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12-1 - Thermo 01 - Notes
Mechanical Engineering (Colegio de San Juan de Letran)
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PSE MODULE 12.1 (SOLUTION)
So,
V2
W=
1. Water enters the heater at 25C and leaves at 80C. What is the
temperature change in F ?
A. 55
B. 99
*
V1
π
W=
C. 21
D. 65

 ( kD )  2 D dD 
s
2
s
→ where: k =
s
V1
V2

π
k ( D s 3 ) dDs =
2
2

V1
Solution :
ΔF =

V2
PdV =
1.3
 120 
P
120
=
Ds
1
  1  ( D )dD = 87.5 kJ
3
s
s
1.0
Ans. C
9
9
( ΔC ) = (80 - 25 ) = 99 F°
5
5
6. Convert specific heat of 0.23 Btu/lb-F to kJ/kg-K.
A. 1.036
B. 0.963*
Ans. B
C. 0.089
D. 0.722
Solution :
2. Find the enthalpy of Helium if its internal energy is 200 kJ/kg.
A. 144 kJ/kg
B. 223.42 kJ/kg
C. 333.42 kJ/kg *
D. 166 kJ/kg
0.23
Btu  4.187 kJ kg-k 
kJ

 = 0.963
lb -°F  1Btu lb-°F 
kg - k
Ans. B
Solution :
k=
ΔH
ΔU
7. An ideal gas with molecular weight of 7.1 kg/kg mol is compressed
ΔH
200
ΔH = 333.4 kJ kg
 1.667 =
from 600 kPa and 280 deg K to a final specific volume of 0.5 m3/kg.
Calculate the work of compression in KJ/kg if the pressure given is
620 + 150V + 95V2?
A.32.8
C.35.6
D.28.7
B.33.6 *
Ans. C
3. A pressure gage registers 50 psig in a region where the barometer is
14.25 psia. Find the absolute pressure in psia.
C 151.325
A. 64.25 *
B. 443
D. 35.75
Solution :
V2
V2

W = PdV =
Solution :
V1
 (620 +150V+ 95V ) dV
2
V1
Determining V1 :
P = Pg + Patm
abs
P = 50 psig + 14.25 psia
abs
P = 64.25 psia
abs
V1 =
RT (8.314 7.1kJ kg-K )( 280 K )
=
= 0.5465 m3 kg
P
600 kPa
0.5
W=
Ans. A
 (620 +150V+ 95V ) dV = − 33.66 kJ kg
2
0.5465
4. A condenser vacuum gauge reads 715 mm Hg when the barometer
stands at 757 mm Hg. What is the absolute pressure in the condenser in
kPa.
A. 196.25 kPa
C. 5.60 kPa *
B. 100.92 kpa
D. 95.33 kPa
Solution :
Ans. B
8. Air is compressed adiabatically from 30°C to 100°C. If mass of air
being compressed is 5 kg. Find the change in entropy.
A. 1.039 kJ/°K
C. 0 *
B. 0.746 kJ/°K
D. 1.245 kJ/kg
Solution :
Adiabatic: Δs = 0
P = Patm + Pvac
abs
 101.325 kPa 
P = ( 757 mmHg - 715 mmHg ) 

abs
 760 mmHg 
P = 5.6 kPaa
abs
Ans. C
9. One kilogram of water (Cp = 4.2 kJ / kg.K) is heated by 300 Btu of
energy. What is the change in temperature, in K ?
A. 17.9 K
C. 73.8 K
B. 71.4 K
D. 75.4 K *
Ans. C
5. An elastic sphere containing gas at 120 kPa has a diameter of 1.0 m.
Heating the sphere causes it to expand to a diameter of 1.3 m. During
the process the pressure is proportional to the sphere diameter.
Calculate the work done by the gas in KJ.
A.41.7
C.87.5 *
B.30.6
D.55.4
Solution :
Q = mcpΔT
300 Btu (1.055kJ Btu ) = (1 kg )( 4.2 kJ kg - k )( ΔT )
ΔT = 75.4 k
Ans. D
Solution :
P Ds
10. If the F scale is twice the C scale, what is the reading in the
 P = kD s
Fahrenheit scale ?
4
4 D  π 3
Also: Vsphere = πr 3 = π  s  = D s
3
3  2  6
π 2
dVsphere = D dD s
2 s
3
A. 160
B. 320 *
C. 140
D. 280
Solution :
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°F = 2°C
Alternate Solution:
9
Note: °F = °C + 32
5
9
2°C = °C + 32
5
°C = 160 °C
Mode 32
x ( °C ) y ( °F )
0
32
100 212
°F = 320 °F
160y = 320 ( Twice )
Ans. B
with helium inside at 101 kPa and 320 K surrounded by air at 101 Kpa
and 298.15 K ?
A. 17.38 kN
C. 18.73 kN
B. 5.28 kN *
D. 8.25 kN
n-1
T2  P2  n
= 
T1  P1 
1.25−1
t 2 + 273  1 2 P1  1.25
=

32 + 273  P1 

kJ   1.4 − 1.25 
Q = (1 kg )  0.7186

 ( −7.482 − 32 ) K= 17.023 kJ
kg-K   1 − 1.25 

Ans. C
15. The following expressions relate to a particular gaseous mass: PV =
Solution :
Fnet = BF − Wg =  air Vballon − mballon g = Vg (  air −  He )
where:
air =
ρ He =
air
R
T
air air
ρ He
R HeTHe
=
101 kPa
= 1.18 kg m3

kJ 
 0.287
 ( 298.15 K )
kg-K 

=
101 kPa
= 0.1519 kg m3
kJ 

 8.314 kg-K 

 ( 320 K )
4






3
 4  D 
Fnet =  π   ( g ) ( ρ air − ρ He )
 3  2 
3
kg 
 4  10 m 
2 
Fnet =  π 
 9.81m s 1.18 − 0.1519 3 
 3  2 
m 

Fnet = 5282.80 N = 5.28 kN
(
)
Ans. B
95T, h = 120 + 0.60T where these units obtain in psf, V in ft3/lb, T in
R and h in Btu/lb. If the specific heats are temperature dependent
only, find Cp and Cv.
A. 0.6 Btu/lbR, 0.48 Btu/lb *
B. 0.60 Btu/lbR, 0.7 Btu/lbR
C. 0.5 Btu/lbR, 0.50 Btu/lbR
D. 0.50 Btu/lbR, 0.48 Btu/lbR
Solution :
Comparing to PV = RT, and h = CpΔT
R= 95 and Cp = 0.60 Btu lb-°F
Thus,
ft-lbf
95
Btu
lbm
-°F
Cv = Cp - R = 0.60
lb -°F 778 ft - lbf
1Btu
Btu
= 0.48
lb -°F
Ans. A
16. Calculate the change in enthalpy as 1 kg of nitrogen is heated from
12. If a block of copper weighing 0.50 lb is dropped from a height of 100
ft into the tank, what is the change in internal energy of the water?
A. 0.08575 Btu
C. 0.09543 Btu
B. 0.0645 Btu *
D. 0.0465 Btu
Solution :
1000 K to 1500 K, assuming the nitrogen is an ideal gas at a constant
pressure. The temperature dependent specific heat of nitrogen is C p =
39.06 – 512.79 T-1.5 + 1072.7 T-2 – 820.4T-3 where Cp is in kJ/kg-mol,
and T is in K.
A. 600 kJ
C. 800 kJ
B. 697.27 kJ *
D. 897.27 kJ
Solution :
mgΔz
k
( 0.50 lbm ) ( 32.2 ft s2 ) (100 ft )
ΔU = ΔPE
where: ΔPE =

lb m -ft  
ft-lbf 
778
 32.2

2 
lb
-s
Btu 



f
T2
= 0.06427 Btu

ΔH = C p dT
T1
1500
ΔH =
Ans. B
 (39.06 - 512.79T + 1072.7 T − 820.4T ) dT
−1.5
−2
−3
1000
13. Convert water pressure of 50 kN/m2 in equivalent meter – head of
water.
A. 5.1 *
B. 6.1
ΔH = 19524.406
kJ  1 kgmol 

 (1kg )
kgmol  28 kg 
ΔH = 697.3 kJ
C. 7.1
D. 4.1
Ans. B
17. For a certain ideal gas, R = 0.277 kJ/kg-K and k = 1.384. What are the
Solution :
values of Cp and Cv ?
A. 0.9984 kJ/kg-K, 0.7213 kJ/kg-K *
B. 0.7124 kJ/kg-K, 0.8124 kJ/kg-K
C. 1 kJ/kg-K , 0.8124 kJ/kg-K
D. 0.9984 kJ/kg-K, 0.6124 kJ/kg-K
2
h=
k − n
where: Cn = Cv 

1 − n 
t 2 = − 7.482 °C
11. What is the lifting force in kN for a 10 m diameter spherical balloon
ΔU =
Q = mCn ΔT
P
50 kN m
=
= 5.1 m
 9.81kN m 2
Ans. A
Solution :
14. Consider 1 kg of air at 32°C that expanded by a reversible polytropic
process with n = 1.25 until the pressure is halved. Determine the heat
transfer. Specific heat at constant volume for air is 0.7186 kJ/kg.K.
A. 17.02 kJ heat rejected
C. 17.02 kJ heat added *
B. 7.07 kJ heat rejected
D. 7.07 kJ heat added
Solution :
1.384 ( 0.277 )
kR
=
= 0.9984 kJ kg-K
k −1
1.384 − 1
R
0.277
Cv =
=
= 0.7214 kJ kg-K
k − 1 1.38401
Cp =
Ans. A
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18. An air with mass of 0.454 kg and an unknown mass of CO2 occupy an
85 liters tank at 2068.44 Kpaa. If the partial pressure of the CO 2 is
344.74 kPaa, determine its mass.
C. 0.183 kg
A. 0.138 kg *
B. 0.238 kg
D. 0.283 kg
Btu 

Δh = CpΔT =  0.2483
 ( 200 − 500 °R )
lbm − °R 

Btu
Δh = − 74.49
lbm
Ans. A
Solution :
23. What is the resulting pressure when one pound of air at 15 psia and
P
n
co2 = co2 → where: n = m
P
n
MW
AIR
AIR
P
m
MW
co2 =
co2
co2
P
m
MW
AIR
AIR
AIR
mCO2 44
344.74
=
2068.44 − 344.74 0.454 kg/29
mCO2 = 0.138 kg
200 F is heated at constant volume to 800F ?
A. 15 psia
C. 36.4 psia
B. 28.6 psia *
D. 52.1 psia
Solution :
P1
P
= 2
T1
T2
15 psia
P2
=
200 + 460 800 + 460
P2 = 28.64 psia
Ans. A
19. After series of state changes, the pressure and volume of 2.286 kg of
Nitrogen (Cp = 1.0414 kJ/kg-K and Cv = 0.7442 kJ/kg-K) are each
doubled. What is S ?
A. 2.807 kJ/kg-K *
C. 2.987 kJ/kg-K
B. 2.268 kJ/kg-K
D. 3.407 kJ/kg-K
Solution :
Ans. B
24. What horsepower is required to isothermally compress 800 ft 3 of air
per minute from 14.7 psia to 120 psia?
A. 28 hp
C. 256 hp
D. 17,000hp
B. 108 hp *
Solution :
V 
P 
Δs = mCp ln  2  + mC vln  2 
V 
P 
 1
 1
Δs = 2.268 (1.0414 ) ln 2 + 2.268 ( 0.7442 ) ln 2
W = P1V1 ln ( P1 P2 )

ft 3   14.7   144 in 2  
W = (14.7psia )  800
 ln 

 x 
min   120   1ft 2  

Δs = 2.8071 kJ/kg-K
W = -3555621.56
Ans. A
20. The specific gas constant of oxygen is R = 0.25983 kJ/kgK. If a 2 m3
tank contains 40 kg of oxygen at 40 °C, what is the gage pressure in
the tank?
A. 61 kPa
C. 110 kPa
B. 160 kPa
D. 1.53 MPa *
Solution :

ft - lbf 
1hp

 = -107.75 hp
min  33000ft - lbf min 
Ans. B
25. How much work is necessary to compress air in an insulated cylinder
from 0.20 m3 to 0.01 m3 . Use T1 = 20C and P1 = 100 kPa
A. 113.4 kJ
C. 110.1 kJ
B. 121.4 kJ
D. 115.7 kJ *
Solution :
PV = mRT
)
)(

kJ 
Pg + 101.325 kPa 2m3 = ( 40 kg )  0.25983
 ( 40 + 273 K )
kg - k 

 1 mPa 
Pg = 1525.21 kPa 
 = 1.53 mPa
 1000 kPa 
(
Ans. D
21. A gas bubble rising from the ocean floor is 1 inch in diameter at a
depth of 50 feet. Given that sp. gr. of seawater is 1.03, the buoyant
force in lbs being exerted on the bubble at this instant is nearest to:
A. 0.014
C.0.076
D.0.14
B. 0.020 *
Solution :
BF = Wdisplaced
( )
BF =  V = ( SG )  H O V
→ where: V =
2
4 3
πr
3

lb   4  0.5 
ft  = 0.0195 lb
BF = (1.03)  62.4 3   π 
ft   3  12 

3
Ans. B
22. Determine the change in enthalpy per Ibm of nitrogen gas as its
temperature changes from 500 F to 200 F. ( Cp = 0.2483 Btu / IbmR)
A. -74.49 Btu/Ibm *
C. -68.47 Btu/Ibm
B. -72.68 Btu/Ibm
D. -84.48 Btu/Ibm
Solution :
Determining
P ,
2
P Vk = P Vk
1 1
2 2
(
(100 kPa ) 0.20 m3
) = P2 (0.01m3 )
1.4
1.4
P = 6628.91 kPa
2
P V − PV
1 1 = ( 6628.91)( 0.01) − (100 )( 0.20 ) = − 115.72 kJ
W= 2 2
1− k
1 − 1.4
Ans. D
26. A heat engine is operated between temperature limits of 1370C and
260C. Engine is supplied with 14,142 kJ/kwh. Find the Carnot cycle
efficiency in percent.
A. 70.10
C. 67.56 *
B. 65.05
D. 69.32
Solution :
T
260 + 273
e = 1− L = 1−
= 0.6756  67.56%
TH
1370 + 273
Ans. C
27. A closed vessel contains air at a pressure of 160 kN/m2 gauge and
temperature of 30C . The air is heated at constant volume to 60C
with the atmospheric pressure of 759 mm Hg. What is the final gauge
pressure ?
A. 174
C. 167
B. 186 *
D. 172
Solution :
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32. A Steam expands adiabatically in a turbine from 2000 kPa, 400C to
P
P
1 = 2
T
T
1
2
 759 
 759 
160 kPa + 
 (101.325 kPa ) P2 + 
 (101.325 kPa )
 760 
 760 
=
30 + 273 K
60 + 273 K
P = 185.86 kPa
2
Ans. B
28. What is the temperature in C of 2 liters of water at 30C after 500
Calories of heat have been added ?
A. 35.70
C. 38
B. 30.25 *
D. 39.75
400 kPa, 250C. What is the effectiveness of the process in percent
assuming an atmospheric temperature of 15C. Neglect changes in
kinetic and potential energy.
Properties of steam:
At 2000 kPa and 400C ( h = 3 247.6 kJ/kg, S = 7.1271 kJ/kg-K)
and at 400 kPa and 250C ( h = 2 964.2 kJ/kg , S = 7.3789 kJ/kgK)
A. 82
B. 84
Solution :
∈=
Solution :
(
Q = mC T − T
2 1
)
∈=
where: m =  V
(
500 Cal = (1000g L )( 2L )(1cal g-K ) T − 30
2
T = 30.25°C
2
Ans. B
)
C. 80 *
D. 86
Δh
Δh − TΔs
3247.6 − 2964.2
( 3247.6 − 2964.2 ) − (15 + 273)( 7.1271 − 7.3789 )
∈ = 79.625%
Ans. C
33. Steam enters the superheaters of a boiler at a pressure of 25 bar and
29. A volume of 450 cm3 of air is measured at a pressure of 740 mm Hg
absolute and a temperature of 20C. What is the volume in cm3 at 760
mm Hg absolute and 0C ?
A. 516.12
C. 620.76
B. 408.25 *
D. 375.85
Solution :
dryness of 0.98 and leaves at the same pressure at a temperature of
370C. Calculate the heat energy supplied in the superheaters.
Properties of steam:
At 25 bar and 370C ( h = 3171.8 kJ/kg ) and at 25 bar and 0.98 dryness
( hf = 962.11 kJ/kg , hfg = 1841.01 kJ/kg ).
A. 407.46
B. 408.57
C. 405.51 *
D. 406.54
Solution :
PV P V
1 1= 2 2
T
T
1
2
(
( 740 mmHgabs ) 450 cm3
20 + 273 K
V = 408.25 cm3
2
Ans. B
Q = h −h
1
4
) = (760 mmHgabs) V2
→ where: h = h + xh
4 f
fg
h = 962.11 + 0.98 (1841.01) kJ kg
4
h = 2766.3kJ kg
4
Q = 3171.8 − 2766.3 = 405.5kJ kg
0 + 273 K
Ans. C
30. Assuming compression is according to the law PV = constant.
Calculate the initial volume of gas at a pressure of 2 bar which will
occupy a volume of 6 m3 when it is compressed to a pressure of 42
bar.
C. 130 m3
A. 126 m3 *
B. 120 m3
D. 136 m3
34. The thermal efficiency of a particular engine operating on an ideal
cycle is 35%. Calculate the heat supplied per 1200 watt-hr of work
developed in kJ.
A. 12 343 *
C. 14 218
B. 10 216
D. 11 108
Solution :
Solution :
( )
( )
 ( 42 bar ) 6m3 = ( 2 bar ) V
2
3
V = 126 m
2
PV =P V
1 1
2 2
Ans. A
31. A steam condenser receives 10 kg per second of steam with an
enthalpy of 2,570 kJ/kg . Steam condenses into liquid and leaves with
an enthalpy of 160 kJ/kg. Cooling water passes through the condenser
with temperature increases from 13C to 24C. Calculate the cooling
water flowrate in kg/s.
A. 533
C. 523 *
B. 518
D. 528
Solution :
Wnet
Qs
1200 W - hr  3600 s  1 kJ 
0.35 =



Q
 1hr  1000 J 
Q = 12 342.86 kJ
e=
Ans. A
35. Determine the average Cp value in kJ/kg-K of a gas if 522 kJ of heat is
necessary to raise the temperature from 300 K to 800 K making the
pressure constant.
A. 1.440
C. 1.038
B. 1.044 *
D. 1.026
Solution :
(
Q
= Qwater
cond
)
(
ms h − h = mw C t
−t
1 2
2w 1w
Q = mCp ( ΔT )
)

kg 
kJ 
kJ 
10  2570 -160  = m w  4.187
 ( 24 − 13 ) K
s 
kg 
kg
-K

( )
 522 kJ = (1 kg ) Cp (800 − 300 ) K
Cp = 1.044
kJ
kg - K
Ans. B
mw = 523.26 kg s
36. A perfect gas has a value of R = 58.8 ft-lb/lb-°R and k = 1.26. If 20
Ans. C
Btu are added to 10 lbs of the gas at constant volume when initial
temperature is 90°F, find the final temperature.
A. 97°F *
C. 144°F
B. 107°F
D. 175°F
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Solution :
(
Finding n from PV n = C :
P Vn = P Vn
1 1
2 2
n
n
( 2070 kPa )(170 ) = ( 275 kPa )(850 )
)
Q = mc v t − t
f
i
R
58.8
ft - lbf
where: C v =
=
= 226.154
k − 1 1.26 − 1
lbm -°R
ft - lbf 
 778 ft - lbf 

20 Btu 
 = 10 lbs  226.154
 t − 90°F
Btu
lbm -°R  f



t = 96.88°F
f
Ans. A
(
n = 1.2542
)
P V − PV
1 1
W= 2 2
1− n
( 275 kPaa )(850 ) − ( 2070 kPaa )(170 )
W=
1000 /m3 1 − 1.2542
)(
(
)
W = 464.82 kJ
37. If atmospheric air 14.7 psia and 60F at sea level, what is the pressure
at 12000 ft altitude if air is compressible. Note: @ 60F ; the density of
air is 0.0763 lbm/ft3 ; P1 = 14.7 psia
A. 5.467 psia
C. 8.342 psia
B. 9.53 psia *
D. 2.346 psia
Solution :
Δh
P = P e RT
1
2
41. Air is flowing through a 100 mm I.D. pipe at the rate of 6 kg/min. The
air pressure and temperature are 500 kPa and 30 C respectively. Find
the velocity of the air in m/s.
A. 3.81
C. 2.18 *
B. 2.81
D. 3.18
Solution :
V = A → where: V is determined from PV = mRT
12000 ft
(53.342)(60 + 460)

( 500 kPa ) V = ( 6 kg min )  0.287
14.7 psia = P e
2
P = 9.54 psia
2
Ans. B

V = 0.0174 m3 s
kJ 
 1 min 
 (30 + 273 K ) 

kg-K 
 60 s 
Then,
38. If air is at a pressure of 3200 lb/ft2 and at a temperature of 800 R,
what is the specific volume?
A. 9.8 ft3 /lb
B. 13.33 ft3 /lb *
C. 11.2 ft3 /lb
D. 15.8 ft3 /lb
0.0174 m3 s =
2
π  100 mm 

 
4  1000 mm/m 
 = 2.214 m s
Ans. C
Solution :
PV = RT
(
Ans. A
42. A quantity of 55 m3 of water passes through a heat exchanger and
)
lbf - ft
3200lb ft 2 ( V ) = 53.342
(800 °R )
lbm -°R
V = 13.34ft 3 lbm
Ans. B
absorbs 2,800,000 kJ. The exit temperature is 95 C. The entrance
water temperature in C is nearest to
A. 49
C. 68
B. 56
D. 83 *
Solution :
39. Given 280 liters of a gas at 63.5 cm Hg. The gas has a specific heat at
constant pressure of 0.847 kJ/kg K and a specific heat at constant
volume of 0.659 kJ/kg K. Which of the following most nearly equals
the volume the gas would occupy at a final pressure of 5 atm if the
process is adiabatic?
A. 62 liters
C. 77 liters
B. 70 liters *
D. 82 liters
Q = mCΔT
where : m = ρV
(
)(
)

kJ 
2 800 000 kJ = 1000 kg m3 55 m3  4.187
 ( 95 − t )
kg °C 

t = 82.84 °C
Ans. D
43. A gas company buys gas at 620 kPa gauge and 24 C and sells it at
9.65 cm of water pressure and –2 C. Disregarding the losses in the
distribution, which of the following most nearly equals the number of
cubic meters sold for each cubic meter purchased?
A. 2.3
C. 6.4 *
B. 4.1
D. 7.1
Solution :
P V k = P Vk
1 1
2 2
Cp
0.847
where: k =
=
= 1.2853
Cv
0.659
( 63.5 cmHg )( 280 )
1.2853
Solution :
76 cmHg  1.2853

=  5atm x
V
1atm 

V = 69.6
Ans. B
40. If 0.5 kg of nitrogen with an initial volume of 170 liters and a pressure
of 2,070 kPaa expand in accordance with the law PVn = constant to a
final volume of 850 liters and a pressure of 275 kPaa, which of the
following most nearly equals the work done by the gas?
A. 465 kJ *
C. 481 kJ
B. 472 kJ
D. 489 kJ
Solution :
 mRT 
 P 
m3sold

sold
=
3
mRT


m
purchased
 P 

purchased
Then,
-2 + 273 k

kN   9.65 cm 
 +101.325 kPa
 9.81 3  
m3
m  100cm m 
sold
= 
24 + 273 k
m3
purchased
620 kPa + 101.325 kPa
Thus,
m3
sold = 6.44
3
m
purchased
Ans. C
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44. A closed container of helium dropped from a balloon 4 km above. Find
PV = mRT
the temperature difference when the container hits the ground.
A. 18.6 C
C. 12.6 C *
B. 24.5 C
D. 30.2 C
( 2200 +14.7 psia ) 2.3 ft3 
Solution :
m = 28.665 lbm
(

2
lbf-ft 
)  144ft2in  = m  1545
 ( 70 + 460°R )
32 lbm-°R 
Ans. C
ΔU = ΔPE
mc vΔT = mgΔz
where Cv =
Cv =
R
R
=
k -1 MW ( k - 1)
8.314
= 3.1162
4 (1.667 - 1)
Then,
m 

9.81m s2 ) ( 4 km ) 1000

(

kJ 
km 

3.116
ΔT =

(
kg-k 

ΔT = 12.59°C
)
1000 J kJ
49. How many gallons per minute of cooling water are required to
removed 100,000 kJ/hr of heat from a diesel engine if the designed
temperature rise of the water is 20 °C?
A. 2.56
C. 5.26 *
B. 6.25
D. 4.25
Solution :
Q = mcpΔT
100000
where m = ρV

kJ 
lbm   0.454 kg 
kJ 
 60 min 
=  8.33

 ( 20K ) 
 ( V )  4.187

hr 
gal   1 lbm 
kg - k 
 1 hr 

V = 5.263 gpm
Ans. C
Ans. C
45. In a particular cycle, 350 MJ of heat are transferred into the system
each cycle. The heat transferred out the system is 297.5 MJ per cycle.
What is the thermal efficiency of the cycle?
A. 1.6%
C. 5.0%
B. 7.5%
D. 15% *
Solution :
50. Twenty grams of oxygen gas (O2) are compressed at a constant
temperature of 30 C to 5  of their original volume. What work is
done on the system ?
A. 824 Cal
C. 944 Cal
B. 1124 Cal *
D. 1144 Cal
Solution :
Q − Qout 350 − 297.5
e t = in
=
= 0.15
Q
350
in
Ans. D
V 
W = mRT1 ln  2 
 V1 
46. Steam flows into a turbine at the rate of 10 kg/s and 10 kW of heat are
lost from the turbine. Ignoring elevation and kinetic energy effects,
calculate the power output from the turbine.
Note: h1 = 2739.0 kJ/kg and h2 = 2300.5 kJ/kg
A. 4605 kW
B. 4973 kW
C. 4375 kW *
D. 4000 kW
Solution :
Ans. B
51. A certain mass of gas having an initial pressure of 17,237 kPa and an
P = ms ( Δh ) - QL
0
P = (10 kg s )( 2739 - 2300.5 kJ kg ) - 10 kW
0
P = 4375 kW
0
Ans. C
47. The enthalpy of air is increased by 139.586 kJ/kg in a compressor. The
rate of air flow is 16.42 kg/min. The power input is 48.2 kW. Which of
the following values most nearly equals the heat loss from the
compressor in kW?
A. – 10
*
C. –9.95
B. 10.2
D. 9.95
Solution :
Q = ΔH + W
Q = 16.42
kJ 

8.314
 0.05 V1 
 20 g  
kg-K 
 ( 30 + 273 k ) ln 
W =


32

 1000g kg  
 V1 




 1 kcal 
W = - 4.72 kJ 

 4.187 kJ 
W =-1.1265 kcal = -1126.5 cal
kg 
kJ   1 min 
139.586 
 - 48.2 kW
min 
kg   60 s 
Q = -10 kW
Ans. A
48. An oxygen cylinder of volume 2.3 ft3 has a pressure of 2200 psig and
is at 70F. Determine the mass of oxygen in the cylinder.
A. 26.66 lbs
C. 28.66 lbs *
B. 26.86 lbs
D. 28.88 lbs
Solution :
initial volume of 0.057 m3 is expanded isothermally to volume of
0.113 m3. What is the work done by the gas in kJ?
A. 500
C. 672 *
B. 564
D. 600
Solution :
V 
 0.113 
W = P V ln  2  = (17237 kPa ) 0.057m3 ln 

1 1 V 
 0.057 
 1
W = 672.37 kJ
)
(
Ans. C
52. Air having an initial pressure of 6,516 kPa and an initial volume of
0.113 m3 is compressed adiabatically to a final volume of 0.057 m3 .
Calculate the work done by the gas as it compresses to a final pressure
of 17, 237 kPa.
A. - 615.5 kJ *
C. -157.9 kJ
B. - 197.5 kJ
D. - 179.5 kJ
Solution :
P V − PV
1 1
W = 2 2
1− k
(
)
(
(17237 kPa ) 0.057 m3 − ( 6516 kPa ) 0.113 m3
W =
)
1 − 1.4
W = − 615.5025 kJ
Ans. A
53. Gas is enclosed in a cylinder with a weighted piston as the top
boundary. The gas is heated and expands from a volume of 0.04 m3 to
0.10 m3 at a constant pressure of 200 kPa. Calculate the work done by
the system.
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A. 8 kJ
B. 10 kJ
C. 12 kJ *
D. 14 kJ
T = 454.67°R
2
Converting to K,
Solution :
(
)
)
(
W = P V − V = 200 kPa 0.10 m3 − 0.04 m3 = 12 kJ
2 1
T = 454.67°R = 252.26 K
2
Ans. D
Ans. C
58. With three different quantities x, y, and z of the same kind of liquid
54. Gas is enclosed in a cylinder with a weighted piston as the top
boundary. The gas is heated and expands from a volume of 0.04 m3 to
0.10 m3 . The pressure varies such that PV = constant, and the initial
pressure is 200 kPa. Calculate the work done by the system.
A. 6.80 kJ
C. 9.59 kJ
B. 7.33 kJ *
D. 17.33 kJ
Solution :
of temperatures 9, 21 and 38C respectively, it is found that when x
and y are mixed together the resultant temperature is 17C and when y
and z are mixed together the resultant temperature is 28C. Find the
resultant temperature if x and z were mixed.
A. 29.87C
C. 20.85C
B. 25.92C*
D. 24.86C
Solution :
V 
 0.10 
W = P V ln  2  = ( 200 kPa ) ( 0.04 m3 ) ln 
 = 7.33 kJ
1 1 V 
 0.04 
 1
Mixture of x and y: Q x = Qy
m x (17 − 9 ) = ( 21 − 17 ) m y
Ans. B
m x = 0.5m y
55. Nitrogen is expanded isentropically. Its temperature changes from 620
→ ①
Mixture of y and z: Q y = Qz
F to 60 F. Find the pressure ratio (p1 / p2).
A. 0.08
C. 26.2
D. 3547
B. 12.9 *
m y ( 28 − 21) = ( 38 − 28 ) m z
Solution :
Mixture of x and z: Q x = Qz
mz = 0.70m y → ②
m x ( t − 9 ) = mz ( 38 − t )
k-1
P k
T
1 = 1 
→ note : k N 2 = 1.399 1.4
P 
T
2  2
1.4 -1
620 + 460  P1  1.4
= 
60 + 460  P 
 2
P
1 = 12.91
P
2
Ans. B
0.5m y ( t − 9 ) = 0.70m y ( 38 − t )
t = 25.92C
Ans. B
59. An air bubble rises from the bottom of a well where the temperature is
25C, to the surface where the temperature is 27C. Find the percent
increase in the volume of the bubble if the depth of the well is 5 m.
Atmospheric pressure is 101,528 Pascals.
A. 49.3 *
C. 41.3
B. 56.7
D. 38.6
56. An engine takes in 11,000 Btu/min of heat from a hot body while
giving an output of 110 hp. What is the thermal efficiency of this
engine?
A. 44.2
C. 52.4
B. 42.4 *
D. 54.2
Solution :
Solution :
V − V
1
%V = 2
V
1
( 0.287 kJ kg - K )( 25 + 273) K = 0.568m3 kg
where : V =
1
(101.528 + 9.81(5) kPa )
V =
2
W
(110 hp )( 33000ft-lb hp-min )
e t = out =
Btu 
W

in
11000
 ( 778ft-lb Btu )
min 

e t = 0.4242  42.42%
( 0.287 kJ kg-K )( 27 + 273k ) = 0.848m3 kg
101.528 kPa
0.848 − 0.568
%V =
= 0.4931  49.31%
0.568
Ans. A
Ans. B
60. An ideal gas is 45 psig and 80F is heated in a closed container to
130F. What is the final pressure?
57. Air expanded adiabatically from an initial absolute pressure of 31026
kPa to a final absolute pressure of 1705 psi. If the initial temperature of
the air is 600 R, calculate its final temperature.
A. 455 R
C. 252.6 R
B. 252.6 K
D. A and B *
Solution :
A. 54 psia
B. 65 psia *
C. 75 psia
D. 43 psia
Solution :
P P
1= 2
T T
1 2

P
45 + 14.7
2
=
80 + 460 130 + 460
P = 65.23 psia
2
k-1
P  k
T
2 = 2 
P 
T
1  1
Ans. B
61. A large mining company was provided with a 3000 cm3 of compressed
1.4-1
101.325 kPa  1.4

1705
psi
.

T
14.7 psia 
2 =
600°R 
31026 kpa





air tank. Air pressure in the tank drops from 700 kPa to 180 kPa while
the temperature remains unchanged at 28C. What percentage has the
mass of air in the tank been reduced ?
A. 74 *
C. 76
B. 72
D. 78
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Solution :
65. If 6 liters of a gas at a pressure of 100 kPaa are compressed reversibly
according to PV2 = C, until the volume becomes 2 liters. Find the final
pressure.
A. 600 kPaa
C. 900 kPaa *
B. 800 kPaa
D. 1000 kPaa
m −m
2
%m= 1
m
1

PV
m = 1 1=
1 RT
1
3
1m  
 
 100 cm  
( 700 kPa ) 3000 cm3x 


kJ 
 0.287
 ( 28+273 K )
kg-K 


Solution :
= 0.02431 kg
3
1m  
 
 100 cm  
(180 kPa ) 3000 cm3x 
P V

m = 2 2 =
2
RT
0.287 kJ kg-K )( 28 + 273 K )
(
2
0.02431 − 0.006251
%m=
= 0.743  74.3%
0.02431
= 0.006251 kg
Ans. A
P V2 = P V2
1 1
2 2
2
2
(100 kPa )( 6 L ) = P2 ( 2 L )
P = 900 kPa
2
Ans. C
66. If 10 lbs of water evaporated at atmospheric pressure until a volume of
288.5 ft3 is occupied, how much work is done?
A. 610,358 ft-lb
C. – 610,358 ft-lb *
B. 0
D. 550,000 ft-lb
62. A closed vessel contains air at a pressure of 140 Kpag and temperature
of 20C. Find the final gauge pressure if the air is heated at constant
volume to 40C. Take the atmospheric pressure as 759 mm Hg.
C. 136.46
A. 156.46 *
B. 146.46
D. 126.46
Solution :
Solution :
(
W = P V −V
2 1
) At 14.7 psia,
lb 
in 2  
 144
 288.5 − (10 )( 0.016714 ) ft3 

2


in 
ft 2  
W = 610343 ft-lbf ( on )
W = 14.7
P P
1= 2
T T
1 2
 101.325 kPa 
 101.325 
140 kPag + 759 mmHg 
 Pg + 759 
 kPa
 760 mmHg  = 2
 760 
( 20 + 273) K
( 40 + 273) K
Pg = 156.46 kPag
2
Ans. A
Ans. C
67. What pressure is column of water 100 cm high equivalent to ?
A. 9807 dynes/cm2
B. 9807 N/m2
C. 0.10 Bar
D. 9810 N/m2*
Solution :
63. Water substance at 70 bar and 65C enters a boiler tube of constant
inside diameter of 25 mm. The water leaves the boiler tube at 50 bar
and 700K at velocity of 100 m/s. Calculate the inlet volume flow in
li./sec. At 70 bar & 65C, 1 = 0.001017 m3/kg, at 50 bar & 700 K, 2
= 0.06081 m3/kg.
A. 0.75
C. 0.82 *
B. 0.64
D. 0.96
 100 cm 
P =  h = 9.81kN m3 

 100cm m 
P = 9810 N m 2
Ans. D
68. Work done by a substance in reversible nonflow manner in accordance
Solution :
ρ A υ =ρ A υ
1 11 2 2 2
υ
υ
1 = 2
v
v
1
2
υ
100
1
=
0.001017 0.06081
υ = 1.67 m s
1
 1000 L 
π
π
2
V = A υ = D2 υ = ( 0.025 m ) (1.67 m s ) 

11 4 1 1 4
 1 m3 
V = 0.821m3 s
with V = 100/P ft3, where P is in psia. Evaluate the work done on or
by the substance as the pressure increases from 10 psia to 100 psia.
A. 33 157.22 ft-lb
C. 43 157.22 ft-lb
D. – 43, 157.22 ft-lb
B. –33 157.22 ft-lb*
Solution :
At P = 10 psia, V = 10 ft 3
1
1
At P = 100 psia, V = 1ft 3
2
2
1
 100   144 
W= 
 
 dV = −33,157.22 ft-lb
 V   1 
10

Ans. B
Ans. C
69. Assume 8 lb of a substance receive 240 Btu of heat at constant volume
64. Determine the average constant pressure specific heat of steam at 10
kPa and 45.8C. Note: From steam table, at 47.7C, h = 2588.1 kJ/kg
and at 43.8C, h = 2581.1 kJ/kg.
C. 30.57 kJ/kgC
A. 1.79 kJ/kgC *
B. 10.28 kJ/kgC
D. 100.1 kJ/kgC
Solution :
and undergo a temperature change of 150F. Determine the average
specific heat of the substance during the process.
A. 0.5 Btu/lbF
C. 0.40 Btu/lbF
B. 0.3 Btu/lbF
D. 0.20 Btu/lbF *
Solution :
Δh 2588.1- 2581.1
=
Δt
47.7- 43.8
Cp = 1.795kJ kg -K
Cp =
Q = mCvΔT
Ans. A
C v = 0.2
240 Btu = (8 lb ) ( C v ) (150°F )
Btu
lb-°F
Ans. D
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70. Determine the specific weight of air at 760 mmHg absolute and 22°C.
A. 1.134 kg/m3
B. 1.416 kg/m3
C. 1.197 kg/m3 *
D. 1.276 kg/m3
Solution :
P = ρRT
 101.325 kPa 

kJ 
760 mmHg 
 = ρ  0.287
 ( 22 + 273 ) K
760
mmHg
kg-K




3
ρ = 1.20 kg m
Ans. C
71. A one cubic container contains a mixture of gases composed of 0.02
kg-mol of oxygen and 0.04 kg-mol of helium at a pressure of 220 kPa.
What is the temperature of this ideal gas mixture in degrees Kelvin?
C.404
A.441 *
B.450
D.360
Solution :
PV = nRT
( 220 kPa ) (1m3 ) = ( 0.02 + 0.04 kgmol )  8.314

kJ 
 (T )
kmol-K 
T = 441 K
Solution :
(
W = RTln V V
2 1
) → Note: V2 = 14 V1
1 V 
Btu 

4 1
W =  0.4968
 ( 68 + 460 °R ) ln 
lbm-°R 
V 

 1 
W = - 363.64 Btu lbm
Ans. A
76. Twenty grams of oxygen gas (O2) are compressed at a constant
temperature of 30C to 5% of their original volume. What work is
done on the system? Use R of air , 0.0619 Cal/gm-K
A. 824 Cal
C. 944 Cal
B. 924 Cal
D. 1124 Cal *
Solution :
(
W =  PdV = mRTln V V
2 1
)

Cal 
W = ( 20 g )  0.0619
 ( 30 + 273) ln ( 0.05 )
g-K


W = -1123.74 Cal
Ans. D
77. Gas is enclosed in a cylinder with a weighted piston as the top
Ans. A
72. A pressure gage registers 50 psig in a region where the barometer
reads 14.8 psia. Find the absolute pressure in kPa.
A. 766.66 kPa
C. 446.66 kPa *
B. 558.66 kPa
D. 326.66 kPa
boundary. The gas is heated and expands from a volume of 0.04 m3 to
0.10 m3 at a constant pressure of 200 kPa. Calculate the work done by
the system.
A. 8 kJ
C. 12 kJ *
B. 10 kJ
D. 14 kJ
Solution :
(
Solution :
W = PΔV = 200 kPa 0.10 − 0.04 m3
Pa = Pg + Patm
W = 12 kJ
)
Ans. C
 101.325 
Pa = ( 50 +14.8 ) 
 kPa
 14.7 
Pa = 446.66 kPa
78. A piston-cylinder system contains a gas which expands under a
Ans. C
73. 10 BTU (10 kJ) are transferred in a process where a piston compresses a
spring and in so doing does 1500 ft-Ibf ( 2000 J) of work. Find the change
in internal energy of the system.
C. 21 kJ
A. 8 kJ *
B. 12 kJ
D. 5 kJ
constant pressure of 1200 lb/ft2. If the piston is displaced 1 ft during
the process, and the piston diameter is 2 ft. What is the work done by
the gas on the piston ?
A. 1768 ft-lb
C. 3768 ft-lb *
B. 2387 ft-lb
D. 4000 ft-lb
Solution :
W =  PdV = PΔV = PAΔL

lbf   π 
2
W = 1200
   ( 2 ft ) (1 ft )
ft 2   4 

W = 3769.91 ft-lb
Solution :
Q = U + W
U = Q − W = 10 − 2 = 8kJ
Ans. C
Ans. A
79. Gas is enclosed in a cylinder with a weighted piston as the top
temperature changes from 100 F to 120 F. Use Cv = 0.157 Btu/lb R.
A. 14.7 Btu
C. 16.8 Btu
B. 15.7 Btu *
D. 147 Btu
boundary. The gas is heated and expands from a volume of 0.04 m3 to
0.10 m3. The pressure varies such that PV = constant and the initial
pressure is 200 kPa. Calculate the work done by the system.
A. 6.80 kJ
C. 9.59 kJ
B. 7.33 kJ*
D. 12 kJ
Solution :
Solution :
74. Find the change in internal energy of 5 lb of oxygen gas when the
ΔU = mC vΔT
W = P1V2 ln
Btu 

ΔU = ( 5 lb )  0.157
 (120 - 100°R )
lb-°R 

ΔU = 15.7 Btu
V2
 0.10 
= ( 200 kPa ) ( 0.04 m 3 ) ln 

V1
 0.04 
W = 7.33 kJ
Ans. B
Ans. B
80. Ammonia weighing 22 kg is confined inside a cylinder equipped with
75. Helium (R = 0.4968 Btu/lbmR) is compressed isothermally from 14.7
psia and 68F. The compression ratio is 4. Calculate the work done by
the gas.
A. – 364 Btu/lbm *
C. – 187 Btu/lbm
B. – 145 Btu/lbm
D. – 46.7 Btu/lbm
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a piston has an initial pressure of 413 kPa at 38C. If 2900 kJ of heat
is added to the ammonia until its pressure and temperature are 413 kPa
and 100C, respectively. What is the amount of work done by the fluid
in kJ ?
A. 667 *
C. 420
B. 304
D. 502
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Solution :
Solution :
(
)
(
W = P V − V = mR T − T
2 1
2 1
)
 8.314 kJ 
W = 22 kg 
 (100 - 38 ) K = 667.08 kJ
 3 + 14 kg - K 
Ans. A
81. What is the total required heating energy in raising the temperature of
a given amount of water when the energy supplied is 1000 kwh with
heat losses of 25% ?
A. 1000
C. 1333 *
B. 1500
D. 1250
V 
Δs = Rln  2 
V 
 1
where V =1 4 V
2
1
Btu   1 

Δs =  0.4961
 n 
lb
-°R   4 

Δs = -0.688Btu lb -°R
Ans. A
86. Steam at the rate of 500 kg/hr is produced by a steady flow system
Let: Q = heating energy
boiler from feedwater entering at 40C. Find the rate at which heat is
transformed in kCal/hr if the enthalpy of steam is 600 kCal/kg and of
steam 50 kCal/kg.
A. 275,000 kCal/hr *
C. 375,000 kCal/hr
B. 175,000 kCal/hr
D. 475,000 kCal/hr
Q − 0.25Q = 1000
Solution :
Solution :
Q = 1333.33 kWh
Q = mΔh
Ans. C
Q = ( 500 kg hr )( 600 - 50 kcal kg )
82. What is the rise water temperature of water dropping over a 200 foot
waterfall and setting in a basin below? Neglect all friction and assume the
initial velocity is negligible.
C. 0.256 C
A. – 0.140 C *
B. – 0.918 C
D. 0.429 C
Q = 275 000 kcal hr
Ans. A
87. During the polytropic process of an ideal gas, the state changes from
138 kPa and 5C to 827 kPa and 171C. Find the value of n .
A. 1.354 *
C. 1.345
B. 1.253
D. 1.234
Solution :
Q = U + W → where:Q = 0 (adiabatic)
Solution :
U = − W
mCpT = mgh
n −1
n −1
P  n
T
171 + 273  827  n
2 =  2

=

P 
T
5 + 273  138 
1  1
n = 1.354
− gh − ( 9.81 m / s ) ( 60m )
T =
=
= −0.140C
Cp
4.19 x103 J / kg .K
2
Ans. A
Ans. A
83. Water enters the condenser at 30C and leaves at 60C. What is the
temperature difference in F?
A. 16.67
B. 48.67
Solution :
F =
88. For an ideal gas, what is the specific molar entropy change during an
isothermal process in which the pressure changes from 200 kPa to 150
kPa ?
C. 3.39 J/mol-K
A. 2.39 J/mol-K*
B. 1.39 J/mol-K
D. 4.39 J/mol-K
C. 54 *
D. 22
Solution :
9
9
( C ) = ( 60 − 30 ) = 54 F°
5
5
P  
J
  200 
Δs = R ln  1  =  8314
 ln 

P  
kmol-K   150 
 2
Δs = 2391.79 J kmol-K = 2.39 kJ kmol-K
Ans. C
84. A cylinder and piston arrangement contains saturated water vapor at
110C. The vapor is compressed in a reversible adiabatic process until
the pressure is 1.6 Mpa. Determine the work done by the system per
kg of water. At 110C, S1 = 7.2387 kJ/kg-K, U1 = 2518.1 kJ/kg and at
1.6 Mpa, S2 = 7.2374 kJ/kg-K, U2 = 2950.1 kJ/kg, T2 = 400C.
A. – 500 kJ/kg
C. – 632 kJ/kg
B. – 432 kJ/kg *
D. – 700 kJ/kg
Δs = 2.39 J mol-K
Ans. A
89. An ideal gas mixture consists of 2 kmol of N2 and 6 Kmol of CO2.
The mass fraction of CO2 is:
A.0.175
B.0.250
C.0.825 *
D.0.650
Solution :
Solution :
W = -ΔU
(
n MWx
% mx = x
ΣnMW
)
W = - 2950.1 - 2518.1 kJ kg
%m
%m
W = -432kJ kg
6 ( 44 )
=
CO2 2 ( 28 ) + 6 ( 44 )
CO2
= 0.825
Ans. C
Ans. B
90. An ideal gas mixture consists of 2 kmol of N2 and 6 kmol of Co2. The
85. Helium is compressed isothermally from 14.7 psia and 68F. The
compression ratio is 4. Calculate the change in entropy of the gas
given that RHelium = 0.4961 Btu/lbmR.
A. – 0.688 Btu/lbmR *
C. 0.658 Btu/lbmR
B. – 2.76 Btu/lbmR
D. 2.76 Btu/lbmR
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apparent gas constant of mixture is:
A.0.208 *
C.0.531
B.0.925
D.0.251
Solution :
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n N + n CO2 = nT → where: nT =
2
2 kmol + 6 kmol =
MT
MWT
( 2 kmol )( 28) + ( 6 kmol )( 44 )
MWT
MWT = 40 kmol
R=
R
8.314 kJ
kJ
=
= 0.208
MW
40 kg-k
kg-k
Ans. A
Solution :
PV = mRT
lbf - ft 

2
 1545 lb -°R 


144
in
m
 ( 200 + 460°R )
= m
( 20 psia ) (10 ft 2 ) 
2 
 12 + 32 
 1 ft 




m = 1.243 lbm
Ans. C
91. A certain gas at 101.325 kPa and 16C whose volume is 2.83 m3 are
compressed into a storage vessel of 0.31 m3 capacity. Before
admission, the storage vessel contained the gas at a pressure and
temperature of 137.8 kPa and 24C. After admission, the pressure has
increased to 1171.8 kPa. What should be the final temperature of the
gas in the vessel in Kelvin ?
A. 298.0
C. 180
B. 319.0 *
D. 420
Solution :
95. What is the approximate value of temperature of water having enthalpy
of 208 Btu/lb?
A. 138.67°C
B. 115.55°C *
Solution :
Δh = CpΔT
208
mB + m
=m
A
Added
P V
PBVB PAdded VAdded
+
= A A
R BTB R
T
R T
A A
Added Added
(137.8)( 0.31) + (101.325 )( 2.83) = (1171.8 )( 0.31)
24 + 273
16 + 273
T
A
T = 319.76 K
A
Ans. B
C. 258.67°C
D. 68.67°C
Btu  1.055 kJ   1 lb 
kJ
( T-0 )
 = 4.187


lb  1 Btu   0.454 kg 
kg-k
T = 115.44 °C
Ans. B
96. An ideal gas at 0.80 atmospheres and 87°C 0.450 liter. How many
moles are in the sample? (R = 0.0821 liter-atm/mole-K)
A. 0.0002 mole
C. 0.0122 mole *
B. 0.0278 mole
D. 0.0091 mole
Solution :
92. If the specific heat at constant pressure for CO2 is given as 0.201
Btu/lb-R, what is the value of the specific heat at constant volume ?
A. 0.156 Btu/lb-R *
C. 0.435 Btu/lb-R
B. 0.365 Btu/lb-R
D. 0.435 Btu/lb-R
PV = nRT
( 0.8 atm )( 0.450 L ) = n  0.0821

L - atm 
 (87 + 273 k )
mole - k 
n = 0.0122 mole
Ans. C
Solution :
Cp =
97. Two kilogram of air in a rigid tank changes its temperature from 32°C
kR
k-1
 1545 ft - lbf  1 Btu 
k


Btu
44 lbm -°R  778 ft - lbf 
0.201
= 
lb -°R
k −1
k = 1.2896
where : Cp = kC v
to 150°C. Find the work done during the process.
A. 246
C. 175
B. 180
D. 0 *
Solution :
Rigid Tank
W=0
Btu
0.201
= (1.2896 ) C v
lb °R
Btu
C v = 0.156
lb -°R
Ans. D
98. Nitrogen (k = 1.4) is expanded isentropically. Its temperature changes
from 620F to 60F. Find the pressure ratio (P1/P2) .
A. 0.08
C. 26.2
B. 12.91 *
D. 35.47
Ans. A
93. An amount of 4000 Btu of heat is transferred from a reservoir at 800
deg.F to a reservoir at 200 deg. F. Find the entropy change of the
system.
A. 2.89 *
C. 1.34
B. 3.24
D. 3.21
Solution :
Q
Q
1 − 1
T
T
1
2
4000
4000
Δs =
−
800 + 460 200 + 460
Δs = − 2.89
Δs =
Solution :
k-1
P  k
T
1 =  1
P 
T
2  2
P P = 12.9
1 2
Ans. B
1.4-1
620 + 460  P1  1.4
= 

60 + 460  P 
 2
99. In an isentropic process, P1 = 200 psi , P2 = 300 psi and T1 = 700R.
Find T2 using k = 1.4.
A. 576R
B. 680R
C. 786R *
D. 590R
Solution :
Ans. A
94. Find the mass of carbon dioxide having a pressure of 20 psia at 200°F
with 10 ft3 volume.
A. 1.04 lbs
B. 1.14 lbs
C. 1.24 lbs *
D. 1.34 lbs
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k-1
1.4-1
P  k
T
T
 300  1.4
2 =  2
2
=


P 
T
700 °R  200 
1  1
T = 785.98°R
2
Ans. C
100. Nitrogen is expanded isentropically.Its temperature changes from
620F to 60F. The volumetric ratio is (V2/V1) = 6.22 and the value of
R for nitrogen is 0.0787 Btu/lbmR. What is the work done by the
gas?
A. – 100.18 Btu/lbm
C. 110.18 Btu/lbm *
B. 120.27 Btu/lbm
D. –120.27 Btu/lbm
Solution :
(
)
Btu
0.0787
( 60 − 620°R )
R T −T
P V −P V
2
1
lbm
-°R
2
2
1
1
W=
=
=
1− k
1 −k
1 − 1.4
W = 110.18 Btu lbm
Ans. C
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