Thermodynamics Questions and Answers – Basic Concepts

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This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Basic Concepts”.

1. One kg of diatomic Oxygen is present in a 500 L tank. Find the specific volume on both mass and mole basis.
a) 0.6 m3/kg , 0.260 m3/mole
b) 0.5 m3/kg , 0.0160 m3/mole
c) 0.56 m3/kg , 0.0215 m3/mole
d) 0.7 m3/kg , 0.0325 m3/mole
View Answer

Answer: b
Explanation: The specific volume on mass basis = 0.5/1 = 0.5 m3/kg
specific volume on mole basis= 0.5/(1/0.032) = 0.0160 m3/mole.
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2. A piston/cylinder with a cross-sectional area of 0.01 m^2 is resting on the stops. With an outside pressure of 100 kPa, what should be the water pressure to lift the piston?
a) 178kPa
b) 188kPa
c) 198kPa
d) 208kPa
View Answer

Answer: c
Explanation: Pw = Po + mg/A = 100000 + (100*9.8/0.01) = 198kPa.

3. A large exhaust fan in a lab room keeps the pressure inside at 10 cm water relative vacuum to the hallway? What is the net force acting on the door measuring 1.9 m by 1.1 m?
a) 2020 N
b) 2030 N
c) 2040 N
d) 2050 N
View Answer

Answer: d
Explanation: Net force acting on the door = Gauge pressure*area
= 0.1cm water*1.9*1.1 = 2050 N.
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4. A 5 m long vertical tube having cross sectional area 200 cm^2 is placed in a water. It is filled with 15°C water, with the bottom closed and the top open to 100 kPa atmosphere. How much water is present in tube?
a) 99.9 kg
b) 109.9 kg
c) 89.9 kg
d) 79.9 kg
View Answer

Answer: a
Explanation: m = ρ V = V/v = AH/v
= 200 × 10^(−4) × 5/0.001001 = 99.9 kg.

5. A 5 m long vertical tube having cross sectional area 200 cm2 is placed in a water. It is filled with 15°C water, with the bottom closed and the top open to 100 kPa atmosphere. What is the pressure at the bottom of tube?
a) 119 kPa
b) 129 kPa
c) 139 kPa
d) 149 kPa
View Answer

Answer: d
Explanation: ∆P = ρ gH = gH/v = 9.80665 × 5/0.001001
= 48.98 kPa
P(total) = P(top) + ∆P = 149 kPa.
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6. Find the pressure of water at 200°C and having specific volume of 1.5 m3/kg.
a) 0.9578 m3/kg
b) 0.8578 m3/kg
c) 0.7578 m3/kg
d) 0.6578 m3/kg
View Answer

Answer: a
Explanation: The state is superheated vapour between 100 and 150 kPa.
v = 1.3136 + (0.8689 – 1.3136)(140 − 100)/(150 – 100)
= 1.3136 + ( − 0.4447) × 0.8 = 0.9578 m3/kg.

7. A 5m^3 container is filled with 840 kg of granite (density is 2400 kg/m^3) and the rest of the volume is air (density is 1.15 kg/m^3). Find the mass of air present in the container.
a) 9.3475 kg
b) 8.3475 kg
c) 6.3475 kg
d) 5.3475 kg
View Answer

Answer: d
Explanation: Mass of the air (mair) = ρairVair = ρair (Vtotal – Vgranite)
= ρair (Vtotal – (m/ρ)granite) = 1.15*(5 – 840/2400) = 5.3475 kg.
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8. A 100 m tall building receives superheated steam at 200 kPa at ground and leaves saturated vapour from the top at 125 kPa by losing 110 kJ/kg of heat. What should be the minimum inlet temperature at the ground of the building so that no steam will condense inside the pipe at steady state?
a) 363.54°C
b) 263.54°C
c) 163.54°C
d) none of the mentioned
View Answer

Answer: c
Explanation: FLOT for steam flow results: q + h(ground) = h(top) + gZtop.
h(top) = [email protected] kPa = 2685.35 kJ/kg
H(ground) = 2685.35 + (9.80665*100)/1000 – (–110) = 2796.33 kJ/kg
Minimum temperature at the ground of the building T(ground)
= 163.54°C.

9. The pressure gauge on an air tank shows 60 kPa when the diver is 8 m down in the ocean. At what depth will the gauge pressure be zero?
a) 34.118 m
b) 24.118 m
c) 14.118 m
d) none of the mentioned
View Answer

Answer: b
Explanation: Pressure at 10 m depth = Patm + ρgh = 101.325 + 1000*9.80665*8/1000 = 179.778 kPa
Absolute pressure of the air in the tank = 179.778 + 60 = 239.778 kPa
Depth at which gauge pressure is zero (H) = (239.778 – 101.325)*1000/(1000*9.80665) = 14.118 m.
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10. A piston-cylinder device initially contains air at 150 kPa and 27°C. At this state, the volume is 400 litre. The mass of the piston is such that a 350 kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine the final temperature.
a) 1400 K
b) 400 K
c) 500 K
d) 1500 K
View Answer

Answer: a
Explanation: Final temperature (T3)= 1400 K.–> P1V1/T1=P3V3/T3.

11. A piston-cylinder device initially contains air at 150 kPa and 27°C. At this state, the volume is 400 litre. The mass of the piston is such that a 350 kPa pressure is required to move it. The air is now heated until its volume has doubled. Determine work done by the air.
a) 120 kJ
b) 130 kJ
c) 100 kJ
d) 140 kJ
View Answer

Answer: d
Explanation: Work done = P3(V3- V1) = 140 kJ.

12. Find the change in u for carbon dioxide between 600 K and 1200 K for a constant Cv0 value.
a) 291.8 kJ/kg
b) 391.8 kJ/kg
c) 491.8 kJ/kg
d) 591.8 kJ/kg
View Answer

Answer: b
Explanation: Δu = Cv0 ΔT = 0.653 ×(1200–600) = 391.8 kJ/kg.

13. Calculate the change in enthalpy of carbon dioxide from 30 to 1500°C at 100 kPa at constant specific heat.
a) 2237.7 kJ/kg
b) 1637.7 kJ/kg
c) 1237.7 kJ/kg
d) 2337.7 kJ/kg
View Answer

Answer: c
Explanation: Δh = CpΔT = 0.842 (1500 – 30) = 1237.7 kJ/kg.

14. A sealed rigid vessel has volume of 1 m3 and contains 2 kg of water at 100°C. The vessel is now heated. If a safety pressure valve is installed, at what pressure should the valve be set to have a maximum temperature of 200°C ?
a) 431.3 kPa
b) 531.3 kPa
c) 631.3 kPa
d) 731.3 kPa
View Answer

Answer: a
Explanation: Initial specific volume (v1) = 1 m3/2 kg = 0.5 m3/kg
Interpolating, pressure for the same specific volume at 200°C
= 400 + {(0.53422-0.5)/(0.53422-0.42492)}*(500-400) = 431.3 kPa.

15. A system undergoing change in state from A to B along path ‘X’ receives 100 J heat and does 40 J work. It returns to state A from B along path ‘Y’ with work input of 30 J. Calculate the heat transfer involved along the path ‘Y’.
a) – 60 J
b) 60 J
c) – 90 J
d) 90 J
View Answer

Answer: c
Explanation: ∆UX = AQB – AWB = 100 – 40 = 60 J
Heat transfer involved along the ‘Y’ (BQA) = ∆UY + BWA = – ∆UX + BWA
= – 60 – 30 = – 90 J.

Sanfoundry Global Education & Learning Series – Thermodynamics.
To practice all areas of Thermodynamics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter