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JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - JEE MCQ


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30 Questions MCQ Test - JEE Advanced 2018 Question Paper with Solutions (20th May - Evening)

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*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 1

A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving along the x-axis. Its kinetic energy K changes with time as dK/dt = γt, where γ is a positive constant of appropriate dimensions. Which of the following statements is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 1

The kinetic energy of the particle is given by,

Integrate both sides of the above equation.

As γ and m are constant; therefore, v ∝t.
The acceleration of the particle is calculated as,

The force on the particle is given by,
F = ma

F = constant
Find the distance of the particle from the origin.

Thus, distance does not increase linearly with time.

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 2

Consider a thin square plate floating on a viscous liquid in a large tank. The height h of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity μ0​. Which of the following statements is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 2

The resistive force on the plate is viscous force, therefore,

From the equation,

The tangential shear stress on the floor is given by,

From the equation, 

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*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 3

An infinitely long thin non-conducting wire is parallel to the z-axis and carries a uniform line charge density λ. It pierces a thin non-conducting spherical shell of radius R in such a way that the arc PQ subtends an angle 120° at the centre O of the spherical shell, as shown in the figure. The permittivity of free space is ε0. Which of the following statements is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 3

Consider the diagram.

Use trigonometry to find PQ.
PQ = 2PM
= 2Rsin60°

The charge enclosed by the portion of wire PQ is given by,

Apply Gauss law,

This flux points radially outward from the wire so that the electric field in the direction of z-component is zero and also it is normal to the wire, not to the spherical shell.

JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 4

A wire is bent in the shape of a right angled triangle and is placed in front of a concave mirror of focal length f, as shown in the figure. Which of the figures shown in the four options qualitatively represent(s) the shape of the image of the bent wire? (These figures are not to scale.)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 4

The image of line AB is behind the mirror.
Use the mirror formula,

Thus, the line AB is perpendicular to the principle axis behind the mirror. Also the point C is at focus point; therefore, its image is formed at infinity.

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 5

In a radioactive decay chain, nucleus decays to  nucleus. Let Nα and Nβ be the number of α and β particles, respectively, emitted in this decay process. Which of the following statements is (are) true?

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 5

Let the number of alpha particles emitted be x and the number of beta particles emitted be y in the decay process.
The reaction for the radioactive decay is shown below:
Equate the mass numbers on both sides of the equation, 232=212+4x+0y
4x=20
x=5
Equate the atomic numbers on both sides of the equation, 90=82+2x-y
2x-y=8
y=2

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 6

In an experiment to measure the speed of sound by a resonating air column, a tuning fork of frequency 500Hz is used. The length of the air column is varied by changing the level of water in the resonance tube. Two successive resonances are heard at air columns of length 50.7cm and 83.9cm. Which of the following statements is (are) true? 

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 6

The difference of two successive resonances is equal to half of the wavelength, thus,

The level 50.7 cm is corresponding to third harmonic. If end correction is e, then,

The speed of sound is calculated as,
v=fλ
= 500 x 66.4
= 332.00 m/s

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 7

A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass m = 0.4kg is at rest on this surface. An impulse of 1.0Ns is applied to the block at time to t = 0 so that it starts moving along the x-axis with a velocity v(t) = v0e–t/τ, where v0 is a constant and τ = 4s. The displacement of the block, in metres, at t = τ is_____________. Take e−1 = 0.37.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 7

 At time t = 0, due to impulse, the block moves with velocity v0.
Thus, find v0.

= 1/0.4
= 2.5 m/s
The velocity along the x-axis is calculated as,

Integrate both sides of the above equation.

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 8

A ball is projected from the ground at an angle of 45° with the horizontal surface. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal surface. The maximum height it reaches after the bounce, in meters is ___________.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 8

Consider the diagram showing the given condition for the ball.

For the angle 45°, the maximum height is calculated as,


The velocity of the ball becomes u' after hitting the ground and it loses half of its kinetic energy at that time, that is,

The maximum height for the angle 30° is calculated below,



= 30 m

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 9

A particle, of mass 10−3 kg and charge 1.0 C is initially at rest. At time t = 0 , the particle comes under the influence of an electric field  where E0 = 1.0 NC−1 and ω = 103 rad s−1. Consider the effect of only the electrical force on the particle. Then the maximum speed, in ms−1, attained by the particle at subsequent times is ____________.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 9

The velocity of the charged particle is maximum when the electrostatic force becomes zero. Thus,


Due to electric field, charge particles start to be in motion, therefore,


Simplify the above equation,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 10

A moving coil galvanometer has 50 turns and each turn has an area 2 × 10−4 m2. The magnetic field produced by the magnet inside the galvanometer is 0.02 T. The torsional constant of the suspension wire is 10−4 N m rad−1. When a current flows through the galvanometer, a full scale deflection occurs if the coil rotates by 0.2 rad. The resistance of the coil of the galvanometer is 50Ω . This galvanometer is to be converted into an ammeter capable of measuring current in the range 0−1.0A. For this purpose, a shunt resistance is to be added in parallel to the galvanometer. The value of this shunt resistance, in ohms, is __________.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 10

The moment of the deflecting couple is equal to the moment of the restoring couple. That is,



The resistance is connected in shunt with the ammeter for the galvanometer as follows,

From the above circuit the value of S is calculated below,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 11

A steel wire of diameter 0.5 mm and Young’s modulus 2 ×1011 N m−2 carries a load of mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale division which coincides with a main scale division is __________. Take g = 10 ms−2 and π = 3.2 .


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 11

According to Hooke’s law,


The least count of the vernier caliper is 
Therefore, the third division of vernier scale coincides with the main scale.

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 12

One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R = 8.0 j mol−1 K−1, the decrease in its internal energy, in joule, is__________.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 12

As per the adiabatic process,

For mono-atomic gas, the adiabatic index is given as,

Thus, the final temperature of the gas is calculated below,

The decrease in internal energy for the mono-atomic gas is given as,

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 13

In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The frequency of light is just above the threshold frequency so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is 100%. A potential difference of 500 V is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force F= n ×10−4N due to the impact of the electrons. The value of n is __________. Mass of the electron me = 9 x 10-31kg and 1.0 eV = 1.6 ×10−19 j.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 13

Let N be the number of photons emitted per second, then, P = Nhv 
For negligible kinetic energy, the work function is equal to the energy of incident photon. Thus,


The change in kinetic energy is calculated as,
K = qV

As the work efficiency is 100% , thus, the number of electrons is equal to the number of photons per second.
The force experienced by the anode electrons is given as,

n = 24

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 14

Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The ionization energy of the hydrogen atom is 13.6 eV The value of Z is __________.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 14

Calculate the transition energy for the emitted photon from
n = 2 to n = 1.
   ...(1)
Calculate the transition energy for the emitted photon from n = 3 to n = 2.
  ...(2)
Equate the equations (1) and (2), to find the value of z.

z2 = 9
z = 3

JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 15

The electric field E is measured at a point P (0,0,d) generated due to various charge distributions and the dependence of E on d is found to be different for different charge distributions. List-I contains different relations between E and d. List-II describes different electric charge distributions, along with their locations. Match the functions in List-I with the related charge distributions in List-II.

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 15

(1) For a point charge, the electric field is given as,

(2) For a dipole, the electric field is given as,

(3) For a line charge, the electric field is given as,

(4) Net electric field due to two infinite wires is given as,

(5) The electric field due to sheet is given as,

Hence, E is independent of d.

JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 16

A planet of mass M, has two natural satellites with masses m1 and m2. The radii of their circular orbits are R1 and R2 respectively. Ignore the gravitational force between the satellites. Define v1, L1,K1 and T1 to be, respectively, the orbital speed, angular momentum, kinetic energy and time period of revolution of satellite 1; and v2, L2,K2 and T2 to be the corresponding quantities of satellite 2. Given m1/m2 = 2 and R1/R2 = 1/4, match the ratios in List-I to the numbers in List-II.

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 16

(P) The speed of the natural satellite is given as,

(Q) Angular momentum of the satellite is given as,
L = mvR
Therefore,

(R) The kinetic energy of the satellite is given by,

Therefore,


(S) The Time period of the satellite is given by,

JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 17

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV -diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II.

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 17

(P) From the thermodynamic process shown, the process (1) represents adiabatic process. The heat transfer is zero in case of adiabatic process that is no heat is exchanged between the gas and its surroundings.
(Q) The second thermodynamic process is isobaric process. Pressure is constant in isobaric process. Thus, the work done by the gas is given as,

(R) The third thermodynamics process is isochoric process. Volume is constant in this process. The work done by the gas is zero.
W = 0
(S) The fourth thermodynamics process is isothermal process. Temperature is constant in this process. ∆T= 0

JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 18

In the List-I below, four different paths of a particle are given as functions of time. In these functions, α and β are positive constants of appropriate dimensions and α ≠ β . In each case, the force acting on the particle is either zero or conservative. In List-II, five physical quantities of the particle are mentioned:  is the linear momentum,  is the angular momentum about the origin, K is the kinetic energy, U is the potential energy and E is the total energy. Match each path in List-I with those quantities in List-II, which are conserved for that path.

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 18

(P) The velocity of the particle is given by,

= constant
The linear momentum is given by,
= constant
The angular momentum of the particle is given by,

The relation between the force and the potential energy is,

U = constant
The kinetic energy of the particle is,

= constant
Total energy is also constant.
(Q) The velocity of the particle is given by,

So, both the linear momentum and the kinetic energy are not conserved.
Linear acceleration of the particle is given by,

Force on the particle is calculated as,

The potential energy of the particle is,

Angular momentum of the particle is,

Total energy is also constant because force is central. (R) The velocity of the particle is given by,

Thus, linear momentum is not conserved.
Angular momentum of the particle is,

But the magnitude of the velocity is constant.

Kinetic energy of the particle is,

Linear acceleration of the particle is given as,

The potential energy of the particle is,

U = constant
Total energy is also constant as force is central.
(S) The velocity of the particle is,

So, linear momentum, angular momentum, and kinetic energy are not conserved.
Linear acceleration of the particle is given as,

Force on the particle is,

The potential energy of the particle is,

Total energy is also constant as force is central.

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 19

​The correct option(s) regarding the complex 

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 19

(A) The following figure shows the structures of isomers formed.


Therefore, it has two geometrical isomers.
(B) The following figure shows the structures of isomers formed.

Therefore, it has three geometrical isomers.
(C) The oxidation state of Cobalt is +3. It is d6 in configuration and as it is attached to strong field ligand. Therefore, it is diamagnetic.
(D) Ammonia is a strong ligand in comparison to water. Therefore, CFSE is lower in the former case and thus, it will absorb light of longer wavelength.

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 20

The correct option(s) to distinguish nitrate salts of Mn2+ and Cu2+ taken separately is (are)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 20

(A) Both the ions Mn2+ and Cu2+ cannot be distinguished as they both impart green colour to the flame.
(B) Cu2+ belongs to group (II) and thus, on passing H2S through acidic medium it forms precipitate of CuS.
(C) Both the ions Mn2+ and Cu2+ form precipitate on passing H2S through basic medium.
(D) According to the electrochemical series the value of standard reduction potential for Cu2+ is greater than that of Mn2+.

JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 21

Aniline reacts with mixed acid (conc. HNO3 and conc. H2SO4) at 288 K to give P (51%), Q (47%) and R (2%). The major product(s) of the following reaction sequence is (are)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 21

The following figure shows the correct reaction sequence.

JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 22

The Fischer presentation of D-glucose is given below.

The correct structure(s) of β-L-glucopyranose is (are)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 22

The following figure shows the structure of β − L− glucopyranose


JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 23

For a first order reaction A(g) → 2B(g) + C(g) at constant volume and 300 K, the total pressure at the beginning (t=0) and at time t are P0 and Pt, respectively. Initially, only A is present with concentration [A]0, and t1/3 is the time required for the partial pressure of A to reach 1/3rd of its initial value. The correct option(s) is (are).
(Assume that all these gases behave as ideal gases)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 23

The final pressures are calculated below.

The first order rate equation is given as,

The rate constant is not dependent on the concentration. Hence, the following figure shows the graph obtained.

*Multiple options can be correct
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 24

For a reaction, the plots of [A] and [P] with time at temperatures T1 and T2 are given below.
If T2 >T1, the correct statement(s) is (are)
(Assume ∆Hθ and ∆Sθ are independent of temperature and ratio of lnK at T1 to lnK at T2 is greater than T2/T1. Here H, S, G and K are enthalpy, entropy, Gibbs energy and equilibrium constant, respectively.)

Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 24

As the temperature increases the concentration decreases. That is the given reaction is exothermic in nature. Therefore, ∆HΘ < 0 .
It is given that,

With increase in temperature, the value of change in Gibbs free energy increases, this occurs when ∆SΘ< 0 .

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 25

The total number of compounds having at least one bridging oxo group among the molecules given below is ____.
N2O3, N2O5, P4O6, H4P2O5, H5P3O10, H2S2O3, H2S2O5


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 25

The structures corresponding to the given compounds are shown below.







The compounds containing A − O− A or A =O are known as oxo compounds. Thus, the compounds containing at least one oxo group among the molecules is six .

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 26

Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnace such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is ____. (Atomic weights in g mol−1: O = 16, S = 32, Pb = 207)


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 26

The given reactions are,
2PbS+3O2 → 2PbO + 2SO2
2PbO + PbS → 3Pb SO2 
The number of moles of lead produced by 3 moles of oxygen is 3.
Mass of 1 mole of oxygen is 16 g .
Thus, the mass of 3 moles of oxygen is calculated below.
M= 3 x 16 g
= 96 g
Mass of 1 mole of lead is 207.2 g.
Thus, find the mass of 3 moles of lead.

Mass of lead produced by 96 g oxygen is 621.6 g .
Thus, the mass of lead produced by 1 g oxygen is calculated below,

Mass of lead produced by 1000 g oxygen is calculated below,

Thus, the mass of lead produced by 1 kg oxygen is 6.475 kg .

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 27

To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely converted to KMnO4 using the reaction,
MnCl2 + K2S2O8 + H2O → KMnO4 + H2SO4 + HCl (equation not balanced).
Few drops of concentrated HCl were added to this solution and gently warmed. Further, oxalic acid (225 mg) was added in portions till the colour of the permanganate ion disappeared. The quantity of MnCl2 (in mg) present in the initial solution is ____.
(Atomic weights in g mol-1: Mn = 55, Cl = 35.5)


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 27

The given reactions are.

Number of millimoles of oxalic acid are calculated below,

The number of millimoles of KMnO4 that react with oxalic acid is 1.
The number of millimoles of MnCl2 that are required initially is 1.
Thus, calculate the mass of  MnCl2 that is required initially.
M=1x126 mg
= 126 mg

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 28

For the given compound X, the total number of optically active stereoisomers is ____.


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 28

In the given diagram the solid wedge represents the configuration and geometry is fixed. While the dashed wedge shows that the configuration and geometry at the specified carbon is not fixed.
The number of stereocentres is three.
Thus, the number of stereoisomers is calculated below.
2n = 23
=8
There also occurs the formation of one meso-compound. Therefore, the total number of optically active stereoisomers is 8 −1= 7 .

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 29

In the following reaction sequence, the amount of D (in g) formed from 10 moles of acetophenone is ____.
(Atomic weights in g mol-1: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%) corresponding to the product in each step is given in the parenthesis)


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 29

The following figure shows the correct sequence of reaction.

Thus, the moles of D formed are calculated below,

The molar mass for tribromoaniline is 330 g / mol .
Mass of D formed is given as,
m = 1.5 mol 330 g / mol
= 495 g

*Answer can only contain numeric values
JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 30

The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas contains 1 mole % of water vapour as impurity. The water vapour oxidises 
copper as per the reaction given below:
2Cu(s) + H2O(g) → Cu2O(s) + H2(g)
is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at 1250 K. The value of ln () is ____.
(Given: total pressure = 1 bar,
R (universal gas constant) = 8 J K-1mol-1, ln (10) = 2.3. Cu(s) and Cu2O(s) are mutually immiscible.
At 1250 K: 

Gibbs energy)


Detailed Solution for JEE Advanced 2018 Question Paper with Solutions (20th May - Evening) - Question 30

The given equations are shown below.


Subtract the equation (2) from (1).

The relation between Gibbs free energy and equilibrium constant is given below.


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