This set of Thermodynamics Multiple Choice Questions & Answers (MCQs) focuses on “Pure Substance”.
1. Which of the following represents the specific volume during phase transition.
a) Vf-Vg
b) Vg-Vf
c) Vf+Vg
d) none of the mentioned
View Answer
Explanation: Here Vg is the specific volume of the saturated vapour and Vf is the specific volume of the saturated liquid.
2. At critical point, value of Vg-Vf is
a) two
b) one
c) zero
d) infinity
View Answer
Explanation: As pressure increases, there is a decrease in Vg-Vf and at critical point its value becomes zero.
3. Above the critical point, the isotherms are continuous curves.
a) true
b) false
View Answer
Explanation: These continuous curves approach equilateral hyperbolas at large volumes and low pressures.
4. A rigid tank contains 50 kg of saturated liquid water at 90°C. Determine the pressure in the tank and the volume of the tank.
a) 0.0518 m3
b) 0.0618 m3
c) 0.0718 m3
d) 0.0818 m3
View Answer
Explanation: P = Psat@90 C = 70.183 kPa
v = vf@90 C = 0.001036 m3/kg
Total volume of the tank = mv = (50kg)( 0.001036 m3/kg)
= 0.0518 m3.
5. A piston –cylinder device contains 0.06m3 of saturated water vapour at 350 kPa pressure. Determine the temperature and mass of the vapour inside the cylinder.
a) 0.104 kg
b) 0.124 kg
c) 0.134 kg
d) 0.114 kg
View Answer
Explanation: T = Tsat@350kPa = 138.86°C
v = vg@350kPa = 0.52422 m3/kg
m = V/v = 0.06 m3/0.52422 m3/kg = 0.114 kg.
6. A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in the liquid form and the rest is in the vapour form, determine the pressure in the tank.
a) 60.183 kPa
b) 70.183 kPa
c) 80.183 kPa
d) 90.183 kPa
View Answer
Explanation: P = Psat@90°C = 70.183 kPa.
7. A rigid tank contains 10 kg of water at 90°C. If 8 kg of the water is in the liquid form and the rest is in the vapour form, determine the volume of the tank.
a) 1.73 m3
b) 2.73 m3
c) 3.73 m3
d) 4.73 m3
View Answer
Explanation: P = Psat@90°C = 70.183 kPa
@ 90°C, vf = 0.001036 m3/kg and vg = 2.3593 m3/kg
V = Vf + Vg = mf vf + mg vg = 4.73 m3.
8. An 80 litre vessel contains 4 kg of R-134a at a pressure of 160 kPa. Determine the temperature.
a) -10.60°C
b) -13.60°C
c) -15.60°C
d) -19.60°C
View Answer
Explanation: v = V/m = 0.080 m3/4 kg = 0.02 m3/kg
@ 160kPa, vf = 0.0007437 m3/kg; vg = 0.12348 m3/kg
vf < v < vg Therefore T = Tsat@160kPa = -15.60°C.
9. An 80 litre vessel contains 4 kg of R-134a at a pressure of 160 kPa. Determine the quality.
a) 0.127
b) 0.137
c) 0.147
d) 0.157
View Answer
Explanation: v = V/m = 0.080 m3/4 kg = 0.02 m3/kg
@ 160kPa, vf = 0.0007437 m3/kg; vg = 0.12348 m3/kg.
vf < v < vg
x = (v –vf)/ vfg = 0.157.
10. An 80 litre vessel contains 4 kg of R-134a at a pressure of 160 kPa. Determine the volume occupied by the vapour phase.
a) 0.0775 m3
b) 0.0575 m3
c) 0.0975 m3
d) 0.0375 m3
View Answer
Explanation: v = V/m = 0.080 m3/4 kg = 0.02 m3/kg
@ 160kPa, vf = 0.0007437 m3/kg; vg = 0.12348 m3/kg
vf < v < vg
x = (v –vf)/ vfg = 0.157
mg = x*m(total) = 0.628kg
Vg = mg*vg = 0.0775 m3 or 77.5 litre.
11. Determine the specific volume of R-134a at 1 MPa and 50°C, using ideal gas equation of state.
a) 0.022325 m3/kg
b) 0.024325 m3/kg
c) 0.025325 m3/kg
d) 0.026325 m3/kg
View Answer
Explanation: v = RT/P = (0.0815 kJ/kg.K)* (323 K)/(1000 kPa)
= 0.026325 m3/kg.
Sanfoundry Global Education & Learning Series – Thermodynamics.
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