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 m^{3}

b) 0.0618 m^{3}

c) 0.0718 m^{3}

d) 0.0818 m^{3}

View Answer

Explanation: P = [email protected] C = 70.183 kPa

v = [email protected] C = 0.001036 m

^{3}/kg

Total volume of the tank = mv = (50kg)( 0.001036 m

^{3}/kg)

= 0.0518 m

^{3}.

5. A piston –cylinder device contains 0.06m^{3} 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 = [email protected] = 138.86°C

v = [email protected] = 0.52422 m

^{3}/kg

m = V/v = 0.06 m

^{3}/0.52422 m

^{3}/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 = [email protected]°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 m^{3}

b) 2.73 m^{3}

c) 3.73 m^{3}

d) 4.73 m^{3}

View Answer

Explanation: P = [email protected]°C = 70.183 kPa

@ 90°C, vf = 0.001036 m

^{3}/kg and vg = 2.3593 m

^{3}/kg

V = Vf + Vg = mf vf + mg vg = 4.73 m

^{3}.

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 m

^{3}/4 kg = 0.02 m

^{3}/kg

@ 160kPa, vf = 0.0007437 m

^{3}/kg; vg = 0.12348 m

^{3}/kg

vf < v < vg Therefore T = [email protected] = -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 m

^{3}/4 kg = 0.02 m

^{3}/kg

@ 160kPa, vf = 0.0007437 m

^{3}/kg; vg = 0.12348 m

^{3}/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 m^{3}

b) 0.0575 m^{3}

c) 0.0975 m^{3}

d) 0.0375 m^{3}

View Answer

Explanation: v = V/m = 0.080 m

^{3}/4 kg = 0.02 m

^{3}/kg

@ 160kPa, vf = 0.0007437 m

^{3}/kg; vg = 0.12348 m

^{3}/kg

vf < v < vg

x = (v –vf)/ vfg = 0.157

mg = x*m(total) = 0.628kg

Vg = mg*vg = 0.0775 m

^{3}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 m^{3}/kg

b) 0.024325 m^{3}/kg

c) 0.025325 m^{3}/kg

d) 0.026325 m^{3}/kg

View Answer

Explanation: v = RT/P = (0.0815 kJ/kg.K)* (323 K)/(1000 kPa)

= 0.026325 m

^{3}/kg.

**Sanfoundry Global Education & Learning Series – Thermodynamics.**

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