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GATE Previous Years Solved Papers The Gate Coach

30 The Gate Coach |All Rights Reserved

(Gate 2001)

1. A reasonably general expression for

vapour liquid phase equilibrium a low

modern pressure is 0

i i i i iy P x f where is a vapour

fugacity coefficient i is the liquid activity

coefficient , and o

if is the fugacity of pure

component i. the KI value i i iy K x is

therefore, in general , a function of

(a) temperature only

(b) temperature and pressure only

(c) temperature, pressure liquid

composition xi only

(d) temperature , pressure liquid

composition xi and vapour composition

yi

2. High pressure steam is expanded

adiabatically and reversibly through a well

insulated turbine which produces some

shaft work. If the enthalpy change and

entropy change across the turbine are

represented by H and S . Respectively

for this process:

(a) H 0 & S 0

(b) H 0 & S 0

(c) H 0 & S 0

(d) H 0 & S 0

3. The maxwell relation derived form the

differential expression for the helmohtz free

energy (dA) is

(a) s v

T T

V S

(b) T v

S V

P T

(c) P s

V T

S P

(d) T v

S P

V T

4. At 1000C, water and methyleyclohexane

both have vapour pressures of 1.0 atm. also

at 100 0C the latent heats of vaporization of

these compounds are 40.63 kJ/mol for

water and 31.55 kJ/mol for methyl

cyclohexane. The vapour pressure of water

at 150 0C is 0C , the vapour pressure of meth

cyclohexane would be expected to be

(a) Significantly less than 4.69 atm

(b) Nearly equal to 4.69 atm

(c) Significantly more than 4.69 atm

(d) Indeterminate due to a lack of data

CHAPTER 2

• CHEMICAL ENGINEERING THERMODYNAMICS



5. Air enters an adiabatic compressor at

300 K. the exit temperature for a

compression ratio of 3, assuming air to be

an ideal gas p v(y C / C 7 / 5) and the

process to be reversible is

(a) 300 (3 m2/7) (b) 300 (3 m1/5)

(c) 300 (3 m3/7) (d) 300 (3 m5/7)

(Gate 2002)

6. Which of the following conditions are

satisfied at the critical point by the P-V-T

relation of a real fluid?

(a) 2

2

TT

P P0

V V

(b) 2

2

TT

P P0, 0

V V

(c) 2

2

TT

P P0, 0

V V

(d) 2

2

TT

P P0, 0

V V

7. The number of degrees of freedom for an

azeotropic mixture of ethanol and water in

vapour liquid equilibrium is

(a) 3 (b) 1

(c) 2 (d) 0

8. The partial molar enthalpy of a

component in an ideal binary gas mixture of

compositional temperature T and pressure

P, is a function only of

(a) T (b) T and P

(c) T, P and Z (d) T and Z

9. Which of the following identities can be

most easily used to verify steam table data

for superheated steam?

(a) S V

T / V P / S

(b) S P

T / P V / S

(c) V T

P / T S/ V

(d) P T

V / T S/ P

10. Steam undergoes isentropic expansion

in a turbine form 5000 kPa and 400 0C

(entropy = 6.65 kJ/kg K) to 150 k Pa

(entropy of saturated liquid = 1.4336 kJ/kg

K, entropy of saturated vapour = 7.2234

kJ/kg K). The eixt condition of steam is

(a) superheatedvpaour

(b) partially condensed vapour with quality

of 0.9

(c) saturatedvapour

(d) partially condensed vapour with quality

of 0.1

11. A rigid vessel, containing three moles of

nitrogen gas at 30 0C, is heated to 250 0C .

assume the average heat capacities of

nitrogen to be required, neglecting the heat

capacity of the vessel, is

0 0

p vC 29.1 J / mol C and C 20.8J / mol C. The heat

required neglecting the heat capacity of the

vessel, is

(a) 13728 J (b) 19206 J

(c) 4576 J (d) 12712 J

12. 1 m3 of an ideal gas at 500 K and 1000

kPa expands reversibly to 5 times its

intialvolume in an insulated container. If

the specific heat capacity (at constant

pressue) of the gas is 21J/mol



(a) 35 K (b) 174 K

(c) 274 K (d) 154 K

(Gate 2003)

13. for organic estimation of heat capacity

of a solid compound, one can use

(a) clapeyron’s equation

(b) gibb’s equation

(c) kopp’s rule

(d) Troution’s rule

14. One mole of nitrogen at 8 bar and 600 K

is contained is a piston cylinder

arrangement. It is brought to 1 bar

isothermally against a resisting pressure of 1

bar. The work done (in joules) by the gas is

(a) 30554 (b) 10373

(c) 4988.4 (d) 4364.9

15. For water at 3000C , it has a vapour

pressure 8592.7 kPa and fugacity 6738.9

kPa under these conditions, one mole of

water in liquid phase has a volume 25.28

cm3 and that in vapour phase 391.1 cm3.

Fugacity of water (in kPa at 9000 kpa will

be

(a) 673.89 (b) 6753.5

(c) 7058.3 (d) 90000

16. Heat capacity of air can be

approximately expressed as

CP = 26.693 + 7.365 x 10-3 T

Where Cp is J/ (mol) (K) and T is K. The

heat given off by 1 mole of air when cooled

at 1 atmospheric pressure from 500 0C to –

100 0C is

(a) 10.73 kJ (b) 16.15 kJ

(c) 18.11 kJ (d) 18.33 kJ

17. A solid metallic block weighing 5 kg has

an initial temperature of 500 0C; 40 kg of

water initially at 25 0C is contained in a

perfectly insulated tank. The metallic block

is brought into contact with water. Both of

them come to equilibrium. Specific heat of

block material is 0.4 material is 0.4 kJ kg-1K-

1. Ignoring the effect of expansion and

contraction, and also the heat capacity of

tank, the total entropy change in kJ kg-1K-1 is

(a) – 1.87 (b) 0.0

(c) 1.26 (d) 3.91

18. The following heat engine produces

power of 100,000kW. The heat engine

operates between 00 K and 300 K. it has

thermal efficiency equal to 50% of that of

the Carnot engine for the same temperature.

The rate at which heat is absorbed from the

hot reservoir is

(a) 100,000 kW

(b) 160,000 kW

(c) 200,000 kW

(d) 320,000 kW

19. At 600C, vapour pressures of methanol

and water are 84.562 kPa and 19.953 kPa

respectively an aqueous solution of

methanol at 600C exerts a pressure of

39.223 kPa; the liquid phase andvapour

phase mole fractions of methanol are 0.1686

and 0.5714 respectively. Activity coefficient

of methanol is

(a) 1.572 (b) 1.9398

(c) 3.389 (d) 4.238



20-21 are based on the data supplied

in the paragraph below

One kg of saturated steam at 1000C and

1.01325 bar is contained in a rigid walled

vessel. It has a volume 1.673 m3. It cools to

980C; the saturation pressure is 0.943 bar;

one of water vapour under these conditions

has a volume of 1.789 m-3

20. The amount of water vapour condensed

(in kg) is

(a) 0.0 (b) 0.065

(c) 0.1 (d) 1.0

21. The latent heat of condensation (kJ kg-1)

under these conditions is

(a) 40732 (b) 2676

(c) 2263 (d) 540

(Gate 2004)

22. For an ideal gas mixture undergoing

a reversible gaseous phase chemical

reaction, the equilibrium constant

(A) Independent of pressure

(B) Increases with pressure

(C) Decreases with pressure

(D) Increases / decreases with pressure

depending on the stoichiometric

coefficients of the reaction.

23. As pressure approaches zero, the

ratio of fugacity to pressure (f/P) for a gas

approaches

(A) Zero (B) Unity

(C) Infinity (D) An indeterminate

value

24. A perfectly insulated container of

volume V is divided into two equal halves by

a partition. One side is under vacuum while

the other side has one mole of an ideal gas

(with constant heat capacity) at 298 K. If

the partition is broken, the final

temperature of the gas in the container

(A) Will be greater than 298 K

(B) Will be 298 K

(C) Will be less than 298 K

(D) Can’t be determined

25. The number of degrees of freedom

for an azeotropic mixture in a two

component vapor – liquid equilibria is / are

(A) Zero (B) One

(C) Two (D) Three

26. A car tier of volume 0.057 m3 is

inflated to 300 kPa at 300 K. After the car

is driven for ten hours, the pressure in the

tier increases to 330 kPa. Assume air is an

ideal gas and Cv for air is 21 J/(mol K). The

change in the internal energy of air in the

tier in J/mol is

(A) 380 (B) 630

(C) 760 (D) 880

27. A gas obeys P (V-b) = RT. The work

obtained from reversible isothermal

expansion of one mole of this gas from an

initial molar volume vi to a final molar

volume vf is



(A)

f

i

VRTln

V (B)

f

i

V bRTln

V

(C)

f

i

VRTln

V b (D

f

i

V bRTln

V b

28. A cyclic engine exchanges heat with

two reservoirs maintained at 100 and

3000C, respectively. The maximum work (in

J) that can be obtained from 1000 J of heat

extracted from the hot reservoir is

(A) 349 (B) 651

(C) 667 (D) 1000

29. The vapor pressure of water is given

by sat 5000lnP A

T where A is a constant, Psat is

vapor pressure in atm, and T is temperature

in K. The vapor pressure of water in atm at

50 0C is approximately

(A) 0.07 (B) 0.09

(C) 0.11 (D) 0.13

30. At standard conditions,

N2 + 2O2 2NO2 ∆G0 = 100 kJ/mol

NO + ½ O2 NO2 ∆G0 = – 35 kJ/mol

The standard free energy of formation of

NO in kJ/mol is

(A) 15 (B) 30

(C) 85 (D) 170

(Gate 2005)

31. In van der Waals equation of state,

what are the criteria applied at the critical

point in determine the parameters ‘a’ and

‘b’?

(A) 2

20 0

T T

P P;

V V

(B) 2

20 0

T T

V V;

P P

(C) 2

20 0

V V

P P;

T T

(D) 2

20 0

P P

V V;

V T

32. Which one of the following

statements in TRUE?

(A) Heat can be fully converted into work

(B) Work cannot be fully converted into heat

(C) The efficiency of a heat engine increases

as the temperature of the heat source is

increased while keeping the

temperature of the heat sink fixed.

(D) A cyclic process can be devised whose

sole effect is to transfer heat from a lower

temperature to a higher temperature.

33. A Carnot heat engine cycle is

working with an ideal gas. The work

performed by the gas during the adiabatic

expansion and compression steps, W1 and

W2 respectively, are related as

(A) | W1 | > | W2 | (B) | W1 | < |W2 |

(C) W1 = W2 (D) W1 = – W2

34. The Van-Laar activity coefficient

model for a binary mixture is given by the

form

*

1 *

1

*

2

Aln

xA1

xB

*

2 *

2*

1

ln

1

B

xB

A x

Given γ1 = 1.40, γ2 = 1.25, x1 = 0.25, x2 =

0.75, determine the constants A* and B*,

(A) A* = 0.5, B* = 0.3



(B) A* = 3, B* = 0.5

(C) A* = 0.333, B* = 0.2

(D) A* = 2, B* = 0.333

35. What is the actual power required to

drive a reciprocating air compressor which

has to compress 34 m3 of air per minute

from 1.013 x 105 N/m2 to 4.052 x 105 N/m2?

Assume that PV1.25 is constant, where P is

the pressure and V is the volume, and the

efficiency of the compressor is 85%.

(A) 107.9 kW (B) 200 kW

(C) 82.6 kW (D) 91.7 kW

Linked Answer Questions: 36 and 37

A frictionless cylinder piston assembly

contains an ideal gas. Initially pressure (P1)

= 100 kPa, temperature (T1) = 500 K and

volume (V1) = 700 x 10-6 m3. This system is

supplied with 100 J of heat and pressure is

maintained constant at 100 kPa. The

enthalpy variation is given by H (J/mol) =

30000 + 50 T; where T is the temperature

in K, and the universal gas constant R =

8.314 J/(mole K).

36. The final volume of the gas (V2) in m3 is

(A) 700 x 10-6

(B) 866.32 x 10-6

(C) 934.29 x 10-6

(D) 1000.23x10-6

37. The change in I.E. of the gas in J is

(A) 0 (B) 10

(C) 23.43 (D) 83.37

(Gate 2006)

38. At a given temperature and pressure, a

liquid mixture of benzene and toluene is in

equilibrium with its vapor. The available

degree(s) of freedom is (are)

(A) 0 (B) 1

(C) 2 (D) 3

39. A heat engine operates at 75% of the

maximum possible efficiency. The ratio of

the heat source temperature (in K) to the

heat sink temperature (in K) is 5/3. The

fraction of the heat supplied that is

converted to work is

(A) 0.2 (B) 0.3

(C) 0.4 (D) 0.6

40. For the isentropic expansion of an ideal

gas from the initial conditions T1, T1 to the

final conditions T2, T2, which ONE of the

following relations is valid? (γ = Cp / Cv)

(A) 1 2

2 1

P T

P T

(B) 1

1 2

2 1

P T

P T

(C) 1 1

2 2

P T

P T



(D) 1

1 1

2 2

P T

P T

41. Match the following:

(a) Heat (i) State Function

(b) Internal energy (ii) Path Function

(c) Work

(d) Entropy

(A) (a)-(ii); (b)-(i); (c)-(i); (d)-(i)

(B) (a)-(ii); (b)-(i); (c)-(ii); (d)-(ii)

(C) (a)-(ii); (b)-(ii); (c)-(i); (d)-(i)

(D) (a)-(ii); (b)-(i); (c)-(ii); (d)-(i)

42. For a binary mixture of A and B at

400 K and 1 atm, which ONE of the

following equilibrium states deviates

significantly from ideality?

Given: 2758

6 2 sat

Aln P .T

where,

PAsat = vapor pressure of A, atm;

T = temperature, K

PA = partial pressure of A, atm,

xA = mole fraction of A in liquid; yA =

mole fraction of A in vapor

(A) xA = 0.5; yA = 0.25

(B) xA = 0.5; PA = 0.25

(C) xA = 0.5; PA = 0.5

(D) xA = 0.6; yA = 0.3

43. The molar density of water vapor at the

normal boiling point of water is 33 mol/m3.

The compressibility factor under these

conditions is close to which ONE of the

following? R = 8.314 J/(mol K)

(A) 0.75 (B) 1

(C) 1.25 (D) 1.5

(Gate 2007)

44. If TA and TB are the boiling points of

pure A and pure B respectively and TAB is

that of a non-homogeneous immiscible

mixture of A and B, then

(A) TAB < TA& TB (B) TAB > TA& TB

(C) TA> TAB > TB (D) TB> TAB > TA

45. The state of an ideal gas is changed

from (T1, P1) to T2, P2) in a constant volume

process. To calculate the change in enthalpy

Δh, ALL of the following

properties/variables are required.

(A) CV , P1 , P2

(B) CP , T1 , T2

(C) CP , T1 , T2 , P1 , P2

(D) CV , P1 , P2 , T1 , T2

46. The change in entropy of the

system, ΔSsys, undergoing a cyclic

irreversible process is

(A) Greater than 0

(B) Equal to zero

(C) Less than zero

(D) Equal to the ΔSsurroundings

47. Parameters ‘a’ and ‘b’ in the van der

Waals and other cubic equations of state

represent

(A) a – molecular weight

b – molecular polarity



(B) a – molecular size

b – molecular attraction

(C) a – molecular size

b – molecular speed

(D) a – molecular attraction

b – molecular size

48. If R

i

E

iii mmmm ,,, are molar, partial

molar, residual and excess properties

respectively for a pure species “i”, the

mixture property M of a binary non-ideal

mixture of components 1 and 2, is given by

(A) x1 1m + x2 2m

(B) x1 m1R + x2 m2

R

(C) x1 m1 + x2 m2

(D) x1 m1E + x2 m2

E

49. For the two paths as shown in the

figure, one reversible and one irreversible,

to change the state of the system from a to b,

is

(A) ΔU, Q, W are same

(B) ΔU, is same

(C) Q, W are same

(D) ΔU, Q, are different.

50. For a pure substance, the Maxwell’s

relation obtained from the fundamental

property relation du = Tdz – Pdv is

(A) vs s

P

v

T

(B) Tv v

s

T

P

(C) Ps s

v

P

T

(D) Tv P

s

T

v

51. Which of the following represents the

Carnot cycle (ideal engine)?

52. 2 kg of steam in a piston-cylinder device

at 400 kPa and 175 °C undergoes a



mechanically reversible, isothermal

compression to a final pressure such that

the steam becomes just saturated. What is

the work, W, required for the process?

Data:

T = 175°C, P = 400 kPa – v = 0.503 m3/kg, u

= 2606 kJ/kg, s = 7.055 kJ/kg-K

T = 175°C, satd. vapor – v = 0.216 m3/kg, u

= 2579 kJ/kg, s = 6.622 kJ/kg-K

(A) 0 kJ (B) 230 kJ

(C) 334 kJ (D) 388 kJ

53. Vapor phase hydration of C2H4 to

ethanol by the following reaction

C2H4 (g) + H2O (g) ↔ C2H5 OH (g)

Attains equilibrium at 400 K and 3 bar. The

standard Gibbs free energy change of

reaction at these conditions is Δg° = 4000

J/mol. For 2 moles of an equimolar feed of

ethylene and steam, the equation in terms of

the extent of reaction ε (in mols) at

equilibrium is

(A)

2

(2 )0.3 0

(1 )

(B)

2(1 )0.9 0

(2 )

(C)

2

0.3 0(1 )

(D)

2

(2 )0.9 0

(1 )

Linked Answer Questions 54 & 55:

A methanol-water vapor liquid system is at

equilibrium at 60°C and 60 kPa. The mole

fraction of methanol in liquid is 0.5 and in

vapor is 0.8. Vapor pressure of methanol

and water at 60°C are 85 kPa and 20 kPa

respectively. Assuming vapor phase to be

an ideal gas mixture

54. What is the activity coefficient of water

in the liquid phase?

(A) 0.3 (B) 1.2

(C) 1.6 (D) 7.5

55. What is the excess Gibbs free energy

(gE, in J/mol) of the liquid mixture?

(A) 9.7 (B) 388

(C) 422 (D) 3227

Linked Answer Questions 56 & 57:

56. A perfectly insulated cylinder of

volume 0.6 m3 is initially divided into two

parts by a thin, frictionless piston, as shown

in the figure. The smaller part of volume

0.2 m3 has ideal gas at 6 bar pressure and

100°C. The other part is evacuated.

vacuum0.2 m3

stopper

At certain instant of time t, the stopper is

removed and the piston moves out freely to

the other end. The final temperature is

(A) –140 °C (B) –33 °C

(C) 33°C (D) 100°C

57. The cylinder insulation is now

removed and the piston is pushed back to



restore the system to its initial state. If this

is to be achieved only by doing work on the

system (no heat addition, only heat removal

allowed), what is the minimum work

required?

(A) 3.4 kJ (B) 107 kJ

(C) 132 kJ (D) 240 kJ

(Gate 2008)

58.For a Carnot refrigerator operating

between 40°C and 25°C, the coefficient of

performance is

(A) 1 (B) 1.67

(C) 19.88 (D) 39.74

59. The work done by one mole of a van

der Waals fluid undergoing reversible

isothermal expansion from initial volume Vi

to final volume Vf is

(A)

i

f

V

VRT ln

(B) lnf

i

V bRT

V b

(C) 1 1

lnf

i f i

V bRT a

V b V V

(D) 1 1

lnf

i f i

V bRT a

V b V V

60. The molar volume (v) of a binary

mixture, of species 1 and 2 having mole

fractions x1 and x2 respectively is given by

v = 220 x1 + 180 x2 + x1 x2 (90 x1 + 50 x2).

The partial molar volume of species 2 at x2 =

0.3 is

(A) 183.06 (B) 212.34

(C) 229.54 (D) 256.26

61. The standard Gibbs free energy

change and enthalpy change at 25°C for the

liquid phase reaction CH3COOH (1) +

C2H5OH (1) CH2COOC2H5(1) + H2O (1) are

given as ΔG°298 = - 4650 J/mol and ΔH°298 = -

3640 J/mol. If the solution is ideal and

enthalpy change is assumed to be constant,

the equilibrium constant at 95°C is

(A) 0.65 (B) 4.94

(C) 6.54 (D) 8.65

Linked Answer Questions 62 and 63:

A binary mixture containing species 1 and 2

forms an azeotropic at 105.4°C and 1.013

bar. The liquid phase mole fraction of

component 1 (x1) of this azeotropic is 0.62.

At 105.4°C, the pure component vapor

pressures for species 1 and 2 are 0.878 bar

and 0.665 bar, respectively. Assume that

the vapor phase is an ideal gas mixture. The

van – Laar constants, A and B, are given by

the expressions:

62. The activity coefficients ( γ1, γ2 )

under these conditions are

(A) (0.88, 0.66) (B) (1.15, 1.52),



(C) (1.52, 1.15) (D) (1.52, 0.88)

63. The Van Laar constants (A, B) are

(A) (0.92, 0.87) (B) (1.00, 1.21),

(C) (1.12, 1.00) (D) (1.52, 1.15)

(Gate 2009)

64. An ideal gas at temperature T1 and

pressure P1 is compressed isothermally to

pressure P2 (> P1) in a closed system. Which

ONE of the following is TRUE for internal

energy (U) and Gibbs free energy (G) of the

gas at the two states?

(A) U1 = U2, G1> G2

(B) U1 = U2, G1< G2

(C) U1> U2, G1 = G2

(D) U1< U2, G1 = G2

65. For a binary mixture at constant

temperature and pressure, which ONE of

the following relations between activity

coefficient (γi) and mole fraction (xi) is

thermodynamically consistent?

(A) 2

111 21ln xx , 2

122

1ln x

(B) 2

111 21ln xx , 2

12ln x

(C)2

111 21ln xx , 2

122

1ln x

(D) 2

111 21ln xx , 2

12ln x

Common Data Questions66 and 67:

An ideal gas with molar heat capacity

5

2pC R (where R = 8.314 J / mol.K) is

compressed adiabatically from 1 bar and

300 K to pressure P2 in a closed system. The

final temperature after compression is 600

K and the mechanical efficiency of

compression is 50%.

66. The work required for compression

(in kJ/mol) is

(A) 3.74 (B) 6.24

(C) 7.48 (D) 12.48

67. The final pressure P2 (in bar) is

(A) 23/4 (B) 25/4

(C) 23/2 (D) 25/2

(Gate 2010)

68. An equi-molar liquid mixture of

species 1 and 2 is in equilibrium with its

vapor at 400 K. At this temperature, the

vapor pressures of the species are = 180 kPa

and = 120 kPa. Assuming that Raoult’s law

is valid, the value of y1 is

(A) 0.30 (B) 0.41

(C) 0.50 (D) 0.60

69. A new linear temperature scale,

denoted by °S, has been developed,

where the freezing point of water is 200°S

and the boiling point is 400°S. On this

scale, 500°S corresponds, in degrees

Celsius, to

(A) 100°C (B) 125°C

(C) 150°C (D) 300°C

70. A saturated liquid at 1500 kPa and

500 K, with an enthalpy of 750 kJ / kg, is

throttled to a liquid – vapor mixture at 150

kPa and 300 K. At the exit conditions, the

enthalpy of the saturated liquid is 500 kJ /

kg and the enthalpy of the saturated vapor is



2500 kJ/kg. The percentage of the original

liquid, which vaporizes, is

(A) 87.5 % (B) 67%

(C) 12.5% (D) 10%

71. At constant T and P, the molar

density of a binary mixture is given by ρ = 1

+ x2, where x2 is the mole fraction of

component-2. The partial molar volume at

infinite dilution for component-1, is

(A) 0.75 (B) 1.0

(C) 2.0 (D) 4.0

(Gate 2011)

72. Minimum work (W) required

separating a binary gas mixture at a

temperature T0 and pressure P0 is,

1 2

0 1 2

21

ˆ ˆf fW RT y ln y ln

f f

where y1 and y2 are mole fractions. fpure,1 and

fpure,2 are fugacity of pure species at T0 and

P0 and f1 and f2 are fugacity of species in the

mixture at T0 , P0 and y1.

If the mixture is ideal then W is

(A) 0

(B) – RT0 [y1ln y1 + y2ln y2]

(C) RT0 [y1ln y1 + y2lny2 ]

(D) RT0

73. The partial molar enthalpies of

mixing (in J / mol) for benzene (component

1) and cyclo-hexane (component 2) at 300 K

and 1 bar are given by and, where x1 and x2

are the mole fractions. When ONE mole of

benzene is added to TWO moles of cyclo-

hexane, the enthalpy change (in J) is

(A) 3600 (B) 2400

(C) 2000 (D) 800

74. One mole of methane is contained in

a leak proof piston – cylinder assembly at 8

bar and 1000 K. The gas undergoes

isothermal expansion to 4 bar under

reversible conditions. Methane can be

considered as an ideal gas under these

conditions. The value of universal gas

constant is 8.314 J.mol–1 K–1. The heat

transferred (in kJ) during the process is

(A) 11.52 (B) 5.76

(C) 4.15 (D) 2.38

75. Two systems are available for

compressing 6 m3 / hr of ambient air to 10

bar. The first one uses a single stage

compressor (K1) and the second one uses a

multistage compressor with inter – stage

cooling (K2). Which ONE of the following

statements is INCORRECT?

(A) K2 will have knockout pots in between

the stages

(B) Discharge temperature of K1 will be

higher than that of K2

(C) K2 will consume more power than K1

(D) Cost of K2 will be more than that of K1

76. Consider a binary mixture of methyl

ethyl ketone (component 1) and toluene

(component 2). At 323 K the activity

coefficients γ1 and γ2 are given by

2

1 2 1 2 2 1ln x 4 x ,

1

2

2 1 2 2 2ln x 4 x Where x1 and x2 are

the mole fractions in the liquid mixture, and

Ψ1 and Ψ2 are parameters independent of

composition. At the same temperature, the

infinite dilution activity coefficients, and are

given by 1

ln 0.4 and

ln 0.2 . The vapor



pressure of methyl ethyl ketone and toluene

at 323 K are 36.9 and 12.3 kPa respectively.

Assuming that the vapor phase is ideal, the

equilibrium pressure (in kPa) of a liquid

mixture containing 90 mol % toluene is

(A) 19 (B) 18

(C) 16 (D) 15

(Gate 2012)

77. In a throttling process, the pressure of

an ideal gas reduces by 50 % if P vC and C are

the heat capacities at constant pressure and

constant volume, respectively P VC / C ,

the specific volume will change by a factor of

(A) 2 (B) 12 /

(C) 12

/

(D) 0.5

78. If the temperature of saturated water

is increased infinitesimally at constant

entropy the resulting state of water will be

(A) Liquid

(B) Liquid- vapor coexistence

(C) Saturated vapor

(D) Solid

79. In a parallel flow heat operating

under steady state hot liquid enters at a

temperature h.inT and leaves at a

temperature h,out.T liquid enters at a

temperature c,inT and leaves at a

temperature c,outT Neglect any heat loss from

the heat exchanger to the surrounding id

h,in C,inT T then for a given time interval,

which ONE of the following statements is

true?

(A) Entropy gained by the cold stream is

GREATER than entropy lost by the hot

stream

(B) Entropy gained by the cold stream is

EQUAL to the entropy lost by the hot

stream

(C) Entropy gained by the cold stream is

LESS than the entropy lost by the hot

stream

(D) Entropy gained by the cod stream is

ZERO

80. For an exothermic irreversible

reaction which one of the following correctly

describes the dependence of the equilibrium

constant (K) with temperature (T) and

pressure (p)?

(A) K is independent of T and P

(B) K increase with an increase in T and P

(C) K increases with T and decreases with

P

(D) K decreases with an increase in T and is

independent of P

81. An insulated, evacuated container is

connected to a supply line of an ideal gas at

pressure s,P temperature sT and specific

volume sv . The container is filled with the

gas until the pressure in the container

reaches sP there is no heat transfer between

the supply line to the container, and kinetic

and potential energies are negligible. If

1PC and C are the heat capacities at constant

pressure and constant volume respectively

P vC / C , then the final temperature of

gas in the container is



(A) sT (B)

sT

(C) sI T

(D) 1 sT /

82. Consider a binary liquid mixture at

constant temperature T and pressure P. if

the enthalpy change of mixing

1 2 1 25 H x x , where x and x are the mole

fraction of species 1 and 2 respectively and

the entropy change of mixing

1 2 2 S x In +x In x (with R = 8.314 J /

mol.K) then the minimum value of the

Gibbs free energy change of mixing at 300 K

occurs when

(A) 1

0x

(B) 1

0 2x .

(C) 1

0 4x .

(D) 1

0 5x .

(GATE 2013)

83. A gaseous system contains H2, I2,

and HI, which participate in the gas-phase

reaction

2HI H2 + I2

At a state of reaction equilibrium, the

number of thermodynamic degrees of

freedom is_______

84. The thermodynamic state of a closed

system containing a pure fluid changes from

(T1, p1) to (T2, p2), where T and p denote the

temperature and pressure, respectively. Let

Q denote the heat absorbed (> 0 if absorbed

by the system) and W the work done (> 0 if

done by the system). Neglect changes in

kinetic and potential energies. Which one of

the following is CORRECT?

(A) Q is path-independent and W is path-

dependent

(B) Q is path-dependent and W is path-

independent

(C) (Q − W) is path-independent

(D) (Q+W) is path-independent

85. An equation of state is explicit in

pressure p and cubic in the specific volume

v. At the critical point ‘c’, the isotherm

passing through ‘c’ satisfies

(A) 0,02

2

v

p

v

p

(B) 0,02

2

v

p

v

p

(C) 0,02

2

v

p

v

p

(D) 0,02

2

v

p

v

p

86. The units of the isothermal

compressibility are

(A) 3m (B)

1Pa

(C) 13 Pam (D)

13 Pam

87. In a process occurring in a closed

system F, the heat transferred from F to the

surroundings E is 600 J. If the temperature

of E is 300 K and that of F is in the range

380 – 400 K, the entropy changes of the

surroundings (∆SE) and system (∆SF), in

J/K, are given by

(A) 2,2 FE SS

(B) 2,2 FE SS

(C) 2,2 FE SS

(D) 2,2 FE SS

88. A binary liquid mixture is in

equilibrium with its vapor at a temperature



T = 300 K. The liquid mole fraction x1 of

species 1 is 0.4 and the molar excess Gibbs

free energy is 200 J/mol. The value of the

universal gas constant is 8.314 J/mol-K, and

i γ denotes the liquid-phase activity

coefficient of species i. If ln (γ1) = 0.09, then

the value of ln(γ2), up to 2 digits after the

decimal point, is

(Gate 2014)

89. From the following list, identify the

properties which are equal in both vapour

and liquid phases at equilibrium

P. Density

Q. Temperature

R. Chemical potential

S. Enthalpy

(A) P and Q only (B) Q and R only

(C) R and S only (D) P and S only

90. In a closed system, the isentropic

expansion of an ideal gas with constant

specific heats is represented by

91. Match the following

Group 1 Group 2

(P)

ijnPTin

G

,,

I. Arrhenius

equation

(Q)

ijnVSin

G

,,

II. Reaction

Equilibrium

constant

(R)

RT

Greaction

0

exp III. Chemical

Potential

(S) 0, PTiidn IV. Gibbs –

Duhem equation

(A) Q-III, R-I, S-II

(B) Q-III, R-II, S-IV

(C) P-III, R-II, S-IV

(D) P-III, R-IV, S-I

92. Which ONE of the following is

CORRECT for an ideal gas in a closed

system?

(A) VS S

UnRV

V

U

(B) PS S

HnRP

P

H

(C) PS S

HnRV

V

U

(D) VS S

UnRP

P

H

93. Consider a binary liquid mixture at

equilibrium with its vapour at 250 C

Antoine equation for this system is given as

log10pisat A -

B

t c where t is in 0C and P in

Torr.



The Antoine constant (A, B, and C) for the

system are guven in the following table

Component A B C

1 7.0 1210 230

2 6.5 1206 223

The vapor phase is assumed to be ideal and

the activity coefficients (γi) for the non ideal

liquid phjase are given by

In (γ1) = x22[ 2-0.6 x1]

In (γ2) = x12 [1.7 + 0.6x2]

If the mole fraction of component 1 in liquid

phase (x1) is 0.1 then the mole fraction of

component 1 in vapor phase (y1) is ………

(Gate 2015)

94. If v, u, s and g represent respectively

the molar volume, molar internal energy,

molar entropy and molar Gibbs free energy,

then match the entries in the left and right

columns below and choose the correct

option.

(P) S

u

v

(I) Temperature

(Q) T

g

P

(II) Pressure

(R) P

g

T

(III) V

(S) V

u

s

(IV) S

(A) P - II, Q - III, R - IV, S - I

(B) P - II, Q - IV, R - III, S - I

(C) P - I, Q - IV, R - II, S - III

(D) P - III, Q - II, R - IV, S - I

95. For a gas phase cracking reaction A

→B + C at 300OC, the Gibbs free energy of

the reaction at this temperature is

2750 /OG J mol . The pressure is 1 bar

and the gas phase can be assumed to be

ideal. The universal gas constant R =

8.314J/mol. K. The fractional molar

conversion of A at equilibrium is:

(A) 0.44 (B) 0.50

(C) 0.64 (D) 0.80

96. For a pure liquid, the rate of change

of vapour pressure with temperature is 0.1

bar/K in the temperature range of 300 to

350 K. if the boiling point of the liquid at 2

bar is 320 K, the temperature (in K) at

which it will boil at 1 bar (up to one decimal

place) is ______

97. Three identical closed systems of a

pure gas are taken from an initial

temperature and pressure (T1, P1) to a final

state (T2, P2), each by a different path.

Which of the following in Always true for

the three systems? (∆ represents the change

between the initial and final states: U, S, G,

Q and W are internal energy, entropy, Gibbs

free energy, heat added and work done,

respectively.)

(A) ∆U, ∆S, Q are same

(B) W, ∆U, ∆G are same

(C) ∆S, W, Q are same

(D) ∆G, ∆U, ∆S are same

98. Which of the following can change if

only the catalyst is changed for a reaction

system?

(A) Enthalpy of reaction

(B) Activation energy



(C) Free energy of the reaction

(D) Equilibrium constant

99. An ideal gas is initially at a pressure

of 0.1 MPa and a total volume of 2m3. It is

first compressed to 1MPa by a reversible

adiabatic process and then cooled at

constant pressure to a final volume of

0.2m3. The total work done (in kJ) on the

gas for the entire process (up to one decimal

place) is ______

Data: R = 8.314 J/mol-K; heat capacity at

constant pressure (CP) = 2.5R

100. Given that molar residual Gibbs free

energy, gR, and molar residual volume VR,

are related as

R RPg v

dP,RT RT0

Find gR at T = 27 0C and P = 0.2MPa. The

gas may be assumed to follow the viral

equation of state, Z = 1+BP/RT, where B= -

10 – 4 m3/mol at the given conditions

(R = 8.314 J/mol-K). The value of gR in

J/mol is:

(A) 0.08 (B) −2.4

(C) 20 (D) −20

101. A binary mixture of components (1)

and (2) forms an azeotrope at 1300C and x1

= 0.3. the liquid phase non-ideality is

described by In 21 1 22

Ax and In , are the

activity coefficient and x1 , x2 are the liquid

phase mole fractions. For both

components, the fugacity coefficients are

0.9 at the isotropic composition. Saturated

vapor pressures are at 1300C are P1sat = 70

bar, P2sat= 30 bar

The total pressure in bars for the above

azeotropic system (up to two decimal

places) is

(Gate 2016)

102. The partial molar enthalpy (in

kJ/mol) of species 1 in a binary mixture is

given by2

21

2

21 100602 xxxh , where

21 and xx are the mole fractions of species 1

and 2, respectively. The partial molar

enthalpy (in kJ/mol, rounded off to the first

decimal place) of species 1 at infinite

dilution is ________

103. A binary liquid mixture of benzene

and toluene contains 20 mol% of benzene.

At 350 K the vapour pressures of pure

benzene and pure toluene are 92 kPa and 35

kPa, respectively. The mixture follows

Raoult’s law. The equilibrium vapour phase

mole fraction (rounded off to the second

decimal place) of benzene in contact with

this liquid mixture at 350 K is _____

104. An ideal gas is adiabatically and

irreversibly compressed from 3 bar and 300

K to 6 bar in a closed system. The work

required for the irreversible compression is

1.5 times the work that is required for

reversible compression from the same

initial temperature and pressure to the same

final pressure. The molar heat capacity of

the gas at constant volume is 30 J mol‒1 K‒1

(assumed to be independent of

temperature); universal gas constant, R is

8.314 J mol‒1 K‒1; ratio of molar heat

capacities is 1.277. The temperature (in K,

rounded off to the first decimal place) of the

gas at the final state in the irreversible

compression case is _______

105. A gas obeying the Clausius equation

of state is isothermally compressed from 5

MPa to 15 MPa in a closed system at 400 K.



The Clausius equation of state is

)(Tbv

RTP

where P is the pressure, T is

the temperature, v is the molar volume and

R is the universal gas constant. The

parameter b in the above equation varies

with temperature as TbbTb 10)( with 0b

= 4 × 10‒5 m3mol‒1 and 1b = 1.35 × 10‒7

m3mol‒1K‒1. The effect of pressure on the

molar enthalpy (h) at a constant

temperature is given by

T

vTv

P

h

T

. Let hi and hf denote

the initial and final molar enthalpies,

respectively. The change in the molar

enthalpy (in J mol‒1, rounded off to the first

decimal place) for this process is ______

106. A binary system at a constant

pressure with species ‘1’ and ‘2’ is described

by the two-suffix Margules equation,

213 xxRT

gE

,where Eg is the molar excess

Gibbs free energy, R is the universal gas

constant, T is the temperature and 21 and xx

are the mole fractions of species 1 and 2,

respectively.

At a temperature T, RT

g1 = 1 and RT

g2 = 2,

where 1g and 2g are the molar Gibbs free

energies of pure species 1 and 2,

respectively. At the same temperature, g

represents the molar Gibbs free energy of

the mixture. For a binary mixture with 40

mole % of species 1, the value (rounded off

to the second decimal place) of RT

g is

(Gate 2017)

107. The volumetric properties of two

gases M and N described by the generalized

compressibility chart which expresses the

compressibility factor (Z) as a function of

reduced pressure and reduced temperature

only. The operating pressure (P) and

temperature (T) of two gases M and N

along with their critical properties (PC,Tc)

are given in the table below.

Gas P

(bar)

T (K) PC

(bar)

TC

(K)

M 25 300 75 150

N 75 1000 225 500

ZM and ZN are the compressibility factor of

the gases M and N under the given

operating conditions respectively.

The relation between ZM and ZN is

(a) 8M NZ Z (b) 3

M NZ Z

(c) M NZ Z

(d) 0 333

M NZ . Z

108. Water is heated at atmospheric

pressure from 400C to 800 C using two

different process. In process I, the heating is

done by a source at 800C. In process II, the

water is first heated form 400C to 600C by a

source at 600C, and then from 600C to 800 C

by another source at 800C. Identify the

correct statement.

(a) Enthalpy change of water in process I is

greater than enthalpy change in process II

(B) Enthalpy change of water in process II is

greater than enthalpy change in process I

(c) Process I is closer to reversibility

(d) Process II is closer to reversibility

109. The vapour pressure of a pure

substance at a temperature T is 30 bar. The



actual and ideal gas values of g /RT for the

saturated vapour at this temperature T and

30 bar are 7.0 and 7.7 respectively. Here, g

is the molar Gibbs free energy and R is the

universal gas constant the fugacity of the

saturated liquid at these conditions,

rounded to 1 decimal place is ___bar

110. The pressure of a liquid is increased

isothermal compressibility. the molar

volume of the liquid decreases from 50.45

10-6 m3/mol to 48 10-6 m3/mol during this

process. The isothermal compressibility of

the liquid is 10-9 Pa-1, which can be assumed

to be independent of pressure The change in

the molar Gibbs free energy of the liquid,

rounded to nearest integer, is ____J/mol.

111. A sparingly soluble gas (solute) is in

equilibrium with a solvent at 10 bar. The

mole fraction of the solvent in the gas phase

is 0.01 At the operating temperature and

pressure, the fugacity coefficient of the

solute in the gas phase and the Henry’s law

constant are 0.92 and 1000 bar,

respectively. Assume that the liquid phase

obeys Henry’s law. The mole percentage of

the salute in the liquid phase rounded to 2

decimal places, is ____

(Gate 2018)

112.Under isothermal condition, a vertical

tube of length L = 100 m contains a gas of

molecular weight equal to 60. The pressure

and temperature at the top of the tube are

100 kPa and 250C respectively. Consider the

universal gas constant and acceleration due

to gravity as 8.314 J mol-1 K-1 and 9.81 m s-

2respectively. If the gas is ideal, the pressure

(in kPa) at the bottom of the tube will be

______

113. Consider the following properties: (P)

temperature (Q) specific gravity (R)

chemical potential (S) volume. The option

which lists ALL the intensive properties is

(A) P (B) P and Q

(C) P, Q, and R (D) P, Q, R and S

114. In a closed piston-cylinder system,

methane was observed to obey the following

equation of state P(V-nb) = nRT where b =

0.029 m3/mol. The temperature and volume

are 500°C and 5 m3 respectively for 100

moles of methane. At this state of the

system, the isobaric rate of change of

temperature with volume is (0C/m3) _____

(rounded off to second decimal place).

115. G denotes the Gibbs free energy of a

binary mixture, nT denotes the total number

of moles present in the system, 1 is the

chemical potential of the component

( 1 1 20 and ), ix is the mole

fraction of the component. The correct

variation ofG

nT(J/mol) at constant temperature

and pressure is given by



Answer Key Table

1 C

2 B

3 D

4 A

5 A

6 C

7 B

8 D

9 D

10 B

11 B

12 B

13 C

14 B

15 B

16 C

17 1.249 = 1.26

18 D

19 A

20 B

21 B

22 A

23 B

24 B

25 B

26 B

27 D

28 A

29 D

30 C

31 A

32 C

33 D

34 B

35 A

36 B

37 D

38 C

39 B

40 B

41 D

42 C

43 B

44 A

45 B

46 A

47 D

48 A

49 B

50 A

51 C

52 C

53 C

54 B

55 C

56 D

57 C

58 C

59 D

60 B



61 B

62 B

63 C

64 B

65 D

66 C

67 D

68 D

69 C

70 C

71 A

72 B

73 D

74 B

75 C

76 C

77 B

78 A

79 A

80 D

81 A

82 A

83 3

84 C

85 D

86 B

87 A

88 0.0736

89 B

90 D

91 C

92 D

93 0.693

94 A

95 D

96 310 K

97 D

98 B

99 750 kJ

100 D

101 27.54

102 -58

103 0.396

104 373 K

105 400 J /mol

106 1.64

107 C

108 D

109 14.9

110 2511.4

111 0.91

112 102.375

113 C

114 368.116



Solutions

(Gate 2001)

1. Sol. C

0

i i i i, y P x f

ii

i

yK

x = equilibrium constant form

equation

0

i ii

i

fK

P

Where: I = liquid phase activity coefficient =

f (T, p, xi)

0

if = fugacity of pure component = f (T, p)

i = Vapopur fugacity coefficient = f (T, P)

Ki = f (T , P, xi)

2. Sol. B

Given adiabatic and reversible means

isentropic

So, S 0

sH W (shaft work) 0

3. Sol. D

Notification a – molar Helmholtz free energy

U – Molar internal energy

g – molargibbs free energy

M – Molar flow rates

a = U – Ts

da = dU – T.ds - s dT

da = (T ds – Pdv ) – T ds – s d T

da = - P dv – sdT

Now T v

a aP and s

V T

T v Tv T

a a s

T V V T V

Therefore, V T

P s

T V

“a” is a state function

T V V T

a a

T V V T

4. Sol.A

Clausius – clapeyron equation

2

1

vapvap

T

vap1 2

T

P H 1 1In

R T TP

For methyl cyclohexane

0

0

vap vap

150 MCH

vap

100

P H 1 1In

R 373 423P

For water



0

0

vap vap

150 w

vap

100

(P ) H 1 1In

(P ) R 373 423

Divide equation 0

0

0

0

vap vap

150 MCH MCH

vap vap

w100 MCH

vap

150 W

vap

100 W

(P ) HIn

(P ) H

(P )In

(P )

Given

00

0

vap VAP vap

W100 w100 MCH

vap vap

MCH 150 W

P (p ) 1 atm H 40.63 kJ / mol

H 31.55 kJ / mol (P ) 4.69 atm

00

0

0

vap vap

150 w150 MCH

vap

150 MCH

vap

150 MCH

31.55In P In (P )

40.63

31.55In P In (4.69)

40.63

In P 3.320

5. Sol. A

For adiabatic process

1

2 2

1 1

T P

T P

Given; 2

1

P3

P

1

7T 300 K

5

So,

71

5

7 2

2 5 72

T3 T 300 3

300

(Gate 2002)

6. Sol. C

At critical point 2

2

TT

P P0

V V

Critical point is a state at which no phase

boundaries exist

7. Sol. B

Gibbs phase rule

F = C – P + 2 –r-s

Given

C = No. of component = 2 (water and ethanol)

P = No. of phase = 2 (liquid and vapour)

r =No. of independent chemical reaction=0

s = 1, for a zero tropic mixture

So, F = 2 – 2 + 2 - 0 – 1 = 1

8. Sol. (D)

9. Sol. (D)

Values of V, S, U, h are given for differential

temperature at constant pressure therefore

TP

V S

T P

Can be used easily for verify steam data for

super heated steam.

10. Sol. (b)

Isentropic means adiabatic reversible

in out

i v

S S

6.65 S x(S S )

6.65 1.4336 x(7.2234 1.4336)

x 0.90

Mixture of saturated liquid and volume

mixture with quality 0.90



11. Sol. B

Heat required for constant volume

Q = n CP (T2 – T1)

= 3 x 29.1 x (250 – 30) = 19206 J

12. Sol. B

For adiabatic process

2 2 7 2 2 4

2 1

1 2

p P

v P

H S O ( ) H O( ) 2H SO ( )

T VSo,

T V

C C 21r 1.655

V C R 21 8.314

r 1

12 1

2

(1.655 1)

2 1

VT T

V

1500 V 5 V (given)

5

174.24 K

(Gate 2003)

13. Sol. C

Kopp’s rule Molar heat capacity of a solid of a

solid substance can be obtained by adding the

particular value for the various atoms in the

formula of the substance

H Li Be B C N O F other

10 21 15 13 8 18 20 26 J / deg mole

14. Sol. B

For isothermal process

Work done = n RT In 1

2

P

P

81 8.314 600 In 10373.08

1

+ ve sign shows work done by the system

15. Sol. B

3

3

6738 9

25 28 9000 8592 7

8314 6 573 15

6753 5

satl

sat

Vfln p p

RTf

putting values,

fln

. kPa

cm. . kPa

mol

kPa.cm. . K

mol.K

f . kPa

16. Sol. C

2

1

s

T

p

T

P

p (T,P) sat

P

ssat

H n C dT

1C 26.693 7.365F F (p) exp V dP

RT

V (P P )F

RT

6 3 3

3

25.28 10 (9000 10 8592.7 10 )6738.9 exp

8.314 573

6753.48 6753.510 T Given

for n 1



s

2

1

P

(T,P) sat

P

Ts

sat

T6 3 3

3

1(26.693 7.365F F (p) exp V dP

RT

V (P P )F

H RT

25.28 10 (9000 10 8592.7 10 )6738.9 exp

8.314 573

6753.48 6753.510 T)dT

2 2

1 1

3T T2

T T

32 2

2 1 2 1

7.365 10H 26.693T I T I

2

7.365 10H 26.693 (T T ) T T

2

0

1

0

2

0

2

32 2

Given ;T 500 C 77 K

T 100 C 173K

T 100 C 173K

7.365 10H 26.693 (173 773) (173 773 )

2

H 18105.987 J

H 18.11 KJ

Here negative sign indicates heat is given off.

17. Sol. (c)

By energy balance

Heat given by block = heat taken by water

Block

water

P i f P f iBlock water

i

i

f f

f

mC T T mC T T

Given; T 500 273 773k;

T 25 273 298K

5 0.4 (773 T ) 40 4.18 (T 298)

T 303.61 K

total Block water

f fP Block P

i i water

S S S

T TmC In mC In

T T

303.61 303.615 0.4 In 40 4.18 In

773 298

1.869 3.118 1.249 1.26

18. Sol. D

Carnot engine efficiency

1 2c

1

T T 800 3000.625

T 800

Thermal efficiency = H c0.5

0.5 0.625 0.3125

H

work done by actual engine W0.3125

Heat absorbed by engine Q

W 100000 kW

100000Q 320000kW

0.3125

19. Sol. A

vap

1 1 1 1y p x p ; ; 1 stands for methanol

activity coefficient

11 vap

1 1

y p 0.5714 39.2231.572

x p 0.1686 84.562

20. Sol. B

Mass of water condensed + mass of water

vaporization = mass of liquid and vapor (1 lg)

Mass of water condensed = 1 – mass of water

vaporization

specific vol. of mix specific vol. of sat. liq1

specific vol. of sat. vap. spefific vol. of sat. liq

mass of liq. vapour

1.673 1 0.0011 1 0.0065 kg

1.789 0.001



21. Sol. B

Clausius – clapeyron equation

dp

dT T V

V specific volume change

vV V

3 3

v

v

v

v

V 1.789m / kg and V 0.001 m / kg

V .V

V V

dp

dT TV

Assume satdvap. As an ideal gas

v

RTV

P

dP P

dT T RT

(Gate 2004)

22. (A)

For an ideal gas mixture undergoing a

reversible gas phase chemical reaction,

Consider a A+b B cC dD

The equilibrium constant is given by,

c d

c Dp a b

A b

P .pK

P .p

23. (B)

As the pressure approaches zero the ratio of

fugacity to pressure (f/p) for a gas

approaches.

0

1

Since as P , f p

f / p

Hence (b) is the correct answer.

24. Sol. (B)

It is a case of free expansion,

1 2 1 2

2 1 2 1

2 1

0 0

0 0

298

Q , w and

therefore u u i.e. (T T )

andT T k

Hence(B)iscorrectanswer

25. Sol. (B)

The system consists of two non-reacting

species in two phases, if there were no

zoetrope then

2 2 2 2 0 2 F N r

But is azeotropic mixer, this provides on an

additional equation i ix y

Thus we have to use equation

2

2 2 2 0 1 1

F N r S

F =

Hence (b) is the correct answer.

26. (B)

For ideal gas PV = RT

1 1300Given: p kPa, T =300K

2 2

1 1

2 2

2

2

330

300 300

330

P KPa, T =?

P TFor constant volume process,

P T

T

T

v

= 330

Change in internal energy, U = C T

= 21 (332-300) = 630 J/mol

Change in internal energy = 630 J/mol



27. (D)

P(v b) RT

1 2

f

f

i

i

vv

vv

f

i

RTP

v b

RTw PdV dv

v b

v bRT n Hence

v b

D is the correct answer

28. Sol. (A)

Efficiency of the cycle =

2

1

3731 1 1000

573

349 04

W.D H.S

TQ

T

. J

Hence A is the correct answer

Maximum work obtained 200

1000573

349.04

Hence (A) is the correct answer,

29. Sol. (D)

Given vapor pressure relation is,

0

5000

1 100

5000

373

5000

373

5000 500

373 323

0 1255

SAT

sat

sat

sat

ln P A ,T

we know that

p atm, at T C

ln (1) =A

A =

lnP

P .

30. Sol. (C)

given that

0

2 2 2

0

2 2

0

2 2

20 2 100

135

2

2

2 2 70

N NO G kJ / mol ...(i)

NO O NO G kJ / mol ...(ii)

Multiplying equation (ii) by , we have

NO O NO G kJ / mol ... (iii)

Substracting equation (iii) from (i),w0

2 2 2

0

2

0

2 2 2

0

2 2

2 2 100

2 2 70

2 0 170

2 170

e get

N O NO ; G kJ / mol

NO O NO; G ( kJ / mol)

N O NO ; G kJ / mol

N O NO; G kJ / mol

Thus for one mole of NO, 0 85 / G kJ mol

(Gate 2005)

31. (A)

Vander Waals equation of state at the critical

point (A) Is the correct answer

32. (C)

The efficiency of a heat engine increases as

the temperature of the heat source is

increased while keeping the temperature of

the heat sink fixed. (C) is the correct answer

33. (D)

2 2 3 3

2 3 1

2 3

1

1

p v p vw w

R(T T )



2

4 4 1 1

4 1 2

42

1

H LR(T T )b ac

p v p vw w

4 1

1

R(T T )

1

1

L H

H L

R(T T )

R(T T )

Thus w1 = – w2

Hence (D) is the correct answer

34. Sol. (B)

Putting the given values, we have

2

2

2

2

1 4 13

1 33

31

31 4 5

Aln( . ) A

B

A A ...(i)

B

Band, ln(1.25) B

A

B . B (ii)

A

Solving equations (i) and (ii), we get A=3, B=0.5

35. (A) Power required =

11

21 1

1

5

1 25

3

1

1 25 1 013 10 34 25

25 85 80 4 1

107 88 10

107 88

n

n

Comp

.

pp v

pn

n

. ( . ) .

. .

. w

. kW

(A)is thecorrectanswer

36. Sol. B)

We know that

Ideal Gas Equation of state,

1 1 1p v nRT

3 6

1 1

1

1 2 2 1 2 1

2

2

1 2

1 2

6

1 2

2

1

6

2

100 10 700 100 01684

8 314 500

50

100 0 01684 50 500

618 76

700 10 619

500

866 6 10

p vn . mol

RT .

constant pressure process,

Q n H H n (T T )

. (T )

T . K

For constant pressure,

V V

T T

V TV

T

V . 3m

37. Sol. (D)

We know that

1 2 1 2 2 1

2 1 1 2 2 1

13 16100 100 10 866 26 700 16

83 374

Q w (U U )

U U Q p(V V )

( . )

. J

(D)is thecorrectanswer

(Gate 2006)

38. Sol. (C)

F = C-P+2: Here c=2 (benzene and toluene),

P=2 (liquid and vapour). Therefore, F=2

F=C-P+2-0-0

=2-2+2-0-0=2

39. Sol . (B)

We know that



0 75 1

30 75 1

5

0 3

LT.

T

.

w.

q

(B)is thecorrectanswer

40. (B) we know that

1 1

1 2

1 2

T T

p p

1

1

1 1

2 2

1 1

2 2

rT p

T p

T p

T p

(B)is thecorrectanswer

41. Sol. D

(a) (ii)

(b) (i)

(c) (ii)

(d) (i)

Hence (D) is the correct answer

42. (C)

given that

2758

6 2400

0 49907

0 5 0 5

.

ln .

.

( ) ,

'

o ,

. ; .

maximum

sat

A

sn

A

A A

P

P

The deviation obtained in step A is

Applying Roult s Law and applying it all

options deviation is btained

x P

43. (B) Given:

3

3

3

3

10 030

33

8 314 373 1

1 101325

0 0306

0 03001

0 0306

realgas

idealgas

mrealgas molmol

m

Jmol

idealgas

mmol

vCompressiblity Factor, Z

v

where, v specific volume m / mol

According to the problem,

v .

. KRT atmv

p atm Pa

.

.Z .

.0

(Gate 2007)

44. Sol. (A)

For an immiscible mixture, the boiling point

of the mixture is always less than the boiling

point of each individual component. (See the

concept of steam distillation of high boiling

organic compounds) (A) Is the correct

answer

45. Sol. (B)

46. Sol. (A)



Every process proceeds in such direction that

total entropy change associated with it will be

positive. (A) is the correct answer

47. Sol. (D) is the correct answer 48. Sol. (A) is the correct answer. 49. Sol. (B) is the correct answer 50. Sol. (A) Given,

du = Tds – Pdv

This is an exact differential equation, so

vs

T P

s

(A) Is the correct answer

51. Sol. (C) is the correct answer

52. Sol. (C) We know that

1 2 2 2

2 1

1 2 1 2 2 1

1 2

1 2 1 2

448 6 622 77 055

193 9984

21579 2606 27

193 484 27 166 984

2 166 984

333 968

q T(s s ) ( . . )

. kJ / kg

u u kJ / kg

q w (u u )

kJw . .

kg

W m w ( . )

. kJ

ve sign indicates that work is reqd.

hence, C is the co

rrect answer

53. Sol. (C)

We have:

2 5

2 4 2

2 5

2

0

4000 8 314 400

0 3

0 3

1 1 1 1

2

1

2

p y

y p

C H OH

y

C H H o

p

C H OH

H O

For the given reaction

G RTln k

putting values,

J J. k ln k

mol mol.k

k .

& here we know that

K K .p

K K .p

yWhere, K

y .y

K K .

y

y

2 4

2

1

2

20 9 0

1

C Hy

on solving we get :

( ).

54. Sol. (B)

Vapor liquid equilibrium relationship is given

by

0 2 60 0 5 20

1 2

sat.

i i i i

i

i

y p x p

Hence, for water in liquid phase activity coefficient,

. .

.

55. Sol. (C)

Activity coefficient of Methanol in liquid

phase,



1 1 2 2

8 60 85

0

i

i

E

E

0. = 5

= 1.129.

G x ln x ln RT

G = 8.314 333 .5 {ln 1.2+ln 1.129}

= 420.34

= 422 J/mol

56. Sol. (D)

it is a case of free expansion temperature

remains same i.e. 1000 C.

(D) Is the correct answer

57. Sol. (C)

Isothermal compression,

22

1 2 1 1

1

2

2 2

1

4

0 26 100 0 2

0 6

131 83

vw p v ln b ac

v

vp v ln

v

.. ln

.

. kJ

– ve sign indicates that work is reqd.

(C) Is the correct answer

(Gate 2008)

58. Sol. (C)

29819 87

313 298

L

H L

TCOP .

T T

Hence, (C) Is the correct answer

59. Sol. (D)

Real gas equation is

2

1 2 2

f f

i i

f

i

v V

v v

v

v

ap (V b) RT

V

RT aw PdV dv

V b v

aRTln(v b) a

V

1 1

f

i f i

v bRTln a

v b v v

(D)is thecorrectanswer

60. Sol. (B)

We know that

2 1

1

dV

V V xdx

1 1 1 1 1

2 1

2 3

1 1 1

220 180 1 1 40 50

1

90 10 40 180

V x ( x ) ( x )x ( x )

(x x )

V x x x

2 21

1 1 1

1

90 20 120 x dV

x x xdx

3 2

2 1 180 10 180

hence,

V x x

1 2

2

At x = 0.7 and x = 0.3

V = 212.34

61. Sol. (B)

0

2984650 G J / mol

0

298

0

3640

H J / mol

G RT lnK

G RT lnK

1

2 1 2

2

4550

8 314 298

6 274 298

1 1

6 274 3640 1 1

8 314 298 368

4 94

lnK .

K . at K

K HWe know, ln

k R T T

. ln

K .

K .

62. Sol. (B)



Activity coefficient i

i v

i i

y P

x p

1 1

1

1

2

2

At Azeotrope y

1.013 1.15

0.878

1.013 1.52

0.665

v

v

x

p

p

p

p

63. Sol. (C)

we have

2

0 48 1 521 1 15 1 12

0 62 1 15

. ln( . )A ln( . ) .

. ln( . )

2

0 62 1 151 1 52 1 00

0 48 1 52

. ln( .B ln( . ) .

. ln( . )

(Gate 2009)

64. Sol. (B)

65. Sol. (D)

The thermodynamics consistency of activity

coefficients for a binary mixture can be

cheeked by Gibbs-Duhem equation,

1 2

1 2

1 2

2

1 1 1

1

1 1

1

1

1 1 1

1

1

1 2

0 2 2 2 1

2 1

2

dln dlnx x ( )

dx dx

for all options, ln x x

and we have to find ln :

dlnhence, x x

dx

dlnx x x

dx

1

2

2 1 1

2

2

1 1 1

1

2

1

1

2

2

2 1

1 2 1

2

from equation (1),

dlnx x x

dx

dlnor, x x x

dx

dlnx

dx

ln x

66. Sol. (C)

we have

1 1 2 2

1 2

5

2

3

2

5

3

1

1

8 314 300 600

20 5

3

7482 6

7 4826

p V

v p

p

v

Mech

Mech

C C R

C C R R R

R

C

C

p v p vW.D

(r )

R T T

(r )

. ( )

.

. J / mol

. kJ / mol

Here, –ve sign indicates that work is reqd.

67. Sol. (C)

1 2

1 1

1 2

T T

p p

2

2 2

1 1

5 3

2 3

5

2

6001

30

2

/

/

p T

p T

p

bar



(D) Is the correct answer

(Gate 2010)

68. Sol. (D)

for Binary solution x1 + x2 = 1

For equimolar mixture, x1 = x2 = 0.5

From Raoult’s law,

sat

1 1 1

sat

2 2 2

1 2

sat sat

1 1 2 2

sat

1 1 1

sat

1 1 1

sat

1 11

1

p x p

p x p

p p p

p x p x p

0.5 180 0.5 120

p 150kpa

p x p

y p x p

x p 0.5 180y

p 150

y 0.6

69. Sol. (C)

0

C 0 S 200

100 0 400 200

S 500

C (500 200)1.5

100 (400 200)

C 150 C

70. Sol. (C)

Given:

F = 1 kg

P = 1500 kPa

T = 500 K

(1-x)kg

Hp = 750 kJ/kg,

T = 300 k, P = 150 kPa

Hc = 500 kJ/kg

Take energy balance

Input heat = vapor heat + liquid. Heat

F. HF = x Hv + (1-x) Hl

750 = x 2500 + (1-x) 500

Solve it for x

x = 0.125 =12.5%

71. Sol. (A)

2

1 1

1

V

P X

1 2

1 2

1 2

1

1

1

At infinite dilution, x x

dvV V X

dx x

2

2 2

2

1 1

2 2

1

1 1

1 1

1 1

2 2

1 11

2 2

xx x

d = x

dx x

= x ( )( x )

= 0.75



(Gate 2011)

72. Sol. (B)

For ideal mixture.

i i

0 1 1 2 2

i

according to randall rule for an ideal gas

f y .f

putting this for two component into given equation

W RT y lny y lny

hence, (B) is the correct answer.

73. Sol. (D)

1 21 2mix

H x H x H

1 2

2

1 2

2

2 1

1 1 2 2

1 2 3 1 2 3

3600 1600

3600 400

1 21600 400 800

3 3mix

x x

H X

H x

H

74. (B)

Isothermal expansion

1

1 2 1 2

2

81 8 314 100

4

5762 8

5 762

pQ w nRTln

p

. ln

. J

. kJ

(B) is thecorrectanswer

75. Sol. (C) is the correct answer.

76. Sol. (C) given:

(Gate 2012)

77. Sol. (B)

Throttling process adiabatic process

VP constant

1 1 1

2 2 2

12

1

0 5

2 /

P V P.

P V P

V

V

78. Sol. (A)

At constant entropy, no phase change, hence

state of water is liquid

Hence A is correct answer

79. Sol. (A)

0 system surrs s

0 0M L

L H

Q Q

T T

Q Q

T T

Thus entropy change of cold fluid is

more than that of hot fluid .

Hence(A)is thecorrectanswer

80. Sol. (D)

Equilibrium constant is only being function of

temperature according to Van’t Hoff equation,

(D) is answer

81. Sol. (A)

we have



2

2

1 2 1 1 1 2 2

1 2 1 2

1 1

1

1

0 0 0 0

s

s

p s v

s

Q m h m'u' w m h m"u"

Q ,m' , w ,m

m h m"u"

m m"

h u"

C T C T"

T" γT

Hence(A)is thecorretanswer

82. (A) we have

1 2 1 1 2 25

0

H x x and S R x lnx x lnx

we know that

G H TS

or G H T S S T

at Constant T, T

1

1 2 1 1 2 2

1 2 1 1 2 2

1 1 1 1 1 1

2

1 1 1 1 1

5

5

5 1 1 1

5 5 1 1

G H T S

G x x T R x lnx x lnx

G x x RT x lnx x lnx

G x x RT x lnx x ln x

G x x RT x lnx x ln x

1

0

Gfor G to be minimum

x

thus,

1

1

1

1 1 1

1 1

1

1 1

1 1

1 1 1 1

1

5 10

1 1

1

Gx RT x lnx

x x

RT x ln xx

here,

lnx xx lnx x lnx

x x x

lnx

1 1

1

1

1 1 1

1

1

1

1

1 1 1

1

1

1

1

1

11 1 1

11

1 1 1

1 1

5 10 1 1 1

0 5 101

0 5

ln xx ln x x

x x

xln x

x

ln x

ln x

Gx RT lnx ln x

x

xx RTln

x

on solving, we get x .

(Gate 2013)

83. Sol. 3

The degree of freedom is given as

F = C – P + 2 – r – s = 3 – 1 + 2 – 1 – 0 = 3

84. Sol. (C)

from First Law of Thermodynamics,

Q W U

U = internal energy which solely depends

only on initial and final temperature. It is

independent of path.

85. Sol. (D)

we know that, at critical point

2 2

2

40

2

TT

P P b b ac

V aV

Hence, (D) is correct answer

86. Sol. (B)

87. Sol. (A)

Given that



300

600

6002

300

0

0

2

surr

JE K

surr

universe

E F

E F

JF K

T K, Heat transferred from system

to suurounding,

Q J

Entropy change of surrounding,

QS

T

As we know S

or S S

S S

S

88. Sol. 0.0736

We know that

1 1 2 2

2 1

2

2

1 1 0 4 0 6

2000 4 0 09 0 6

8 314 300

0 0736

E

i i

Gx ln x ln x ln

RT

x x . .

. . . ln.

ln .

(Gate 2014)

89. Sol. (B)

90. Sol. (D)

for isentropic expansion

0

PV cons tant

Taking log on both sides

In P In V

In Pnegative

In V

is positive slope of In P vs InV is negative

Hence, (C) is correct answer

91. Sol. (C)

92. Sol. (D)

93. Sol. 0.6925

Form given Antoine equation log10 = A - B

t C,

calculate vapor pressure of both component

sat

10 1

sat

1

1210log P 7

25 230

P 177.8 torr

sat

10 2

sat

2

1 2

1206for component 2, log P 6.5

25 22.3

P 43.35 torr

Given x 0.11 and x 0.89

2

1 2 1In(y ) x [2 0.6x ]

20.89 [2 0.6x0.11]

1.53

1y 4.63

2

2 1 2

2

2

s s

1 1 1 1 2 2 2 2

s

1 1 1 1

s

2 2 2 2

In(y ) x [1.7 0.6x ]

0.11 [1.7 0.6 0.89]

0.027

y 1.02

from modified Roult 's law component 1 and 2

y P x y P and y P x y P

y P x y P

y P x y P

1

1

y 0.11 4.63 177.8

1 y 0.89 1.02 43.35

0.693

(Gate 2015)

94. Sol. (A)

We know that

s s

v

PT

We have

du Tds pdv

and dg vdp sdT

u uSo, P P

v v

uT

s

g gV and S

p T



95. Sol. (D)

we have

0 1 0 0

1

A B C

at t ,

Let fraction of A connverted

at t t

2

2

2

2

1 1

1 1

1

2750

8 314 573

1 78

1 78 0 801

B CP

A

P

O

Thus,

y yK P

y

K P P

Also,

GlnK

RT

K exp.

K .

so, . .

96. Sol. 310

We have

0 1 0 1

0 1

320 2

2 0 1 320

30

0 1 30

310

dPGiven . dP . dT

dT

P . T c (i)

Given at T K, P bar

. c

c

equation (i) becomes

P . T

T K

97. Sol. (D)

we have

U, S, and G are state variables (point

functions)

So U, S and G

Remain same between two states 1 and 2.

98. Sol. (B)

Catalyst changes the activation energy of the

reaction

Hence, (B) is correct answer.

99. Sol. 750

Given:

1 2 : Reversible adiabatic process

2 3 : Reversible isobaric process

Given

1

5

31

62 3

33

1 1 2 2

15

11 1 667

2 1 62

22

0 1

10

2

10

0 2

2 5 1 5

2 51 667

1 5

1 2

102

10

0 50

a

P v

P v

.

P . MPa

Pa

v m

P P P

V . m

C . R So C . R

.C / C .

.

for process : P V P V

PV V x

P

V . m

Work done on for process 1-2

6 6 61 1 2 2

1 2

1 2

2 3

6

0 1 10 2 10 10 0 5

1 0 667

450

2 3

10 0 2 0 5

300

P V P V . X x x x .w

.

w kJ

work done for process

w P V

X( . . )

kJ

Net work done = 450+350=750 kJ



100. Sol. (D)

Given R R

Pg VdP

RT RT0

1

0

34 6

1

1

11

0 10 0 2 10

20

R g

R

RP

R

R

ZRT RT RTwe know, V V V (Z )

P P P

V (Z )or

RT P

BP (Z ) BAnd Z

RT P RT

g Bso, dp

RT ERT

mg B(P ) x . x Pa

mol

g J / mol

101. Sol. 27.52

The general VLE equation

For component 1

s

1 1 1 1 1

s

2 2 2 2 1

y P x P ............................eq.1

y P x P ............................eq.2

i iat azeotrope y x

Equation 1 divides by Equation 2

s

1 1 1 1 1

s

2 2 2 2 2

y P x P

y P x P

22

21

Ax

Ax

0.9P 70e

0.9P 30e

2 22 1

A(x x )30e

70

2 22

Ax 2.12x0.70.9P 70e 70 e

P . bar27 52

(Gate 2016)

102. Sol.

Infinitely dilute solution of component (1)

1 20 1 x & x

So 2 21 2 60 1 100 0 1 h x ( ) x x( )

1 58h

103. Sol. 0.396

From Raoult’s law

. vi ii

p x p

& we’

i i

i

p pyi

p p

Mol fraction of benzene in vapor phase is

given by0 2 92

0 3960 2 92 0 8 35

V

B B BB V T

B B T Ti

x ppy

x P x pp

..

. .

104. Sol. 373

We know that

1.5irr revW W

(Change in I.E.)irr = 1.5 (change in I.E.)rev

V V

V

C (T T ) . C (T T )

T T . C (T T )

2 1 2 1

2 1 2 1

1 5

1 5

P

PT T . (T T )

1

2

12 1 1 11 5

..

T .

T K

2771 277

2

2

6300 1 5 300 1 73

3

373

105. Sol. 400

We know that

0 1

1 0

( )

RTP

v b

P v b RT

RTv

PRT

b b TP

Rb T b

P



1 0 1 0

T P

h vv T

P T

R Rh T b T b b

P P

0

0

5 6

( )

4 10 (15 5) 10 400 /

T T

f i f i TT

dh b dP

h h b P P

J mol

106. Sol. 1.647

idE ggg

RT

g

RT

g

RT

g idE

1 1 2 2 1 1 2 2

1 2

1 2 1 2 1 1 2 23

3 0 4 0 6 0 4 1 0 6 2

0 4 0 4 0 6 0 6

E id

Ei i i i

E

g g g

RT RT RT

x g RT x lnxg g

RT RT RT

x g x g RT x lnx x lnxg g

RT RT RT

g ggx x x x x lnx x lnx

RT RT RT

g. . . .

RT

. ln( . ) . ln( . )

1 647 g

.RT

(Gate 2017)

107. Sol. (C)

For gas M; P = 25 bar, Pc = 75 bar

25 1

75 3

r

c

PP

P

For Gas N; P = 75 bar, PC = 225 bar

75 1

225 3

rP

For Gas M; T = 300 K, TC = 150 K, Tr =

2C

T

T

For Gas N; T = 1000 K, TC = 500 K, Tr =

2C

T

T

Compressibility factor, Z = k r

r

P

T

1

3

2 61

3

2 6

m

n m n

KK

Z

KK

Z Z Z

108. Sol. (D)

109. Sol. 14.9

7 7 7g g

, .RT RT

As we know

R

i

x

f gexp

P RT

7 7 7

0 7 0 7

30 0 7 14 9

R

i

x

R

ig

R

i

i

f gexp

p RT

g g g.

RT RT RT

fg. p ( . )

RT Pf exp ( . ) . bar



110. Sol. 2511.4

As we know 1

1

T

dv( )

V dp

For isothermal process pv = const

Pdv +Vdp = 0 1

dv V

dp P

From equation

0

1

1

1 110P Pa

P

At constant temperature

111. Sol. 0.91

1000 0 92H bar, . for dilute solution

We can use Henry law for solute Hx = yP

1000 x = 0.92 (1-0.01) (10)

Here, y and x are mole fraction of solute in

gas liquid respectively x = 9.108 x 10-3

Mole percentage of solute in liquid phase =

100 (9.108 10-3) = 0.91

Gate 2018

112. Sol. 102.375

113. Sol (C)

114. Sol. 368.166



115. Sol.

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