Practice Questions with Detailed Solutions
Part 1 (question 1-300)(part 2 question 1 -150 answers with rationales )
PART 1: FOUNDATIONAL CONCEPTS & PROPERTIES
Question 1
A pressure gauge on a tank reads 450 kPa. If the barometric pressure is 755 mmHg, what is the
absolute pressure in the tank? (Density of mercury = 13,600 kg/m³, g = 9.81 m/s²)
Solution:
Given: Pg=450 kPaPg=450 kPa, hHg=0.755 mhHg=0.755 m
Find: PabsPabs
Analysis: Convert atmospheric pressure from mmHg to kPa: Patm=ρghPatm=ρgh.
Calculation:
Patm=(13,600)(9.81)(0.755)=100,720 Pa=100.72 kPaPatm
=(13,600)(9.81)(0.755)=100,720 Pa=100.72 kPa
Pabs=Pg+Patm=450+100.72=550.72 kPaPabs=Pg+Patm=450+100.72=550.72 kPa
Question 2
A vacuum gauge on a condenser reads 60 kPa. If atmospheric pressure is 98 kPa, what is the
absolute pressure in the condenser?
Solution:
Given: Pvac=60 kPaPvac=60 kPa, Patm=98 kPaPatm=98 kPa
Find: PabsPabs
Analysis: For vacuum pressures, Pabs=Patm−PvacPabs=Patm−Pvac.
Calculation: Pabs=98−60=38 kPaPabs=98−60=38 kPa
Question 3
The temperature of a system is 120°C. Express this temperature in Kelvin, Rankine, and
Fahrenheit.
Solution:
,Given: T=120°CT=120°C
Find: TT in K, °R, °F
Analysis: Use conversion formulas.
Calculation:
T(K)=120+273.15=393.15 KT(K)=120+273.15=393.15 K
T(°R)=1.8(393.15)=707.67°RT(°R)=1.8(393.15)=707.67°R
T(°F)=1.8(120)+32=248°FT(°F)=1.8(120)+32=248°F
Question 4
A rigid tank contains a saturated liquid-vapor mixture of water at 100°C. The quality is 0.4.
Determine the specific volume of the mixture. (At 100°C: vf=0.001043 m3/kgvf
=0.001043 m3/kg, vg=1.6729 m3/kgvg=1.6729 m3/kg)
Solution:
Given: T=100°CT=100°C, x=0.4x=0.4
Find: vv
Analysis: Use v=vf+x(vg−vf)v=vf+x(vg−vf).
Calculation:
v=0.001043+0.4(1.6729−0.001043)=0.001043+0.6687=0.6697 m3/kgv=0.001043+0.4(1.6729−0.
001043)=0.001043+0.6687=0.6697 m3/kg
Question 5
A 0.5 m³ rigid tank contains water at 200°C with a quality of 0.6. Determine the mass of water
in the tank. (At 200°C: vf=0.001157 m3/kgvf=0.001157 m3/kg, vg=0.1274 m3/kgvg
=0.1274 m3/kg)
Solution:
Given: V=0.5 m3V=0.5 m3, T=200°CT=200°C, x=0.6x=0.6
Find: mm
Analysis: v=vf+x(vg−vf)v=vf+x(vg−vf). Then m=V/vm=V/v.
Calculation:
v=0.001157+0.6(0.1274−0.001157)=0.001157+0.0757=0.0769 m3/kgv=0.001157+0.6(0.1274−0.
001157)=0.001157+0.0757=0.0769 m3/kg
m=0.5/0.0769=6.50 kgm=0.5/0.0769=6.50 kg
Question 6
, For an ideal gas with R=0.287 kJ/kg⋅KR=0.287 kJ/kg⋅K, determine the specific volume at 300 K
and 200 kPa.
Solution:
Given: T=300 KT=300 K, P=200 kPaP=200 kPa, R=0.287 kJ/kg⋅KR=0.287 kJ/kg⋅K
Find: vv
Analysis: Use the ideal gas law: Pv=RTPv=RT.
Calculation:
v=RT/P=(0.287)(300)/200=0.4305 m3/kgv=RT/P=(0.287)(300)/200=0.4305 m3/kg
Question 7
A rigid tank contains 5 kg of air at 400 kPa and 600 K. Determine the volume of the tank.
(R=0.287 kJ/kg⋅KR=0.287 kJ/kg⋅K)
Solution:
Given: m=5 kgm=5 kg, P=400 kPaP=400 kPa, T=600 KT=600 K
Find: VV
Analysis: Use PV=mRTPV=mRT.
Calculation:
V=mRT/P=(5)(0.287)(600)/400=2.1525 m3V=mRT/P=(5)(0.287)(600)/400=2.1525 m3
Question 8
A gas with a molecular weight of 32 kg/kmol is at 300 K and 200 kPa. Determine the gas
constant and specific volume.
Solution:
Given: M=32 kg/kmolM=32 kg/kmol, T=300 KT=300 K, P=200 kPaP=200 kPa
Find: RR and vv
Analysis: R=Rˉ/MR=Rˉ/M, where Rˉ=8.314 kJ/kmol⋅KRˉ=8.314 kJ/kmol⋅K. Then v=RT/Pv=RT/P.
Calculation:
R=8.314/32=0.2598 kJ/kg⋅KR=8.314/32=0.2598 kJ/kg⋅K
v=(0.2598)(300)/200=0.3897 m3/kgv=(0.2598)(300)/200=0.3897 m3/kg
Question 9
A 2 m³ tank contains nitrogen at 500 kPa and 27°C. Determine the mass of nitrogen.
(MN2=28 kg/kmolMN2=28 kg/kmol)