Chapter 5

43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry · 82 exercises

Problem 1

A mixture of one mole each of \(\mathrm{H}_{2}, \mathrm{He}\) and \(\mathrm{O}_{2}\) each are enclosed in a cylinder of volume \(V\) at temperature \(T\). If the partial pressure of \(\mathrm{H}_{2}\) is \(2 \mathrm{~atm}\), the total pressure of the gases in the cylinder is : (a) 6 atm (b) 38 atm (c) \(14 \mathrm{~atm}\) (d) \(22 \mathrm{~atm}\)

4 step solution

Problem 2

Consider the van der Waals constants, a and \(\mathrm{b}\), for the following gases, Gas \(\quad \mathrm{Ar} \quad \mathrm{Ne} \quad \mathrm{Kr} \quad \mathrm{Xe}\) \(\begin{array}{lllll}\mathrm{a} /\left(\mathrm{atm} \mathrm{dm}^{6} \mathrm{~mol}^{-2}\right) & 1.3 & 0.2 & 5.1 & 4.1\end{array}\) \(\begin{array}{lllll}\mathrm{b} /\left(10^{-2} \mathrm{dm}^{3} \mathrm{~mol}^{-1}\right) & 3.2 & 1.7 & 1.0 & 5.0\end{array}\) Which gas is expected to have the highest critical temperature? [Main April 9, 2019 (I)] (a) \(\mathrm{Kr}\) (b) \(\mathrm{Ne}\) (c) Xe (d) \(\mathrm{Ar}\)

4 step solution

Problem 3

The predominant intermolecular forces present in ethyl acetate, a liquid, are: (a) London dispersion and dipole-dipole (b) hydrogen bonding and London dispersion (c) Dipole-dipole and hydrogen bonding (d) London dispersion, dipole-dipole and hydrogen bonding

4 step solution

Problem 3

At a given temperature \(\mathrm{T}\), gases \(\mathrm{Ne}, \mathrm{Ar}, \mathrm{Xe}\) and \(\mathrm{Kr}\) are found to deviate from ideal gas behaviour. Their equation of state is given as $$ \mathrm{P}=\frac{\mathrm{RT}}{\mathrm{V}-\mathrm{b}} \text { at } \mathrm{T} $$ Here, \(\mathrm{b}\) is the van der Waals constant. Which gas will exhibit steepest increase in the plot of \(Z\) (compression factor) vs \(\mathrm{P}\) ? [Main April 9, 2019 (II)] (a) Xe (b) \(\mathrm{Kr}\) (c) \(\mathrm{Ne}\) (d) Ar

4 step solution

Problem 4

The relative strength of interionic/ intermolecular forces in decreasing order is: (a) dipole-dipole \(>\) ion-dipole \(>\) ion-ion (b) ion-dipole \(>\) ion-ion \(>\) dipole-dipole (c) ion-dipole \(>\) dipole-dipole \(>\) ion-ion (d) ion-ion \(>\) ion-dipole \(>\) dipole-dipole

3 step solution

Problem 4

Initially, the root mean square (rms) velocity of \(\mathrm{N}_{2}\) molecules at certain temperature is \(\mathrm{u}\). If this temperature is doubled and all the nitrogen molecules dissociate into nitrogen atoms, then the rms velocity will be : (a) \(2 \mathrm{u}\) (b) \(14 \mathrm{u}\) (c) \(4 \mathrm{u}\) (d) \(\mathrm{u} / 2\)

6 step solution

Problem 4

At very high pressures, the compressibility factor of one mole of a gas is given by: [Main Online April 9, 2016] (a) \(1+\frac{P b}{R T}\) (b) \(\frac{P b}{R T}\) (c) \(1-\frac{P b}{R T}\) (d) \(1-\frac{b}{(V R T)}\)

3 step solution

Problem 5

\(0.5\) moles of gas \(\mathrm{A}\) and \(x\) moles of gas \(\mathrm{B}\) exert a pressure of \(200 \mathrm{~Pa}\) in a container of volume \(10 \mathrm{~m}^{3}\) at \(1000 \mathrm{~K}\). Given \(\mathrm{R}\) is the gas constant in \(\mathrm{JK}^{-1} \mathrm{~mol}^{-1}, x\) is: (a) \(\frac{2 \mathrm{R}}{4+\mathrm{R}}\) (b) \(\frac{2 \mathrm{R}}{4-\mathrm{R}}\) (c) \(\frac{4+\mathrm{R}}{2 \mathrm{R}}\) (d) \(\frac{4-\mathrm{R}}{2 \mathrm{R}}\)

7 step solution

Problem 5

Which of the following is not an assumption of the kinetic theory of gases? (a) Gas particles have negligible volume. (b) A gas consists of many identical particles which are in continual motion. (c) At high pressure, gas particles are difficult to compress. (d) Collisions of gas particles are perfectly elastic.

3 step solution

Problem 6

An open vessel at \(27^{\circ} \mathrm{C}\) is heated until two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is : (a) \(500^{\circ} \mathrm{C}\) (b) \(500 \mathrm{~K}\) (c) \(750{ }^{\circ} \mathrm{C}\) (d) \(750 \mathrm{~K}\)

5 step solution

Problem 6

The ratio of masses of oxygen and nitrogen in a particular gaseous mixture is \(1: 4\). The ratio of number of their molecules is: (a) \(1: 4\) (b) \(7: 32\) (c) \(1: 8\) (d) \(3: 16\)

5 step solution

Problem 6

If \(Z\) is a compressibility factor, van der Waals equation at low pressure can be written as: [Main 2014] (a) \(Z=1+\frac{R T}{P b}\) (b) \(Z=1-\frac{a}{V R T}\) (c) \(Z=1-\frac{P b}{R T}\) (d) \(Z=1+\frac{P b}{R T}\)

4 step solution

Problem 7

Assuming ideal gas behaviour, the ratio of density of ammonia to that of hydrogen chloride at same temperature and pressure is: (Atomic wt. of \(\mathrm{Cl}=35.5 \mathrm{u}\) ) (a) \(1.46\) (b) \(1.64\) (c) \(0.46\) (d) \(0.64\)

5 step solution

Problem 7

The temperature at which oxygen molecules have the same root mean square speed as helium atoms have at \(300 \mathrm{~K}\) is: Atomic masses: He \(=4 \mathrm{u}, \mathrm{O}=16 \mathrm{u})\) (a) \(300 \mathrm{~K}\) (b) \(600 \mathrm{~K}\) (c) \(1200 \mathrm{~K}\) (d) \(2400 \mathrm{~K}\)

5 step solution

Problem 7

van der Waals equation for a gas is stated as, $$ P=\frac{n R T}{V-n b}-a\left(\frac{n}{V}\right)^{2} $$ [Main Online April 9, 2014] This equation reduces to the perfect gas equation, \(P=\frac{n R T}{V}\) when, (a) temperature is sufficient high and pressure is low. (b) temperature is sufficient low and pressure is high. (c) both temperature and pressure are very high. (d) both temperature and pressure are very low.

4 step solution

Problem 8

At \(300 \mathrm{~K}\), the density of a certain gaseous molecule at 2 bar is double to that of dinitrogen \(\left(\mathrm{N}_{2}\right)\) at 4 bar. The molar mass of gaseous molecule is : (a) \(28 \mathrm{~g} \mathrm{~mol}^{-1}\) (b) \(56 \mathrm{~g} \mathrm{~mol}^{-1}\) (c) \(112 \mathrm{~g} \mathrm{~mol}^{-1}\) (d) \(224 \mathrm{~g} \mathrm{~mol}^{-1}\)

5 step solution

Problem 8

For gaseous state, if most probable speed is denoted by \(\mathrm{C}^{*}\), average speed by \(\overline{\mathrm{C}}\) and mean square speed by \(\mathrm{C}\), then for a large number of molecules the ratios of these speeds are : (a) \(\mathrm{C}^{*}: \overline{\mathrm{C}}: \mathrm{C}=1.225: 1.128: 1\) (b) \(\mathrm{C}^{*}: \overline{\mathrm{C}}: \mathrm{C}=1.128: 1.225: 1\) (c) \(C^{*}: \bar{C}: C=1: 1.128: 1.225\) (d) \(\mathrm{C}^{*}: \overline{\mathrm{C}}: \mathrm{C}=1: 1.225: 1.128\)

3 step solution

Problem 9

Two closed bulbs of equal volume \((V)\) containing an ideal gas initially at pressure \(p_{i}\) and temperature \(T_{I}\) are connected through a narrow tube of negligible volume. The temperature of one of the bulbs is then raised to \(T_{2}\). The final pressure \(p_{f}\) is : (a) \(2 p_{i}\left(\frac{T_{2}}{T_{1}+T_{2}}\right)\) (b) \(2 p_{i}\left(\frac{T_{1} T_{2}}{T_{1}+T_{2}}\right)\) (c) \(p_{i}\left(\frac{T_{1} T_{2}}{T_{1}+T_{2}}\right)\) (d) \(2 p_{i}\left(\frac{T_{1}}{T_{1}+T_{2}}\right)\)

7 step solution

Problem 9

By how many folds the temperature of a gas would increase when the root mean square velocity of the gas molecules in a container of fixed volume is increased from \(5 \times 10^{4} \mathrm{~cm} / \mathrm{s}\) to \(10 \times 10^{4} \mathrm{~cm} / \mathrm{s}\) ? (a) Two (b) Three (c) Six (d) Four

5 step solution

Problem 9

The term that corrects for the attractive forces present in a real gas in the van der Waals equation is \([2009-3 M ;-1]\) (a) \(n b\) (b) \(\frac{a n^{2}}{V^{2}}\) (c) \(-\frac{a n^{2}}{V^{2}}\) (d) \(-n b\)

4 step solution

Problem 10

The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is: (a) London force (b) hydrogen bond (c) ion - ion interaction (d) ion - dipole interaction

6 step solution

Problem 10

When the temperature is increased, surface tension of water (a) increases (b) decreases (c) remains constant (d) shows irregular behaviour

4 step solution

Problem 11

When does a gas deviate the most from its ideal behaviour? (a) At low pressure and low temperature (b) At low pressure and high temperature (c) At high pressure and low temperature (d) At high pressure and high temperature

3 step solution

Problem 11

The compressibility of a gas is less than unity at STP. Therefore, (a) \(V_{m}>22.4\) litres (b) \(V_{m}<22.4\) litres (c) \(V_{m}=22.4\) litres (d) \(V_{m}=44.8\) litres

4 step solution

Problem 12

Sulphur dioxide and oxygen were allowed to diffuse through a porous partition. \(20 \mathrm{dm}^{3}\) of \(\mathrm{SO}_{2}\) diffuses through the porous partition in 60 seconds. The volume of \(\mathrm{O}_{2}\) in \(\mathrm{dm}^{3}\) which diffuses under the similar condition in 30 seconds will be (atomic mass of sulphur \(=32 \mathrm{u}\) ): (a) \(7.09\) (b) \(14.1\) (c) \(10.0\) (d) \(28.2\)

7 step solution

Problem 12

The rms velocity of hydrogen is \(\sqrt{7}\) times the rms velocity of nitrogen. If \(T\) is the temperature of the gas, then (a) \(T\left(\mathrm{H}_{2}\right)=T\left(\mathrm{~N}_{2}\right)\) (b) \(T\left(\mathrm{H}_{2}\right)>T\left(\mathrm{~N}_{2}\right)\) (c) \(T\left(\mathrm{H}_{2}\right)

6 step solution

Problem 12

The compressibility factor for an ideal gas is (a) \(1.5\) (b) \(1.0\) (c) \(2.0\) (d) \(\infty\)

3 step solution

Problem 13

The ratio of the rate of diffusion of helium and methane under identical condition of pressure and temperature will be (a) 4 (b) 2 (c) 1 (d) \(0.5\)

5 step solution

Problem 13

The ratio between the root mean square speed of \(\mathrm{H}_{2}\) at \(50 \mathrm{~K}\) and that of \(\mathrm{O}_{2}\) at \(800 \mathrm{~K}\) is, (a) 4 (b) 2 (c) 1 (d) \(1 / 4\)

7 step solution

Problem 14

Positive deviation from ideal behaviour takes place because of (a) Molecular interaction between atoms and \(P V / n R T>1\) (b) Molecular interaction between atoms and \(P V \ln R T<1\) (c) Finite size of atoms and \(P V / n R T>1\) (d) Finite size of atoms and \(\mathrm{PV} / \mathrm{nRT}<1\)

5 step solution

Problem 14

Longest mean free path stands for : (a) \(\mathrm{H}_{2}\) (b) \(\mathrm{N}_{2}\) (c) \(\mathrm{O}_{2}\) (d) \(\mathrm{Cl}_{2}\)

4 step solution

Problem 14

The values of van der Waals constant ' \(a\) ' for the gases \(\mathrm{O}_{2}, \mathrm{~N}_{2}, \mathrm{NH}_{3}\) and \(\mathrm{CH}_{4}\) are \(1.360,1.390,4.170\) and \(2.253 \mathrm{~L}^{2}\) atm \(\mathrm{mol}^{-2}\) respectively. The gas which can most easily be liquified is : [1989-1 Mark] (a) \(\mathrm{O}_{2}\) (b) \(\mathrm{N}_{2}\) (c) \(\mathrm{NH}_{3}\) (d) \(\mathrm{CH}_{4}\)

4 step solution

Problem 15

According to kinetic theory of gases, for a diatomic molecule (a) the pressure exerted by the gas is proportional to mean velocity of the molecule (b) the pressure exerted by the gas is proportional to the root mean velocity of the molecule (c) the root mean square velocity of the molecule is inversely proportional to the temperature (d) the mean translational kinetic energy of the molecule is proportional to the absolute temperature.

4 step solution

Problem 15

In van der Waals equation of state for a non-ideal gas, the term that accounts for intermolecular forces is [1988-1 Mark] (a) \((V-b)\) (b) \(R T\) (c) \(\left(P+\frac{a}{V^{2}}\right)\) (d) \((R T)^{-1}\)

3 step solution

Problem 16

At \(100^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\), if the density of liquid water is \(1.0 \mathrm{~g} \mathrm{~cm}^{-3}\) and that of water vapour is \(0.0006 \mathrm{~g} \mathrm{~cm}^{-3}\), then the volume occupied by water molecules in 1 litre of steam at that temperature is (a) \(6 \mathrm{~cm}^{3}\) (b) \(60 \mathrm{~cm}^{3}\) (c) \(0.6 \mathrm{~cm}^{3}\) (d) \(0.06 \mathrm{~cm}^{3}\)

3 step solution

Problem 16

The average velocity of an ideal gas molecule at \(27^{\circ} \mathrm{C}\) is \(0.3 \mathrm{~m} / \mathrm{sec}\). The average velocity at \(927^{\circ} \mathrm{C}\) will be: (a) \(0.6 \mathrm{~m} / \mathrm{sec}\) (b) \(0.3 \mathrm{~m} / \mathrm{sec}\) (c) \(0.9 \mathrm{~m} / \mathrm{sec}\) (d) \(3.0 \mathrm{~m} / \mathrm{sec}\)

6 step solution

Problem 17

A gas will approach ideal behaviour at (a) low temperature and low pressure. (b) low temperature and high pressure. (c) high temperature and low pressure. (d) high temperature and high pressure.

4 step solution

Problem 17

Helium atom is two times heavier than a hydrogen molecule. At 298 \(\mathrm{K}\), the average kinetic energy of a helium atom is (a) two times that of a hydrogen molecule. (b) same as that of a hydrogen molecule. (c) four times that of a hydrogen molecule. (d) half that of a hydrogen molecule.

4 step solution

Problem 18

\(X \mathrm{~mL}\) of \(\mathrm{H}_{2}\) gas effuses through a hole in a container in 5 seconds. The time taken for the effusion of the same volume of the gas specified below under identical conditions is : (a) 10 seconds : He (b) 20 seconds : \(\mathrm{O}_{2}\) (c) 25 seconds : \(\mathrm{CO}\) (d) 55 seconds : \(\mathrm{CO}_{2}\)

7 step solution

Problem 18

The ratio of root mean square velocity to average velocity of a gas molecule at a particular temperature is (a) \(1.086: 1\) (b) \(1: 1.086\) (c) \(2: 1.086\) (d) \(1.086: 2\)

4 step solution

Problem 19

One mole of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) at \(300 \mathrm{~K}\) is kept in a closed container under one atmosphere. It is heated to \(600 \mathrm{~K}\) when \(20 \%\) by mass of \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g})\) decomposes to \(\mathrm{NO}_{2}(\mathrm{~g})\). The resultant pressure is : (a) \(1.2 \mathrm{~atm}\) (b) \(2.4 \mathrm{~atm}\) (c) \(2.0 \mathrm{~atm}\) (d) \(1.0 \mathrm{~atm}\)

7 step solution

Problem 19

At \(400 \mathrm{~K}\), the root mean square (rms) speed of a gas \(\mathbf{X}\) (molecular weight \(=40\) ) is equal to the most probable speed of gas \(\mathbf{Y}\) at \(60 \mathrm{~K}\). The molecular weight of the gas \(\mathbf{Y}\) is

5 step solution

Problem 19

A gas described by van der Waals equation [2008-1 Mark] (a) behave similar to an ideal gas in the limit of large molar volumes (b) behaves similar to an ideal gas is in limit of large pressures (c) is characterised by van der Waals coefficients that are dependent on the identity of the gas but are independent of the temperature. (d) has the pressure that is lower than the pressure exerted by the same gas behaving ideally

5 step solution

Problem 20

At constant volume, for a fixed number of moles of a gas the pressure of the gas increases with rise in temperature due to (a) Increase in average molecular speed (b) Increased rate of collisions amongst molecules (c) Increase in molecular attraction (d) Decrease in mean free path

5 step solution

Problem 21

The density of neon will be highest at (a) S.T.P. (b) \(0^{\circ} \mathrm{C}, 2 \mathrm{~atm}\) (c) \(273^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). (d) \(273^{\circ} \mathrm{C}, 2 \mathrm{~atm}\).

5 step solution

Problem 21

Eight gram each of oxygen and hydrogen at \(27^{\circ} \mathrm{C}\) will have the total kinetic energy in the ratio of \(\ldots \ldots \ldots \ldots .\)

6 step solution

Problem 21

Read the following statement and explanation and answer as per the options given below : Assertion : The value of van der Waals'constant ' \(\mathrm{a}\) ' is larger for ammonia than for nitrogen. Reason : Hydrogen bonding is present in ammonia. [1998 - 2 Marks] (a) If both assertion and reason are correct, and reason is the correct explanation of the assertion. (b) If both assertion and reason are correct, but reason is not the correct explanation of the assertion. (c) If assertion is correct but reason is incorrect. (d) If assertion is incorrect but reason is correct.

5 step solution

Problem 22

The rate of diffusion of methane at a given temperature is twice that of a gas \(X\). The molecular weight of \(X\) is (a) \(64.0\) (b) \(32.0\) (c) \(4.0\) (d) \(8.0\)

5 step solution

Problem 23

A bottle of dry ammonia and a bottle of dry hydrogen chloride connected through a long tube are opened simultaneously at both ends the white ammonium chloride ring first formed will be (a) at the centre of the tube. (b) near the hydrogen chloride bottle. (c) near the ammonia bottle. (d) throughout the length of the tube.

4 step solution

Problem 23

The compression factor (compressibility factor) for one mole of a van der Waals gas at \(0^{\circ} \mathrm{C}\) and 100 atmospheric pressure is found to be \(0.5\). Assuming that the volume of a gas molecule is negligible, calculate the van der Waals constant \(a\).

7 step solution

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