1. The relationship { \left( \frac { \partial T }{ \partial p } \right) }_{ h } = 0 holds good for
(a) An ideal gas at any state
(b) An real gas at any state
(c) Any gas at its critical state
(d) Any gas at its inversion point
(1 Mark, 1993)

Ans: a, d
2. During the phase change of a pure substance
(a) dG = 0
(b) dP = 0
(c) dH = 0
(d) dU = 0
(1 Mark, 1993)

Ans: a
Explanation:
Gibbs function: G = H – TS
Change in Gibbs function:
dG = dH – TdS – SdT ………….. (1)
From second TdS relation:
TdS = dH – VdP …………..(2)
From equation (1) and (2)
dG = VdP – SdT…………(3)
During phase change of a pure substance both temperature and pressure remain constant.
Hence, dP = dT = 0 ………..(4)
From equation 3 and 4, dG = 0
3. A vessel of volume 1.0 m^{3} contains a mixture of liquid water and steam in equilibrium at 1.0 bar. Given that 90% of the volume is occupied by the steam, find the dryness fraction of the mixture. Assume, at 1.0 bar, v_{f} = 0.001 m^{3}/kg and v_{g} = 1.7 m^{3}/kg.
(2 Mark, 1993)

Ans: 0.005266
Explanation:
A vessel of volume = 1 m^{3}
Volume of steam (V_{g}) = 90% of 1 m^{3 }= 0.9 m^{3}
Volume occupied by liquid water (V_{f}) = 0.1 m^{3}
Mass of steam (m_{g}) = \frac { { V }_{ g } }{ { v }_{ g } } =\frac { 0.9 }{ 1.7 } = 0.529 kg
Mass of liquid water (m_{f}) = \frac { { V }_{ f } }{ { v }_{ f } } =\frac { 0.1 }{ 0.001 } = 100 kg
Dryness fraction (x) = \frac { { m }_{ g } }{ { m }_{ g }+{ m }_{ f } } =\frac { 0.529 }{ 100+0.529 } = 0.005266
4. In the vicinity of the triple point, the vapour pressures of liquid and solid ammonia are respectively given by
ln P = 15.16 – 3063/T and
ln P = 18.70 – 3754/T
Where p is in atmosphere and T is in Kelvin. What is the temperature at the triple point.
(2 Mark, 1993)

Ans: 195.2 K
Explanation:
At triple point:
15.16 – 3063/T = 18.70 – 3754/T
=> T = 195.2 K
5. At the triple point of a pure substance, the number of degrees of freedom is:
(a) 0
(b) 1
(c) 2
(d) 3
(1 Mark, 1993)

Ans: a
Explanation:
At triple point:
Number of phase (P) = 3
Number of component (C) = 1
Degree of freedom (F) = C + 2 – P = 1 + 2 – 3 = 0
6. Constant pressure lines in the superheated region of the Mollier diagram will have
(a) a positive slope
(b) a negative slope
(c) Zero slope
(d) both positive and negative slope
(1 Mark, 1995)
7. When wet steam flows through a throttle valve and remains wet at exit
(a) its temperature and quality increases
(b) its temperature decreases but quality increases
(c) its temperature increases but quality decreases
(d) its temperature and quality decreases
(1 Mark, 1996)

Ans: b
Statement for Linked Answer Questions 8 & 9:
The following table of properties was printed out for saturated liquid and saturated vapour of ammonia. The titles for only the first two columns are available. All that we know is that the other columns (columns 3 to 8) contain data on specific properties, namely, internal energy (kJ/kg), enthalpy (kJ/kg) and entropy (kJ/kg.K).
8. The specific enthalpy data are in columns
(a) 3 and 7
(b) 3 and 8
(c) 5 and 7
(d) 5 and 8
(2 Mark, 2005)

Ans: d
9. When saturated liquid at 40°C is throttled to –20°C, the quality at exit will be
(a) 0.189
(b) 0.212
(c) 0.231
(d) 0.788
(2 Mark, 2005)

Ans: b
Explanation:
In throttling, Enthalpy remain constant.
Enthalpy at 40^{0}C = Enthalpy at – 20^{0}C
=> 371.43 = 89.05 + x(1418 – 89.05)
=> x = 0.212
10. Given below is an extract from steam tables.
Specific enthalpy of water in kJ/kg at 150 bar and 45°C is
(a) 203.60
(b) 200.53
(c) 196.38
(d) 188.45
(2 Mark, 2006)

Ans: a
Explanation:
Enthalpy at 150 bar and 45^{0}C:
h = { \left( { h }_{ f } \right) }_{ { 45 }^{ 0 }C }+{ \left( { v }_{ f } \right) }_{ { 45 }^{ 0 }C }\times \Delta P
=> h = 188.45+0.001010\times \left( 1500.09593 \right) \times { 10 }^{ 2 } = 203.60 kJ/kg
11. Water has a critical specific volume of 0.003155 m^{3}/kg. A closed and rigid steel tank of volume 0.025m^{3} contains a mixture of water and steam at 0.1MPa. The mass of the mixture is 10 kg. The tank is now slowly heated. The liquid level inside the tank
(a) will rise
(b) will fall
(c) will remain constant
(d) may rise or fall depending on the amount of heat transferred
(1 Mark, 2007)

Ans: a
Explanation:
Specific volume of the mixture (v) = 0.025/10 = 0.0025 m^{3}/kg
0.0025 m^{3}/kg < Critical specific volume (0.003155 m^{3}/kg)
If the tank is slowly heated at constant volume, then the pressure increases.
From the figure, it is seen that the line moving towards saturate liquid curve.
So the water portion increases.
Common Data for Questions 12, 13 and 14:
In the figure shown, the system is a pure substance kept in a piston cylinder arrangement. The system is initially a twophase mixture containing 1 kg of liquid and 0.03 kg of vapour at a pressure of 100 kPa. Initially, the piston rests on a set of stops, as shown in the figure. A pressure of 200kPa is required to exactly balance the weight of the piston and the outside atmospheric pressure. Heat transfer takes place into the system until its volume increases by 50%. Heat transfer to the system occurs in such a manner that the piston, when allowed to move, does so in a very slow (quasi – static / quasi – equilibrium) process. The thermal reservoir from which heat is transferred to the system has a temperature of 400^{0}C. Average temperature of the system boundary can be taken as 175^{0}C. The heat transfer to the system is 1 kJ, during which its entropy increases by 10 J/K.
Specific volume of liquid (v_{f}) and vapour (v_{g}) phases, as well as values of saturation temperatures, are given in the below.
12. At the end of the process, which one of the following situations will be true?
(a) superheated vapour will be left in the system
(b) no vapour will be left in the system
(c) a liquid + vapour mixture will be left in the system
(d) the mixture will exist at a dry saturate vapour state
(2 Mark, 2008)

Ans: a
Explanation:
mass of liquid (m_{f}) = 1 kg
Mass of vapour (m_{g}) = 0.3 kg
Dryness fraction (x) = \frac { { m }_{ g } }{ { m }_{ g }+{ m }_{ f } } =\frac { 0.3 }{ 1+0.3 } = 0.029
At dryness fraction x and pressure 100 kPa:
Specific volume (v_{1}) = v_{f }+ x(v_{g }– v_{f})
=> v_{1 }= 0.001 + 0.029(0.1 – 0.001) = 0.0038 m^{3}/kg
The volume increases 50%, mass remains constant.
v_{2 }= 1.5v_{1} = 1.5 * 0.0038 = 0.0057 m^{3}/kg
Specific volume of saturated vapour = 0.002 m^{3}/kg < v_{2 }
So, steam is in super heated state.
13. The work done by the system during the process is
(a) 0.1 kJ
(b) 0.2 kJ
(c) 0.3 kJ
(d) 0.4 kJ
(2 Mark, 2008)

Ans: d
Explanation:
Work done by the system (W) = mp_{2}(v_{2 }– v_{1})
=> W = 1.03 * 200(0.0057 – 0.0038) = 0.4 kJ
14. The net entropy generation (considering the system and the thermal reservoir together) during the process is closest to
(a) 7.5 J/K
(b) 7.7 J/K
(c) 8.5 J/K
(d) 10 J/K
(2 Mark, 2008)

Ans: c
Explanation:
Entropy increases in the system = { \Delta S }_{ sys } = 10 J/K
Heat transfer to the system = 1 kJ
Change in entropy in the surrounding = { \Delta S }_{ sur } = { \frac { \Delta q }{ T } }=\frac { 1000 }{ 673 } = 1.485 J/K
{ \Delta S }_{ Univ }={ { \Delta S }_{ sys }+{ \Delta S }_{ sur } }=> { \Delta S }_{ Univ }= = 10 – 1.485 = 8.515 J/K
15. The VanderWaals equation of state is \left( p+\frac { a }{ { v }^{ 2 } } \right) \left( vb \right) = RT, where p is pressure, v is specific volume, T is temperature and R is characteristic gas constant. The SI unit of a is
(a) J/kgK
(b) m^{3}/kg
(c) m^{5}/kgs^{2}
(d) Pa/kg
(1 Mark, 2015[2])

Ans: c
Explanation:
Unit of p = Unit of a/v^{2}
=> \frac { N }{ { m }^{ 2 } } =\frac { a }{ { \left( \frac { { m }^{ 3 } }{ kg } \right) }^{ 2 } }
=> \frac { kg.m{ s }^{ 2 } }{ { m }^{ 2 } } =\frac { a }{ { \left( \frac { { m }^{ 3 } }{ kg } \right) }^{ 2 } }
=> a=\frac { { m }^{ 5 } }{ kg{ s }^{ 2 } }
16. A rigid container of volume 0.5 m^{3} contains 1.0 kg of water at 120°C (v_{f} = 0.00106 m^{3}/kg, v_{g} = 0.8908 m^{3}/kg). The state of water is
(a) compressed liquid
(b) saturated liquid
(c) a mixture of saturated liquid and saturated vapour
(d) superheated vapour
(1 Mark, 2015[3])

Ans: c
Explanation:
Specific volume inside the container (v) = V/m = 0.5/1 = 0.5 m^{3}/kg
As v_{f }< v < v_{g}
So,It is a mixture of saturated liquid and saturated vapour.
16. A mixture of ideal gases has the following composition by mass:
If the universal gas constant is 8314 J/kmolK, the characteristic gas constant of the mixture (in J/kgK) is ____.
(2 Mark, 2015[3])

Ans: 275
Explanation:
{ R }_{ C }=\left( \frac { 0.6 }{ 28 } +\frac { 0.3 }{ 32 } +\frac { 0.1 }{ 44 } \right) \times 8314 = 274.86 J/kgK
17. The INCORRECT statement about the characteristics of critical point of a pure substance is that
(a) there is no constant temperature vaporization process
(b) it has point of inflection with zero slope
(c) the ice directly converts from solid phase to vapour phase
(d) saturated liquid and saturated vapour states are identical
(1 Mark, 2016[3])

Ans: c
Explanation:
At critical point, Solid state doesn’t convert to vapour state. Liquid state converts to vapour state.
18. Which one of the following statements is correct for a superheated vapour?
(a) Its pressure is less than the saturation pressure at a given temperature.
(b) Its temperature is less than the saturation temperature at a given pressure.
(c) Its volume is less than the volume of the saturated vapour at a given temperature.
(d) Its enthalpy is less than the enthalpy of the saturated vapour at a given pressure.
(1 Mark, 2018[1])
19. A tank of volume 0.05 m^{3} contains a mixture of saturated water and saturated steam at 200^{0}C. The mass of the liquid present is 8 kg. The entropy (in kJ/kg K) of the mixture is ____ (correct to two decimal places).
Property data for saturated steam and water are:
At 200^{0}C, p_{sat} = 1.5538 MPa
v_{f} = 0.001157 m^{3}/kg, v_{g} = 0.12736 m^{3}/kg
s_{fg} = 4.1014 kJ/kg K, s_{f} = 2.3309 kJ/kg K
(2 Mark, 2018[1])

Ans: 2.488
Explanation:
Volume of liquid:
V_{f} = { m }_{ f }\times { v }_{ f }=8\times 0.001157=0.009256{ m }^{ 3 }
Volume of vapour:
V_{g }= 0.05 – 0.009256 = 0.040744 m^{3}
Mass of vapour:
{ m }_{ g }=\frac { { V }_{ g } }{ { v }_{ g } } =\frac { 0.040744 }{ 0.12736 } =0.3199kgDryness fraction:
x=\frac { { m }_{ g } }{ { m }_{ f }+{ m }_{ g } } =\frac { 0.3199 }{ 8.3199 } =0.03845Entropy of the mixture: s = s_{f} + x. s_{fg }= 2.3309 + 0.03845 \times 4.1014 = 2.488 kg/kgK
20. For a simple compressible system, v, s, p and T are specific volume, specific entropy, pressure and temperature, respectively. As per Maxwell’s relation, { \left( \frac { \partial v }{ \partial s } \right) }_{ P } is equal to
(a) { \left( \frac { \partial s }{ \partial T } \right) }_{ P }
(b) { \left( \frac { \partial p }{ \partial v } \right) }_{ T }
(c) { \left( \frac { \partial T }{ \partial v } \right) }_{ P }
(d) { \left( \frac { \partial T }{ \partial p } \right) }_{ s }
(1 Mark, 2019[2])

Ans: d