MPa and 4 litres volume. An isobaric
process finally brings the gas back to a different equilibrium state at the initial
temperature. Determine:
(i)
the mass of gas in the closed system.
(ii)
the temperature, volume, and pressure of the gas, at each of the four
states.
(iii) the overall change in entropy between the initial and final states.
(iv) the work transfer in each of the three processes and the overall work
transfer.
(Argon: CP= 520.3 Jkg ¹K-¹, R= 208.1 Jkg-¹K-¹)
Chapter 12 Solutions
EBK THERMODYNAMICS: AN ENGINEERING APPR
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