Modern Engineering Thermodynamics

54 CHAPTER 2: Thermodynamic Concepts
25. The equilibrium state of carbon at atmospheric pressure and
temperature is graphite. If diamond is the equilibrium form of
carbon only at very high pressures and temperatures, then why
does diamond exist at atmospheric pressure and temperature?
26. Water in equilibrium at 70.0°F and 14.7 psia is in the liquid
phase. If solid ice is in an equilibrium phase only at 32.0°F or
lower at atmospheric pressure, why does solid ice still exist
when you take it from the freezer and put it on the table at
room temperature?
27. Sketch the following process paths on p–v coordinates starting
from state (p 1 , v 1 ).
a. A constant pressure (isobaric) expansion from (p 1 , v 1 )to
(p 2 , v 2 ), where v 2 = 2v 1 .
b. A constant volume (isochoric) compression from (p 2 , v 2 )to
(p 3 , v 3 ), where p 3 = 2p 2 .
c. A process described by p = p 1 + k(v–v 1 ), where k is a constant,
from (p 3 , v 3 ) back to (p 1 , v 1 ) again.
28. Sketch the following thermodynamic cycle on p − V coordinates
for a substance obeying the ideal gas equation of state, pV = mRT:
a. An isothermal compression (i.e., decreasing volume) from
p 1 , V 1
to p 2 , V 2
:
b. An isobaric (i.e., constant pressure) expansion (i.e.,
increasing volume) from p 2 , V 2
to p 3 , V 3
:
c. An isothermal expansion from p 3 , V 3
to p 4 , V 4
:
d. An isochoric (i.e., constant volume) depressurization from
p 4 , V 4
to p 5 , V 5
:
e. An isobaric compression from p 5 , V 5
back to p 1 , V 1
:
29. A new thermodynamic cycle for an ideal gas is described by the
following processes:
a. An isothermal compression from p 1, V 1
to p 2 , V 2
:
b. An isochoric compression from p 2 , V 2
to p 3 , V 3
:
c. An isobaric expansion from p 3 , V 3
to p 4 , V 4
:
d. An isothermal expansion from p 4 , V 4
to p 5 , V 5
:
e. An isochoric decompression from p 5 , V 5
back to p 1 , V 1
:
Sketch this cycle on pressure-volume coordinates.
30. Convert (a) 20.0°C into R, (b) 1.00°C into °F, (c) 56.0°F
into °C, (d) 253°C into K, and (e) 1892°F into R.
31. Convert (a) 32.0°F into °C, (b) 500. R into °F, (c) 373 K
into °C, (d) 20.0°C into R, and (e) −155°F into K.
32. Convert (a) 12.0°C into °F, (b) 6500. K into °C, (c) 1500. R
into °F, (d) 120.°F into K, and (e) −135°C into K.
33. Convert (a) 8900. R into K, (b) −50.0°C into °F, (c) 3.00 K into
°C, (d) 220.°C into R, and (e) 1.00 × 10 6 °F into °C.
34. Convert the following temperatures into kelvin:(a) 70.0°F,
(b) 70.0°C, (c) 70.0 R, and (d) 70.0° Reaumur. The Reaumur
temperature scale was developed in 1730 by the French scientist
René Antoine Ferchault de Réaumur (1683–1757). The freezing
and boiling points of water at atmospheric pressure are defined
to be 0° and 80.0° Reaumur, respectively.
35. Many historians believe that Gabriel Daniel Fahrenheit (1686–
1736) had established his well-known temperature scale by
1724. It was based on three easily measured fixed points: the
freezing temperature of a mixture of water and ammonium
chloride (0.00°F), the freezing point of pure water (32.0°F), and
the temperature of the human body (96.0°F). Later this scale
was changed to read 212°F at the boiling point of pure water,
which moved the body temperature from 96.0 to 98.6°F. Using
the original Fahrenheit scale (freezing point of water = 32.0°F
and body temperature = 96.0°F), determine
a. The temperature of the boiling point of pure water.
b. The conversion formula between the original Fahrenheit and
the modem Celsius temperature scales.
36.* Convert the following pressures into the proper SI units:
a. 14.7 psia.
b. 5.00 atmospheres absolute.
c. 1.00 × 10 5 dynes/cm 2 absolute.
d. 30.0 lbf/ft 2 gauge.
e. 12.4 poundals/ft 2 absolute.
37. Convert the following pressures into psia:
a. 1000. N/m 2 gauge.
b. 3.00 MPa-absolute.
c. 11.0 Pa-gauge.
d. 20.3 kN/m 2 absolute.
e. 556 GPa-absolute.
38. Will the continuum hypothesis hold for the following
thermodynamic states (and why)?
a. Air at 20.0°C and atmospheric pressure.
b. Liquid water at 70.0°F and 14.7 psia.
c. Steam at 1.00 psia and 100.°F.
d. Steam at 1.00 MPa absolute and 100.C.
e. Air at 1.00 µN/m 2 absolute and 10.0 K.
39.* Convert the following pressures into MPa-absolute:
a. 100. psig.
b. 2,000. kPa-absolute.
c. 14.7 psia.
d. 1.00 Pa gauge.
e. 500. N/m 2 absolute.
40. Convert the following pressures into lbf/ft 2 absolute:
a. 14.7 psia.
b. 100. lbf/ft 2 gauge.
c. 0.200 MPa-absolute.
d. 1200. kPa-gauge.
e. 1500. psig.
41.* Convert the following pressures into N/m 2 absolute:
a. 0.100 MPa-absolute.
b. 14.7 psia.
c. 25.0 psig.
d. 100. Pa-absolute.
e. 100. Pa-gauge.
42. In the late 18th century, it was commonly believed that heat
was some kind of colorless, odorless, weightless fluid. Today we
know that heat is not a fluid (it is primarily an energy transport
due to a temperature difference), but we still have many old
phrases and terms in our everyday and technical language that
imply that heat is a fluid (e.g., heat “pours” out of a hot stove
or heat always “flows” down a temperature gradient and so on).
Discuss whether or not heat can be generated or absorbed, and
using the balance concept, discuss whether or not it is a
conserved quantity.
43. In Example 2.2, the Malthus population law was evaluated and
found to produce an exponential growth or decay in the size of
the population. A more sophisticated population model includes
the effects of birth and death rates that vary linearly with the
instantaneous size of the population as
Birthrate = α 1 − β 1 N