Calculate the root mean square speed of $\mathrm{H}_2$ molecules under following condition. (a) 2 mole of $\mathrm{H}_2$ at $27^{\circ} \mathrm{C}$. (b) 3 mole of $\mathrm{H}_2$ in a 5 lit container at $10^5 \mathrm{~Pa}$.

(a) $1934.24 \mathrm{~m} / \mathrm{sec};$ (b) $250 \mathrm{~m} / \mathrm{sec}$

(a) $1349.44 \mathrm{~m} / \mathrm{sec};$ (b) $500 \mathrm{~m} / \mathrm{sec}$

(a) $1446.19 \mathrm{~m} / \mathrm{sec};$ (b) $390 \mathrm{~m} / \mathrm{sec}$

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JEE • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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Showing 1-30 of 262 questions

1 Medium

1 of 262

Application +4 | -1

A quantity of hydrogen gas occupies a volume of

30 \mathrm{~mL}

at a certain temperature and pressure. What volume would half this mass of hydrogen occupy at triple the absolute temperature and the pressure were one-ninth that of the original gas?

View answer and explanation

Correct option: A

405 \mathrm{~mL}

Option A (405 mL) is correct because it is calculated using the combined gas law, considering the changes in mass, temperature, and pressure.

2 Easy

2 of 262

Understanding +4 | -1

In deriving the kinetic equation we make use of the root mean square speed of the molecules which is:

View answer and explanation

Correct option: C

The square root of the average of the square of the speed of the molecules

Option C is correct because the root mean square speed is defined as the square root of the average of the squares of the speeds of the molecules.

3 Medium

3 of 262

Understanding +4 | -1

Consider the following statements for kinetic molecular theory of gases
a. Particles of a gas move in all possible directions in straight lines.
b. At any particular time, different particles in the gas have different kinetic energies.
c. Collisions of gas molecules are perfectly inelastic in nature.
The correct statement(s) is/are

View answer and explanation

Correct option: C

a and b only

Option C (a and b only) is correct because gas particles move in straight lines and different particles have different kinetic energies at a given time, but collisions are perfectly elastic.

4 Easy

4 of 262

Remembering +4 | -1

The compressibility factor of an ideal gas is

View answer and explanation

Correct option: A

Option A (1) is correct because the compressibility factor (Z) for an ideal gas is always equal to 1.

5 Medium

5 of 262

Understanding +4 | -1

The relative strength of interionic/intermolecular forces in decreasing order is

View answer and explanation

Correct option: B

ion-ion > ion-dipole > dipole-dipole

Option B is correct because ion-ion interactions are the strongest, followed by ion-dipole, and then dipole-dipole interactions due to the magnitude of charges involved.

6 Medium

6 of 262

Remembering +4 | -1

Compressibility factor of an ideal gas is

View answer and explanation

Correct option: B

Equal\mathrm{ }to\mathrm{ }1\;

Option B (

Equal\mathrm{ }to\mathrm{ }1\;

) is correct because the compressibility factor (Z) for an ideal gas is defined as 1.

7 Easy

7 of 262

Understanding +4 | -1

Critical temperature of a gas is the temperature

View answer and explanation

Correct option: B

Above which it cannot be liquefied

Option B (Above which it cannot be liquefied) is correct because the critical temperature is the temperature above which a gas cannot be liquefied by pressure alone.

8 Hard

8 of 262

Application +4 | -1

2.24

litre cylinder containing

{O_2}

gas at

27\,^\circ C

and

2\,\,\mathrm{atm}

is found to develop a leakage. When the leakage was repaired, the pressure dropped to

100\,\,cm

of Hg at

27\,^\circ C

. The number of mole of gas escaped out during leakage is:

View answer and explanation

Correct option: A

0.06

Option A (0.06) is correct because by using the ideal gas law before and after the leakage, the difference in the number of moles is approximately 0.06.

9 Medium

9 of 262

Analyzing +4 | -1

A: Hot air balloon rises up by displacing the denser air of the atmosphere.
R: The given mass of a gas becomes less denser.

View answer and explanation

Correct option: A

If both A and R are correct and R is the correct explanation of A

Option A is correct because the hot air balloon rises due to the heated air inside being less dense than the surrounding cooler air, as described by the relationship between density and temperature.

10 Medium

10 of 262

Understanding +4 | -1

Pressure cooker reduces cooking time because

View answer and explanation

Correct option: C

Boiling point of water is elevated

Option C (Boiling point of water is elevated) is correct because a pressure cooker increases the pressure, which elevates the boiling point of water, leading to faster cooking.

11 Medium

11 of 262

Application +4 | -1

Under identical conditions of temperature the density of a gas

A

is three times that of gas

B

while molecular mass of gas

B

is twice that of

A

. The ratio of pressures of

A

and

B

will be:

View answer and explanation

Correct option: A

Option A (6) is correct because using the ideal gas law in terms of density, the ratio of pressures can be found to be 6.

12 Easy

12 of 262

Understanding +4 | -1

At constant volume, for a fixed number of mole of a gas, the pressure of the gas increases with rise of temperature due to

View answer and explanation

Correct option: A

Increase in average molecular speed

Option A (Increase in average molecular speed) is correct because, at constant volume and number of moles, an increase in temperature leads to an increase in the average kinetic energy of the gas molecules, resulting in more frequent and forceful collisions with the container walls, hence increasing pressure.

13 Medium

13 of 262

Understanding +4 | -1

Statement I: The Joule-Thomson coefficient for an ideal gas is zero

Statement II: There are no intermolecular attractive forces between molecules in an ideal gas.

View answer and explanation

Correct option: A

Both Statement I and Statement II are correct and Statement II is the correct explanation of Statement I

Option A is correct because both statements are true, and the absence of intermolecular forces in an ideal gas directly leads to a zero Joule-Thomson coefficient.

14 Easy

14 of 262

Understanding +4 | -1

The value of Vander Waal’s constant ‘a’ is minimum for

View answer and explanation

Correct option: A

He

Option A (

He

) is correct because the Van der Waals constant 'a' is related to the intermolecular forces, which are weakest for Helium due to its small size and inert nature.

15 Medium

15 of 262

Application +4 | -1

If Z is the compressibility factor, vanderwaal’s equation at low pressure can be written as

View answer and explanation

Correct option: B

Z = 1 - \frac{9}{{V\,RT}}

Option B (

Z = 1 - \frac{9}{{V\,RT}}

) is correct because at low pressure, the van der Waals equation simplifies to this form by considering the attractive forces between molecules.

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16 Easy

16 of 262

Analyzing +4 | -1

Indicate the incorrect statement for equal volumes of

\mathrm{N}_2(\mathrm{g})

and

\mathrm{CO}_2(\mathrm{g})

at 25 degree Celsius and

1 \mathrm{~atm}

View answer and explanation

Correct option: B

The rms speed remains the same for both

\mathrm{N}_2

and

\mathrm{CO}_2

The incorrect statement is B (The rms speed remains the same for both N2 and CO2) because the rms speed depends on the molar mass, which is different for N2 and CO2.

17 Easy

17 of 262

Understanding +4 | -1

Dipole - dipole forces act between the molecules possessing permanent dipole. Ends of dipoles possess ‘particle charges’. The partial charge is

View answer and explanation

Correct option: C

Less than unit electronic charge

Option C is correct because the partial charges in dipole-dipole interactions are less than a unit electronic charge due to the sharing of electrons in a covalent bond.

18 Hard

18 of 262

Application +4 | -1

Calculate the root mean square speed of

\mathrm{H}_2

molecules under following condition.
(a) 2 mole of

\mathrm{H}_2

27^{\circ} \mathrm{C}

.
(b) 3 mole of

\mathrm{H}_2

in a 5 lit container at

10^5 \mathrm{~Pa}

View answer and explanation

Correct option: A

(a)

1934.24 \mathrm{~m} / \mathrm{sec};

(b)

500 \mathrm{~m} / \mathrm{sec}

Option A ((a)

1934.24 \mathrm{~m} / \mathrm{sec};

(b)

500 \mathrm{~m} / \mathrm{sec}

) is correct because it correctly calculates the root mean square speed using the given conditions.

19 Medium

19 of 262

Remembering +4 | -1

The van der Waals’ equation for real gas is:

View answer and explanation

Correct option: C

P = \frac{{nRT}}{{V - nb}} - \frac{{a{n^2}}}{{{V^2}}}

Option D (All of the above) is correct because all the given equations are valid representations of the van der Waals equation for real gases.

20 Medium

20 of 262

Analyzing +4 | -1

A: At very high pressures, compressibility factor is greater than 1.
R: At very high pressure, '

b

' can be neglected in van der Waal's gas equation.

View answer and explanation

Correct option: C

If A is correct but R is incorrect

Option D (The complexes [NiCl4]2- and [Ni(CN)4]2- differ in primary valencies of nickel) is correct because the primary valency (oxidation state) of Ni is +2 in both complexes, so they do NOT differ in primary valencies.

21 Medium

21 of 262

Application +4 | -1

The rate of diffusion of two gases

x

and

y

is 10 ml/s
and 40 ml/s respectively then ratio of their molar
masses is :-

View answer and explanation

Correct option: D

0.667361111

Option D (16:01) is correct because the ratio of the rates of diffusion is inversely proportional to the square root of the ratio of their molar masses, as described by Graham's Law.

22 Medium

22 of 262

Understanding +4 | -1

A comparison is made at Standard Temperature and Pressure (STP) of 0.50 mole of

\mathrm{H}_2(g)

and 1 mole of

\mathrm{He}(g)

. The two gases will

View answer and explanation

Correct option: A

have equal average molecular kinetic energies

Option A (have equal average molecular kinetic energies) is correct because the average kinetic energy of a gas depends only on temperature, and both gases are at the same temperature (STP).

23 Medium

23 of 262

Application +4 | -1

Two flasks

A

and

B

of equal volumes maintained at temperatures

300 \mathrm{~K}

and

600 \mathrm{~K}

contain equal mass of

\mathrm{H}_2

and

\mathrm{CH}_4

respectively. The ratio of total translational kinetic energy of gas in flask

A

to that in flask

B

View answer and explanation

Correct option: C

Option C (4) is correct because the ratio of kinetic energies is directly proportional to the ratio of (moles * temperature), and after calculation, it equals 4.

24 Medium

24 of 262

Analyzing +4 | -1

Assertion (A): The value of van der Waals' constant '

a

' is higher for ammonia than for nitrogen.

Reason (R): Intermolecular hydrogen bonding is present in ammonia.

View answer and explanation

Correct option: A

Assertion is correct, reason is correct; reason is a correct explanation for assertion

Option A is correct. The van der Waals constant 'a' reflects the strength of intermolecular forces, and ammonia exhibits hydrogen bonding, which is a stronger intermolecular force than the London dispersion forces present in nitrogen.

25 Medium

25 of 262

Understanding +4 | -1

During the evaporation of liquid

View answer and explanation

Correct option: B

The temperature of the liquid will fall

Option B is correct. During evaporation, the molecules with the highest kinetic energy escape from the liquid's surface, thus lowering the average kinetic energy and therefore the temperature of the remaining liquid.

26 Medium

26 of 262

Remembering +4 | -1

The interaction energy of London forces between two particles is proportional to

{r^x}

, where

r

is the distance between the particles. The value of

x

View answer and explanation

Correct option: C

-6

Option C (-6) is correct because the interaction energy due to London dispersion forces is inversely proportional to the sixth power of the distance between the particles, hence

{r^{-6}}

27 Medium

27 of 262

Application +4 | -1

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 :

View answer and explanation

Correct option: C

112 \mathrm{~g} \mathrm{~mol}^{-1}

Option C (

112 \mathrm{~g} \mathrm{~mol}^{-1}

) is correct because it is calculated using the ideal gas law and the relationship between density, pressure, molar mass, and temperature.

28 Medium

28 of 262

Application +4 | -1

If the pressure is halved and absolute temperature doubled the volume of the gas will be:

View answer and explanation

Correct option: A

Option A (4) is correct because using the combined gas law, if the pressure is halved and the absolute temperature is doubled, the volume will be 4 times the original volume.

29 Easy

29 of 262

Analyzing +4 | -1

Assertion: The pressure of a gas is directly proportional to the density of gas at a constant temperature and molecular mass.

Reason: The molecular weight of a vapour of a substance can be calculated using the expression

M=\frac{d R T}{P}

, where

M

is the molecular weight of the vapour,

d

is its density and

P

is the pressure of the vapour,

{R}

is the gas constant and

{T}

is the temperature.

View answer and explanation

Correct option: A

Assertion is correct, reason is correct; reason is a correct explanation for assertion.

Option A is correct. The ideal gas law can be rearranged to show that pressure is directly proportional to density at constant temperature and molecular mass. The reason provides the correct formula derived from the ideal gas law.

30 Easy

30 of 262

Remembering +4 | -1

The compressibility factor for an ideal gas is

View answer and explanation

Correct option: C

Option C (1) is correct because the compressibility factor (Z) for an ideal gas is always equal to 1, as it perfectly obeys the ideal gas law.

Revision notes and key takeaways

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Key Takeaways

✓ This set contains 262 single-correct-answer MCQs on STATES OF MATTER.
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