1. Which among the following relations is/are valid only for reversible process undergone by a pure substance?
(a) δQ = dU + δW
(b) TdS = dU + δW
(c) TdS = dU + pdv
(d) δQ = pdv + dU
(1 Mark, 1993)

Ans: d
1. Air expands steadily through a turbine from 6 bar, 800 K to 1 bar, 520 K. During the expansion, heat transfer from air to the surroundings at 300 K is 10 kJ/kg air. Neglect the changes in kinetic and potential energies and evaluate the irreversibility per kg air. Assume air to behave as an ideal gas with C_{p} = 1.0 kJ/kgK and R = 0.3 kJ/kgK.
(2 Mark, 1993)

Ans: 42.023
We know, I={ T }_{ 0 }{ \left( \Delta s \right) }_{ universe }
=> I={ T }_{ 0 }\left\{ { \left( \Delta s \right) }_{ system }+{ \left( \Delta s \right) }_{ surrounding } \right\}
=> I={ T }_{ 0 }\left\{ { c }_{ p }ln\left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) Rln\left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right) +\frac { \delta q }{ { T }_{ 0 } } \right\}
=> I=300\left\{ 1\times ln\left( \frac { 520 }{ 800 } \right) 0.3\times ln\left( \frac { 1 }{ 6 } \right) +\frac { 10 }{ 300 } \right\}
=> I = 42.023 kJ/kg
2. In the previous question, find the actual work and maximum work per kg air.
(2 Mark, 1993)

Ans: 280, 322.023
Maximum Work { (W }_{ actual })={ c }_{ p }\left( \Delta T \right)
=> { W }_{ actual }=1\times \left( 800520 \right) = 280 kJ/kg
We know, { { W }_{ max }=W }_{ actual }+I
=> { W }_{ max } = 280 + 42.023 = 322.023 kJ/kg
2. The slopes of constant volume and constant pressure lines in the T – s diagram are ___and ___ respectively.
(1 Mark, 1994)

Ans: T/ C_{v}, T/ C_{p}
Explanation:
We know: Tds = du + Pdv
For constant volume:
Tds = du
=> Tds = C_{v }dT [As, du = C_{v }dT]
=> \frac { dT }{ ds } =\frac { T }{ { C }_{ v } }
Again: Tds = dh – vdp
For constant Pressure:
Tds = dh
=> Tds = C_{p }dT [As, dh = C_{p }dT]
=> \frac { dT }{ ds } =\frac { T }{ { C }_{ p } }
3. A 1500 W electrical heater is used to heat 20 kg of water (C_{p} = 4186 J/kgK) in an insulated bucket, from a temperature of 30°C to 80°C. If the heater temperature is only infinitesimally larger than the water temperature during the process, the change in entropy for heater is ____ J/K and for water ____ J/K.
(1 Mark, 1994)

Ans: ds_{water }= 12787 J/kg
ds_{heater }= – 12787 J/kg
Explanation:
ds_{water }= mC_{p }ln\frac { { T }_{ 2 } }{ { T }_{ 1 } }
=> ds_{water }= (20)(4186) ln\frac { 353 }{ 303 }
=> ds_{water }= 12787 J/kg
It is given that, the heater temperature is only infinitesimally larger than the water temperature. i.e the process is a reversible process. So, no entropy generation.
Hence, ds_{heater }= – ds_{water }= – 12787 J/kg
4. One kilomole of an ideal gas is throttled from an initial pressure of 0.5 MPa to 0.1 MPa. The initial temperature is 300 K. The entropy change of the universe is:
(a) 13.38 kJ/K
(b) 401.3 kJ/K
(c) 0.0446 kJ/K
(d) 0.0446 kJ/K
(2 Mark, 1995)

Ans: a
Explanation:
Throttling is an isenthalpic process.
As h = f(T). So, the temperature will not change in the process.
The entropy change of the universe (ds) = nR.ln\left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right)
=> ds = { 10 }^{ 3 }\times 8.314\times ln\left( \frac { 0.1 }{ 0.5 } \right)
=> ds = 13.38 kJ/K
5. A heat reservoir at 900 K is brought into contact with the ambient at 300 K for a short time. During this period 9000 kJ of heat is lost by the heat reservoir. The total loss in availability due to this process is:
(a) 18000 kJ
(b) 9000 kJ
(c) 6000 kJ
(d) None of the above
(2 Mark, 1995)

Ans: c
Explanation:
The total loss in availability (LA)= Q\left( 1\frac { { T }_{ 0 } }{ { T }_{ 1 } } \right)
=> LA = 9000\times \left( 1\frac { 300 }{ 900 } \right)
=> LA = 6000 kJ
6. For an ideal gas the expression \left[ T{ \left( \frac { \partial s }{ \partial T } \right) }_{ P }T{ \left( \frac { \partial s }{ \partial T } \right) }_{ V } \right] is always equal to
(a) Zero
(b) C_{p}/ C_{V}
(c) R
(d) RT
(1 Mark, 1997)

Ans: c
Explanation:
Tds = du + pdv
For constant volume process:
Tds = du => Tds = C_{v}dT
=> T{ \left( \frac { \partial s }{ \partial T } \right) }_{ V } = C_{v}
Again, Tds = dh – vdp
For constant pressure process:
Tds = dh => Tds = C_{p}dT
=> T{ \left( \frac { \partial s }{ \partial T } \right) }_{ p } = C_{p}
\left[ T{ \left( \frac { \partial s }{ \partial T } \right) }_{ P }T{ \left( \frac { \partial s }{ \partial T } \right) }_{ V } \right] = C_{p }– C_{v }= R
7. A system undergoes a state change from 1 to 2. According the second law of thermodynamics for the process to be feasible, the entropy change, S_{2} – S_{1} of the system
(a) is positive or zero
(b) is negative or zero
(c) is zero
(d) can be positive, negative or zero
(1 Mark, 1997)

Ans: d
Explanation:
Entropy of a system can be positive, negative or zero.
Entropy of Universe can be positive for irreversible process and zero for reversible process but can not be negative.
8. Availability of a system at any given state is
(a) a property of the system
(b) the maximum work obtainable as the system goes to dead state
(c) the total energy of the system
(d) the maximum useful work obtainable as the system goes to dead state
(1 Mark, 2000)

Ans: d
9. When an ideal gas with constant specific heats is throttled adiabatically, with negligible changes in kinetic and potential energies
(a) \Delta h=0,\quad \Delta T=0
(b) \Delta h>0,\quad \Delta T=0
(c) \Delta h>0,\quad \Delta s>0
(d) \Delta h=0,\quad \Delta s>0
Where h, T and s represent respectively, enthalpy, temperature and entropy.
(2 Mark, 2000)

Ans: d
Explanation:
Throttling is a irreversible isenthalpic process.
So, in this process the entropy increases and the enthalpy remain constant.
10. Considering the relationship TdS = dU + pdV between the entropy (S), internal energy (U), pressure (p), temperature (T) and volume (V), which of the following statements is correct?
(a) It is applicable only for a reversible process
(b) For an irreversible process, TdS > dU + pdV
(c) It is valid only for an ideal gas
(d) It is equivalent to 1 law, for a reversible process
(2 Mark, 2003)

Ans: d
Data for Q.11 – 12 are given below.
Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of 2 bar, 298 K and 1 m^{3}. In a particular process, the gas slowly expands under isothermal condition, until the volume becomes 2 m^{3}. Heat exchange occurs with the atmosphere at 298 K during this process.
11. The work interaction for the Nitrogen gas is
(a) 200 kJ
(b) 138.6 kJ
(c) 2 kJ
(d) 200 kJ
(2 Mark, 2003)

Ans: b
Explanation:
The work interaction (\delta W)= { p }_{ 1 }{ V }_{ 1 }ln\left( \frac { { V }_{ 2 } }{ { V }_{ 1 } } \right)
=> \delta W = (2\times { 10 }^{ 5 })\times 1\times ln\left( \frac { 2 }{ 1 } \right)
=> \delta W = 138.6 kJ
12. The entropy change for the Universe during the process in kJ/K is
(a) 0.4652
(b) 0.0067
(c) 0
(d) 0.6711
(2 Mark, 2003)

Ans: c
Explanation:
As the process is isothermal:
Entropy change for the system = \frac { \delta Q }{ T }
Entropy change for the surrounding = \frac { \delta Q }{ T }
Entropy change for the universe = Entropy change for the system + Entropy change for the surrounding = \frac { \delta Q }{ T } + \left( \frac { \delta Q }{ T } \right) = 0
13. Which of the following relationships is valid only for reversible processes undergone by a closed system of simple compressible substance (neglect changes in kinetic and potential energy)?
(a) δQ = dU + δW
(b) Tds = dU + pdv
(c) Tds = dU + δW
(d) δQ = dU + pdv
(1 Mark, 2007)

Ans: d
14. A cyclic device operates between three thermal reservoirs, as shown in the figure. Heat is transferred to/form the cyclic device. It is assumed that heat transfer between each thermal reservoir and the cyclic device takes place across negligible temperature difference. Interactions between the cyclic device and the respective thermal reservoirs that are shown in the figure are all in the form of heat transfer.
The cyclic device can be
(a) a reversible heat engine
(b) a reversible heat pump or a reversible refrigerator
(c) an irreversible heat engine
(d) an irreversible heat pump or an irreversible refrigerator
(2 Mark, 2008)

Ans: a
Explanation:
\oint { \frac { \delta Q }{ T } } =\frac { { Q }_{ 1 } }{ { T }_{ 1 } } +\frac { { Q }_{ 2 } }{ { T }_{ 2 } } \frac { { Q }_{ 3 } }{ { T }_{ 3 } } =\frac { 100 }{ 1000 } +\frac { 50 }{ 500 } \frac { 60 }{ 300 } =0So, the cyclic device is reversible.
Work one by the cyclic process = Heat interaction = 100 + 50 – 60 = 90 kJ
As the work done is positive, so the cyclic device is a heat engine.
15. 2 moles of oxygen are mixed adiabatically with another 2 moles of oxygen in mixing chamber, so that the final total pressure and temperature of the mixture become same as those of the individual constituents at their initial states. The universal gas constant is given as R. The change in entropy due to mixing, per mole of oxygen, is given by
(a) –Rln2
(b) 0
(c) Rln2
(d) R ln4
(1 Mark, 2008)

Ans: b
Explanation:
{ \left( \Delta s \right) }_{ universe }={ \left( \Delta s \right) }_{ system }+{ \left( \Delta s \right) }_{ surrounding }As the mixing occurs with in the mixing chamber, So { \left( \Delta s \right) }_{ surrounding } = 0
Now, { \left( \Delta s \right) }_{ system } = { C }_{ p }ln\left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) Rln\left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right)
As the temperature and pressure are not changing. So,
{ \left( \Delta s \right) }_{ system } = 0
hence, { \left( \Delta s \right) }_{ universe }={ \left( \Delta s \right) }_{ system }+{ \left( \Delta s \right) }_{ surrounding } = 0 + 0 = 0
17. If a closed system is undergoing an irreversible process, the entropy of the system
(a) Must increase
(b) Always remains constant
(c) Must decrease
(d) Can increase, decrease or remain constant
(1 Mark, 2009)

Ans: d
Explanation:
Entropy of a system can be positive, negative or zero.
Entropy of Universe can be positive for irreversible process and zero for reversible process but can not be negative.
18. One kilogram of water at room temperature is brought into contact with a high temperature thermal reservoir. The entropy change of the universe is
(a) equal to entropy change of the reservoir
(b) equal to entropy change of water
(c) equal to zero
(d) always positive
(1 Mark, 2010)

Ans: d
Explanation:
The entropy of the universe always increases for irreversible processes.
19. Consider the following two processes:
I. A heat source at 1200 K loses 2500 kJ of heat to sink at 800 K.
II. A heat source at 800 K loses 2000 kJ of heat to sink at 500 K.
Which of the following statements is TRUE?
(a) Process I is more irreversible than Process II
(b) Process II is more irreversible than Process I
(c) Irreversibility associated in both the processes is equal
(d) Both the processes are reversible
(2 Mark, 2010)

Ans: b
Explanation:
For process I:
{ \Delta S }_{ I }=\frac { { Q }_{ I } }{ { T }_{ I } } +\frac { { Q }_{ I } }{ { T }_{ 2 } } =\frac { 2500 }{ 1200 } +\frac { 2500 }{ 800 } =1.04kJ/KFor process II:
{ \Delta S }_{ II }=\frac { { Q }_{ II } }{ { T }_{ I } } +\frac { { Q }_{ II } }{ { T }_{ 2 } } =\frac { 2000 }{ 800 } +\frac { 2000 }{ 500 } =1.5kJ/KSince entropy change in process II is greater. Hence process II is more irreversible than process I.
Common Data Questions: 20 & 21
In an experimental setup, air flows between two stations P and Q adiabatically. The direction of flow depends on the pressure and temperature conditions maintained at P and Q. The conditions at station P are 150 kPa and 350 K. The temperature at station Q is 300 K. The following are the properties and relations pertaining to air:
Specific heat at constant pressure, C_{p} = 1.005 kJ /kgK;
Specific heat at constant volume, C_{v} = 0.718 kJ /kgK;
Characteristic gas constant, R = 0.287 kJ /kgK
Enthalpy, h = C_{p}T, Internal energy, u = C_{v}T
20. If the air has to flow from station P to station Q, the maximum possible value of pressure in kPa at station Q is close to
(a) 50
(b) 87
(c) 128
(d) 150
(2 Mark, 2011)

Ans: b
Explanation:
Since the flow between two station is adiabatic. So,
\frac { { T }_{ 2 } }{ { T }_{ 1 } } ={ \left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right) }^{ \frac { \gamma 1 }{ \gamma } }=> \frac { 300 }{ 350 } ={ \left( \frac { { p }_{ 2 } }{ 150 } \right) }^{ \frac { 0.4 }{ 1.4 } }
After solving,
P_{2 }= 87.45 kpa
21. If the pressure at station Q is 50 kPa, the change in entropy (S_{Q} – S_{P}) in kJ/kgK is
(a) 0.155
(b) 0
(c) 0.160
(d) 0.355
(2 Mark, 2011)

Ans: c
Explanation:
S_{Q} – S_{P }= { C }_{ p }ln\frac { { T }_{ 2 } }{ { T }_{ 1 } } Rln\frac { { p }_{ 2 } }{ { p }_{ 1 } }
=> S_{Q} – S_{P }= 1.005\times ln\frac { 300 }{ 350 } 0.287\times ln\frac { 50 }{ 150 } = 0.160 kJ/kgK
22. An ideal gas of mass m and temperature T_{1} undergoes a reversible isothermal process from an initial pressure P_{1} to final pressure P_{2}. The heat loss during the process is Q. The entropy change \Delta S of the gas is
(a) mR ln\left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right)
(b) mR ln\left( \frac { { p }_{ 1 } }{ { p }_{ 2 } } \right)
(c) mR ln\left( \frac { { p }_{ 2 } }{ { p }_{ 1 } } \right)  \frac { Q }{ { T }_{ 1 } }
(d) Zero
(2 Mark, 2012)

Ans: b
Explanation:
S_{2} – S_{1 }= m{ C }_{ p }ln\frac { { T }_{ 2 } }{ { T }_{ 1 } } mRln\frac { { p }_{ 2 } }{ { p }_{ 1 } }
For constant temperature:
S_{2} – S_{1 }= mRln\frac { { p }_{ 2 } }{ { p }_{ 1 } }
=> S_{2} – S_{1 }= mRln\frac { { p }_{ 1 } }{ { p }_{ 2 } }
23. The pressure, temperature and velocity of air flowing in a pipe are 5 bar, 500 K and 50 m/s, respectively. The specific heats of air at constant pressure and at constant volume are 1.005 kJ/kgK and 0.718 kJ/kgK, respectively. Neglect potential energy. If the pressure and temperature of the surroundings are 1 bar and 300 K, respectively, the available energy in kJ/kg of the air stream is
(a) 170
(b) 187
(c) 191
(d) 213
(2 Mark, 2013)

Ans: b
Explanation:
Available energy (\varphi ) = \left( { h }_{ 1 }{ h }_{ 0 } \right) { T }_{ 0 }\left( { s }_{ 1 }{ s }_{ 0 } \right) +\frac { { v }_{ 1 }^{ 2 } }{ 2000 }
Where,
{ s }_{ 1 }{ s }_{ 0 }={ C }_{ p }ln\frac { { T }_{ 1 } }{ { T }_{ 0 } } Rln\frac { { p }_{ 1 } }{ { p }_{ 0 } }=> { s }_{ 1 }{ s }_{ 0 }=1.005\times ln\frac { 500 }{ 300 } 0.287ln\frac { 5 }{ 1 } = 0.05147 kJ/kgK
Now, \varphi =1.005\times \left( 500300 \right) 300\left( 0.05147 \right) +\frac { { 50 }^{ 2 } }{ 2000 } = 187 kJ/kg
24. The maximum theoretical work obtainable, when a system interacts to equilibrium with a reference environment, is called
(a) Entropy
(b) Enthalpy
(c) Exergy
(d) Rothalpy
(2 Mark, 2014[1])

Ans: c
25. Which one of the following pairs of equations describes an irreversible heat engine?
26. An amount of 100 kW of heat is transferred through a wall in steady state. One side of the wall is maintained at 127°C and the other side at 27°C. The entropy generated (in W/K) due to the heat transfer through the wall is _____.
(2 Mark, 2014[3])

Ans: 83.33
27. A source at a temperature of 500 K provides 1000 kJ of heat. The temperature of environment is 27°C. The maximum useful work (in kJ) that can be obtained from the heat source is ____.
(2 Mark, 2014[3])

Ans: 400
Explanation:
W_{max }= Q_{1}(1\frac { { T }_{ 2 } }{ { T }_{ 1 } })
=> W_{max }= Q_{1}(1\frac { 300 }{ 1000 } ) = 400 kJ
28. A closed system contains 10 kg of saturated liquid ammonia at 10°C. Heat addition required to convert the entire liquid into saturated vapour at a constant pressure is 16.2 MJ. If the entropy of the saturated liquid is 0.88 kJ/kg.K, the entropy (in kJ/kg.K) of saturated vapour is ____.
(2 Mark, 2014[4])

Ans: 6.6
Explanation:
We know: dQ = TdS
=> dQ = T.m(s_{g} – s_{f})
=> 16.2\times { 10 }^{ 3 }=283\times 10\left( { s }_{ g }0.88 \right)
=> s_{g }= 6.604 kJ/kgK
29. Two identical metal blocks L and M (specific heat = 0.4 kJ/kg.K), each having a mass of 5 kg, are initially at 313 K. A reversible refrigerator extracts heat from block L and rejects heat to block M until the temperature of block L reaches 293 K. The final temperature (in K) of block M is _____.
(2 Mark, 2014[4])

Ans: 334.36
Explanation:
Given: Initial temperature of both the blocks (T_{i})= 313 K
final temperature of block L (T_{2}) = 293 K
Let the final temperature of block M = T_{f}
As a reversible refrigerator is used, So { \Delta S }_{ gen }=0
=> { \left( \Delta S \right) }_{ L }+{ \left( \Delta S \right) }_{ m }=0
=> mc.ln\frac { { T }_{ 2 } }{ { T }_{ i } } +mc.ln\frac { { T }_{ f } }{ { T }_{ i } } =0
=> \frac { { T }_{ 2 }{ T }_{ f } }{ { T }_{ i }^{ 2 } } =1
=> { T }_{ f }=\frac { { T }_{ i }^{ 2 } }{ { T }_{ 2 } } =\frac { { 313 }^{ 2 } }{ { 293 } } =334.36K
30. One kg of air (R = 287 J/kgK) undergoes an irreversible process between equilibrium state 1 (20°C, 0.9 m^{3}) and equilibrium state 2 (20°C, 0.6 m^{3}). The change in entropy s_{2} − s_{1} (in J/kgK) is
(2 Mark, 2015[2])

Ans: 116.368
Explanation:
As the temperature remain constant:
s_{2} − s_{1 }= Rln\frac { { v }_{ 2 } }{ { v }_{ 1 } }
=> s_{2} − s_{1 }= 287\times ln\frac { 0.6 }{ 0.9 }
=> s_{2} − s_{1 }= 116.368 J/kgK.
31. One side of a wall is maintained at 400 K and the other at 300 K. The rate of heat transfer through the wall is 1000 W and the surrounding temperature is 25°C. Assuming no generation of heat within the wall, the irreversibility (in W) due to heat transfer through the wall is _____.
(2 Mark, 2015[3])

Ans: 248.33
Explanation:
{ \Delta S }_{ system }=\frac { Q }{ { T }_{ 2 } } \frac { Q }{ { T }_{ 1 } }=> { \Delta S }_{ system }=\frac { 1000 }{ 300 } \frac { 1000 }{ 400 } =0.833W/K
Irreversibility (I) = { T }_{ 0 }{ \Delta S }_{ universe }
Where, T_{0 }= Atmospheric temperature
So, I = { T }_{ 0 }\left( { \Delta S }_{ system }+{ \Delta S }_{ surrounding } \right)
=> I = 298\left( 0.833+0 \right) = 248.33 W
32. One kg of an ideal gas (gas constant, R = 400 J/kg.K; Specific heat at constant volume C_{v }= 1000 J/kgK) at 1 bar and 300 K is contained in a sealed rigid cylinder. During an adiabatic process, 100 kJ of work is done on the system by a stirrer. The increase in entropy of the system is ___ J/K.
(2 Mark, 2017[1])

Ans: 287.68
Explanation:
Using first law of thermodynamics:
dQ = dU + dW
Adiabatic process, dQ = 0
Work is one on the system, so dW = 100 kJ
=> 0 = dU – 100
=> dU = 100
=> mC_{v}dT = 100
=> dT = \frac { dW }{ m{ C }_{ v } } =\frac { 100\times { 10 }^{ 3 } }{ 1\times 1000 } = 100
T_{2 }= T1 + dT = 300 + 100 = 400 K
Increase in entropy of the system:
{ S }_{ 2 }{ S }_{ 1 }=m{ C }_{ v }.ln\frac { { T }_{ 2 } }{ { T }_{ 1 } } =1\times 1000\times ln\frac { 400 }{ 300 } =287.68J/K
33. One kg of an ideal gas (gas constant R = 287 J/kg.K) under goes an irreversible process from stateI (1 bar, 300 K) to state 2 (2 bar, 300 K). The change in specific entropy (s_{2} − s_{1}) of the gas (in J/kg.K) in the process is _____.
(2 Mark, 2017[2])

Ans: 198.93
Explanation:
As the temperature is not changing:
{ s }_{ 2 }{ s }_{ 1 }=R.ln\frac { { p }_{ 2 } }{ { p }_{ 1 } }=> { s }_{ 2 }{ s }_{ 1 }=287.ln\frac { 2 }{ 1 } = 198.93 J/kgK
34. An ideal gas undergoes a process from state 1 (T_{1} = 300 K, p_{1} = 100 kPa) to state 2 (T_{2} = 600 K, p_{2} = 500 kPa). The specific heats of the ideal gas are: c_{p} = 1 kJ/kgK and c_{v} = 0.7 kJ/kgK. The change in specific entropy of the ideal gas from state 1 to state 2 (in kJ/kgK) is ___________(correct to two decimal places).
(1 Mark, 2018[1])

Ans: 0.21
Explanation:
R = C_{p }– C_{v }= 1 – 0.7 = 0.3 kJ/kgK
\Delta s={ C }_{ p }.{ ln\left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right)  }R.ln\left( \frac { p_{ 2 } }{ { p }_{ 1 } } \right)=> \Delta s=1\times { ln\left( \frac { 600 }{ 300 } \right)  }0.3\times ln\left( \frac { 500 }{ 100 } \right) = 0.21 kJ/kgK
35. For an ideal gas with constant properties undergoing a quasistatic process, which one of the following represents the change of entropy ( s) from state 1 to 2?
(a) s ={ C }_{ p }ln\left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right)  Rln\left( \frac { { P }_{ 2 } }{ { P }_{ 1 } } \right)
(b) s = { C }_{ v }ln\left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right)  { C }_{ p }ln\left( \frac { { V }_{ 2 } }{ { V }_{ 1 } } \right)
(c) s = { C }_{ p }ln\left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right)  { C }_{ v }ln\left( \frac { P_{ 2 } }{ { P }_{ 1 } } \right)
(d) s = { C }_{ v }ln\left( \frac { { T }_{ 2 } }{ { T }_{ 1 } } \right) + Rln\left( \frac { V_{ 1 } }{ V_{ 2 } } \right)
(1 Mark, 2018[2])

Ans: a
36. Air of mass 1 kg, initially at 300K and 10 bar, is allowed to expand isothermally till it reaches a pressure of 1 bar. Assuming air as an ideal gas with gas constant of 0.287 kJ/kg.K, the change in entropy of air (in kJ/kg.K, round off to two decimal places) is _____.
(1 Mark, 2019[1])

Ans: 0.66 kJ/kgK
Explanation:
As the temperature is not changing:
{ s }_{ 2 }{ s }_{ 1 }=R.ln\frac { { p }_{ 2 } }{ { p }_{ 1 } }=> { s }_{ 2 }{ s }_{ 1 }=0.287.ln\frac { 1 }{ 10 } = 0.66 kJ/kgK
37. Water flowing at the rate of 1 kg/s through a system is heated using an electric heater such that the specific enthalpy of the water increases by 2.50 kJ/kg and the specific entropy increases by 0.007 kJ/kg K. The power input to the electric heater is 2.50 kW. There is no other work or heat interaction between the system and the surroundings, Assuming an ambient temperature of 300 K, the irreversibility rate of the system is ____ kW (round off to two decimal places).
(2 Mark, 2019[2])

Ans: 2.1 kW
Explanation:
{ \left( { \delta W }_{ max } \right) }_{ useful }={ \phi }_{ 1 }{ \phi }_{ 2 }=> { \left( { \delta W }_{ max } \right) }_{ useful }=\dot { m } \left( \Delta h{ T }_{ 0 }\Delta s \right)
=> { \left( { \delta W }_{ max } \right) }_{ useful } =1\times \left( 2.5300\times 0.007 \right) =0.4kW
Irreversibility (I) = Power input{ \left( { \delta W }_{ max } \right) }_{ useful } = 2.5 – 0.4 = 2.1 kW