Chapter 1
1Consider a system whose temperature is 18 C. Express this temperature in R, K, and F.
Given: T= 18°C
Find:
T(°R)
T(°K)
T(°F)
Solution:
•
T(K) = 18 + 273.15 = 291.15 °K
•
T(°F) = 18*1.8 + 32 = 64.4°F
•
T(°R) = 18 + 459.67 = 524.07 °R
Alternatively, we could use:
•
T(°R)= 1.8 * 291.15 =524.07 °R
-------------------------------------------------------------------------------------------------------------------------------------2) The temperature of a system drops by 45°F during a cooling process. Express this drop-in temperature
in K, R, and °C.
Solution:
T(°C) =( T(°F) - 32 )/1.8 =(45 - 32)/ (1.8) =7.222°C
T(K) = T(°C) + 273
= 7.22°C + 273 =280.222 K
T(R) = T(°F) + 460
= 45 + 460 =505 R
3) A vacuum gage connected to a chamber reads 45 kPa at a location where the atmospheric pressure is 92
kPa. Determine the absolute pressure in the chamber.
Solution:
Pabs =Patm - Pvacu
Pabs = 92-45 =47 kpa
4)
The absolute pressure in water at a depth of 5 m is read to be 145 kPa. Determine
(a) the local atmospheric pressure, and
(b) the absolute pressure at a depth of 5 m in a liquid whose specific gravity is 0.85 at the same
location.
Solution:
5)
The basic barometer can be used to measure the height of a building. If the barometric readings at the
top and at the bottom of a building are 730 and 755 mm Hg, respectively, determine the height of the
building. Take the densities of air and mercury to be 1.18 kg/m3 and13,600 kg/m3, respectively
Solution:
6)
A gas contained in a vertical frictionless piston-cylinder device. The piston has a mass of 4kg and a crosssectional area of 35 cm2. A compressed spring above the piston exerts a force of 60N
on the piston. If the atmospheric pressure is 95kPa, determine the pressure inside the
cylinder.
Solution:
7)
A vertical piston–cylinder device contains a gas at a pressure of 100 kPa. The piston
has a mass of 5 kg and a diameter of 12 cm. Pressure of the gas is to be increased by
placing some weights on the piston. Determine the local atmospheric pressure and
the mass of the weights that will double the pressure of the gas inside the cylinder.
Solution:
---------------------------------------------------------------------------------------------------------------------------------------------------------------
8)
The lower half of a 10-m-high cylindrical container is filled with water ( ƿ= 1000 kg/m3) and the upper
half with oil that has a specific gravity of 0.85. Determine the pressure difference between the top and
bottom of the cylinder.
Solution :
Chapter 2
1) In what forms can energy cross the boundaries of a closed system?
• Energy can cross the boundaries of a closed system in the form of heat or work. Energy
transfer across a system boundary due solely to the temperature difference between a system
and its surroundings is called heat.
-------------------------------------------------------------------------------------------------------------2)
A car is accelerated from rest to 85 km/h in 10 s. Would the energy transferred to the car
be different if it were accelerated to the same speed in 5 seconds?
Solution:
• We need to know the mass of the car. Let’s assume that to be 800 kg. We’ll first consider
the energy required to reach 85 km/h
The energy needed is the difference in kinetic energy for the car at rest and at 85 km/h. The
form of energy delivered to accelerate the car is work.
Note that the time for acceleration did not enter into this calculation. When time comes into the
picture, we are dealing with a work rate or power calculation. Consider the following analysis
for acceleration to speed in 10 s & 5 s
3)
Determine the energy required to accelerate a 1300-kgcar from 10 to 60 km/h on an uphill road
with a vertical rise of 40 m
Solution:
4)
Determine the power required for a 2000-kg car to climb a 100-m-long uphill road with a slope
of 30◦ (from horizontal, as shown in Figure 1) in 10s (1) at a constant velocity, (2) from rest to a
final velocity of 20m/s, and (3) from 25m/s to a final velocity of 5m/s. Disregard friction, air
drag, and rolling resistance.
Solution:
5)
A frictionless piston–cylinder device contains 2 kg of nitrogen at 100 kPa and 300 K. Nitrogen is now
compressed slowly according to the relation PV1.4 = constant until it reaches a final temperature of 360 K.
Calculate the work done during this process.
In this case, the path equation is PVn = C with n = 1.4. We can rewrite this equation as P = CV-n,
and integrate this revised path equation to find the work as follows.
W 
V2
 PdV   CV
path
n
dV 
V1


V2
C
C

V n 1 V1 
V21n  V11n 
 n 1
1 n
To reduce this equation further we can use a trick. Since PV n = C, the constant C may be written
either as P1V1n or as P2V2n. By carefully choosing which way we write this constant, we can simplify our result.
W
CV21n  CV11n P2V2nV21n  P1V1nV11n P2V2  P1V1
C

V21n  V11n  


1 n
1 n
1 n
1 n
The equation above is valid for any substance. For an ideal gas, we can obtain a simpler result for the work by using the
ideal gas equation of state, writing each PkVk product as mRTk.
W 
P2V2  P1V1 mRT2  mRT1 mR (T2  T1 )


1 n
1 n
1 n
We can find the gas constant for nitrogen from Table A-1 on page 908; R = 0.2968 kJ/(kg•K). We are given m = 2 kg, T1 =
300 K, T2 = 360 K and n = 1.4. Substituting these values into the equation for the work gives the answer as follows.
W
mR (T2  T1 )

1 n
( 2 kg )
0.2968 kJ
(360 K  300 K )
kg  K
= –89.0 kJ
1  1.4
The negative sign for the work indicates that this is a work input of 89.0 kJ
6)
A mass of 2.4 kg of air at 150 kPa and 12oC is contained in a gas-tight, frictionless piston-cylinder device.
The air is now compressed to a final pressure of 600 kPa. During the process, heat is transferred from
the air such that the temperature inside the cylinder remains constant. Calculate the work input during
this process
Solution:
In this case the path is a constant temperature path. If we assume that air is an ideal gas, we can write the ideal-gas
equation of state, at a constant temperature, as follows:
P
mRT
V
Since mRT is a constant in a constant temperature process, we can write the equation for work as follows.
V2
W 
 PdV  
path
V1
mRT
dV  mRT
V
V2
dV
V
V1
V
V
 mRT ln(V )V2  mRT ln 2
1
 V1
We can find the ratio, V2/V1 as follows for the constant temperature process.



mRT
V2
P
P
 2  1
V1 mRT P2
P1
Substituting P1/P2 in place of V2/V1, using the gas constant for air = 0.2870 kJ/(kg•K) from Table A-1 on page 908, and
using the given values of m = 2.4 kg, and T = 12oC = 285.15 K, gives the following result for the work.
P 
 150 kPa 
0.2870 kJ
  –272 kJ
W  mRT ln 1   ( 2.4 kg )
( 285.15 K ) ln
kg  K
 600 kPa 
 P2 
The negative sign for the work indicates that this is a work input of 272 kJ
--------------------------------------------------------------------------------------------------------------------------------------7)
Nitrogen at an initial state of 300K, 150KPA, and 0.2 m3 cubed is compressed slowly in an isothermal
process to a final pressure of 800KPA. What is the amount of work done.
Solution:
P1V1 = P2V2
(150 kPa) (0.2 m3) = (800 kPa) V2
V2 = 0.0375 m3
W = nRT ln(V2/V1)
nRT = P1V1= (150) (0.2) = 30
W = 30 ln (0.0375/0.2) = -50.2 kJ
sign indicates that the work is done on the system.
---------------------------------------------------------------------------------------------------------------------------------------
8)
A gas is compressed from an initial volume of 0.42 m3 to a final volume of 0.12 m3. During the
quasi-equilibrium process, the pressure changes with volume according to the relation P = aV + b,
where a = -1200kPa/m3 and b = 600 kPa. Calculate the work done during this process (a) by
plotting the process on a P–V diagram and finding the area under the process curve and (b) by
performing the necessary integrations
Solution:
10)
12)
A piston -cylinder device contains 0.5 m3 helium gas at 150 kpa and 20°c. the helium is now
compressed in polytropic process to 400 kpa and 140°c. determine the work done during this
process
Solution:
Helium is an ideal gas with constant specific heats. 2 The cylinder is stationary and thus the
kinetic and potential energy changes are negligible. 3 The thermal energy stored in the cylinder
itself is
negligible. 4 The compression or expansion process is quasi-equilibrium.
Properties The gas constant of helium is R = 2.0769 kPa.m3/kg.K (Table A-1). Also, cv =
3.1156 kJ/kg.K
13)
14
15)
A 4-m x 3 5-m x 3 6-m room is to be heated by a baseboard resistance heater. It is desired
that the resistance heater be able to raise the air temperature in the room from 5 to 258C
within 11 min. Assuming no heat losses from the room and an atmospheric pressure of 100
kPa,
determine the required power of the resistance heater. Assume constant specific heats at
room temperature
Solution:
16)
17)
An insulated piston–cylinder device contains 100 L of air at 400 kPa and 25°C. A paddle
wheel within the cylinder is rotated until 15 kJ of work is done on the air while the
pressure is held constant.
Determine the final temperature of the air.
Solution:
19)
A mass of 15 kg of air in a piston-cylinder device is heated from 25 to 77oC by passing current through
a resistance heater inside the cylinder. The pressure inside the cylinder is held constant at 300 kPa
during the process and a heat loss of 60 kJ occurs.
Determine the electrical energy supply in kWh.
Solution:
The first law for this problem can be written as Q = U + W = U + W e,out + PdV,
where W e,out is the electrical work done by the system.
We expect that this will be negative since we see that there is an electrical energy input to the
system.
We can say that W e,in = -W e,out = U + PdV – Q.
For this constant pressure process, PdV = P1-2 (V2 – V1) where P1-2 = P1 = P2 is the constant
pressure of 300 kPa.
With this work term we can write our first law as follows.
We, in = U + P1-2 (V2 – V1) – Q = U2 – U1 + P2V2 – P1V1 – Q = H2 – H1 – Q = m (h2 – h1) – Q
We can find the specific enthalpy change, h2 – h1 = cpdT,
At the average temperature of (25oC + 77oC)/2 = 51oC = 324.15 K,
we can interpolate to find cp = 1.0065 kJ/(kg∙K).
With this value of cp (and using Q = –60 kJ because it is a heat loss)
we can find the electrical work input as follows.
We,in  h2  h1  Q  mc p (T2  T1 )  Q 

 1 kWh
 1.0065 kJ 


(
15
kg
)
350
.
15
K

298
.
15
K

(

60
kJ
)




 kg  K 



 3600 kJ
We,in = 0.235 kWh .