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) $390 \mathrm{~m} / \mathrm{sec}$

A mixture of 0.50 mole of ${{\rm{H}}_{\rm{2}}}$ and 0.50 mole of ${\rm{S}}{{\rm{O}}_{\rm{2}}}$ is introduced into a 10.00 L container at ${\rm{2}}{{\rm{5}}^{\rm{0}}}{\rm{C}}$. The container has a pinhole leak. After a period of time, the partial pressure of ${{\rm{H}}_2}$ in the remaining mixture

Is less than that of ${\rm{S}}{{\rm{O}}_2}$

Exceeds that of ${\rm{S}}{{\rm{O}}_2}$

Is equal to that of ${\rm{S}}{{\rm{O}}_2}$

Is the same as in the original mixture

Calculate the fraction of $\mathrm{N}_2$ molecules whose speeds are in the range of $\text{U}_{\text{mp}}$ to $\left(\text{U}_\text{mp}+\text{f} \times \text{U}_\text{mp}\right)?$

$\frac{2 \mathrm{f}}{\mathrm{e} \sqrt{\pi}}$

$\frac{\mathrm{f}}{\sqrt{\pi}}$

$\frac{\mathrm{f}}{\mathrm{2e} \sqrt{\pi}}$

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JEE • CHEMISTRY • STATES OF MATTER

States of matter practice · Hard

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Practice format and chapter quiz

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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1 Hard

1 of 21

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.

2 Hard

2 of 21

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.

3 Hard

3 of 21

Remembering +4 | -1

The value of compression factor,

Z

for critical constants is

View answer and explanation

Correct option: D

\frac{3}{8}

Option D (

\frac{3}{8}

) is correct because the compression factor Z at critical conditions is defined as

\frac{p_cV_c}{RT_c}

which equals 3/8 for van der Waals gases.

4 Hard

4 of 21

Application +4 | -1

A gaseous mixture of

\mathrm{O}_2

and

\mathrm{X}

containing

20 \%

(mole %) of

\text{X},

diffused through a small hole in 234 seconds while pure

\mathrm{O}_2

takes 224 seconds to diffuse through the same hole. Molecular weight of

\mathrm{X}

is:

View answer and explanation

Correct option: A

46.6

Option A (46.6) is correct because it is calculated by using Graham's law of diffusion to find the molecular weight of the mixture and then solving for the molecular weight of gas X.

5 Hard

5 of 21

Analyzing +4 | -1

Statement I: Second virial coefficient of

\mathrm{CO}_2

-0.065 \mathrm{~L} \mathrm{~mol}^{-1}

400 \mathrm{~K}

and

30 \mathrm{~atm}

making

\mathrm{CO}_2

more compressible than an ideal gas.

Statement II: Compressibility factor of

\mathrm{CO}_2(Z=0.941)

is less than 1 (

Z=1

for ideal gas) due to the attractive forces between

\mathrm{CO}_2

molecules.

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 C (

LiOH > NaOH > KOH > RbOH > CSOH

(Solubility)) is incorrect because the solubility of group 1 hydroxides increases down the group due to decreasing lattice energy.

6 Hard

6 of 21

Application +4 | -1

If 2 lit. of gas

\text{A}

1.5 \mathrm{~atm}

and 3 lit. of gas

\text{B}

at 2 atm are mixed in a 5 lit. container then find the final pressure, considering all are at same temperature.

View answer and explanation

Correct option: A

1.8 \mathrm{~atm}

Option A (

1.8 \mathrm{~atm}

) is correct because it is calculated by using the ideal gas law to find the number of moles of each gas and then finding the total pressure.

7 Hard

7 of 21

Analyzing +4 | -1

Statement I: Reduced equation of state in terms of reduce: temperature

\theta

, reduced volume

\phi

and reduced pressure x can be applied to solid, gases and liquid.

Statement II:

\left(\pi+\frac{3}{\phi^2}\right)(3 \phi-1)=8 \theta

. Equation is independent of

p, V

and

T

as well as independent of vander Waals' constant

a

and

b

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 D (

\mathrm{N}_2^ +

becomes diamagnetic) is correct because removing an electron from N2, which has an odd number of electrons in its molecular orbitals, will not result in a diamagnetic species.

8 Hard

8 of 21

Analyzing +4 | -1

Consider 1

c{m^3}

sample of air at absolute temperature

{T_0}

at sea-level and another

1\,\,c{m^3}

sample of air at a height where the pressure is one third atmosphere. The absolute temperature

T

of the sample at the height is :

View answer and explanation

Correct option: D

Cannot be determined in terms of

{T_0}

from the above data

Option D is correct because without knowing the mass of the air samples, we cannot determine the relationship between the temperatures using the ideal gas law.

9 Hard

9 of 21

Application +4 | -1

A given mass of helium at STP is allowed to expand to 1.4 times its initial volume by changing the temperature and pressure, then its resultant density would be

View answer and explanation

Correct option: C

0.13 \mathrm{~kg} \mathrm{~m}^{-3}

Option C (

0.13 \mathrm{~kg} \mathrm{~m}^{-3}

) is correct because density is inversely proportional to volume. When the volume increases by 1.4 times, the density decreases by 1.4 times.

10 Hard

10 of 21

Understanding +4 | -1

The equation,

\left[ {{P_r} + \frac{3}{{V_r^2}}} \right]\left[ {3{V_r} - 1} \right] = 8{T_r}:

View answer and explanation

Correct option: D

All of the above

Option D (All of the above) is correct because the given equation represents the law of corresponding states, which relates reduced pressure, volume, and temperature.

11 Hard

11 of 21

Application +4 | -1

At what temperature do the average speed of

\mathrm{CH}_{4(\mathrm{g})}

molecule equal the average speed of

\mathrm{O}_2

molecule at

300 \mathrm{~K}?

View answer and explanation

Correct option: A

150 \mathrm{~K}

Option A (

150 \mathrm{~K}

) is correct because the average speed of a gas is inversely proportional to the square root of its molar mass, and the temperature is adjusted accordingly.

12 Hard

12 of 21

Application +4 | -1

An open vessel containing air is heated from 300 K to 400 K. The fraction of air originally present which goes out of it is:

View answer and explanation

Correct option: B

\frac{1}{4}

Option B (

\frac{1}{4}

) is correct because it represents the fraction of air that escapes the vessel when heated from 300 K to 400 K, calculated using the ideal gas law.

13 Hard

13 of 21

Application +4 | -1

To an evacuated vessel with movable piston under external pressure of 1 atm, 0.1 mole of He and 1.0 mole of an unknown compound (vapour pressure 0.68 atm at

0\,\,^\circ C

) are introduced. Considering the ideal gas behaviour, the total volume (in litre) of the gases at 

0\,^\circ C

  is close to

View answer and explanation

Correct option: C

Option C (7) is correct because the partial pressure of He is calculated, and then using the ideal gas law, the volume is found to be approximately 7L.

14 Hard

14 of 21

Analyzing +4 | -1

A mixture of 0.50 mole of

{\mathrm{H}_\mathrm{2}}

and 0.50 mole of

\mathrm{S}{\mathrm{O}_\mathrm{2}}

is introduced into a 10.00 L container at

\mathrm{2}{\mathrm{5}^\mathrm{0}}\mathrm{C}

. The container has a pinhole leak. After a period of time, the partial pressure of

{\mathrm{H}_2}

in the remaining mixture

View answer and explanation

Correct option: C

Is less than that of

\mathrm{S}{\mathrm{O}_2}

Option C is correct because hydrogen gas effuses faster than sulfur dioxide due to its lower molar mass, leading to a lower partial pressure of hydrogen in the remaining mixture over time.

15 Hard

15 of 21

Application +4 | -1

Calculate the fraction of

\mathrm{N}_2

molecules whose speeds are in the range of

\text{U}_{\text{mp}}

\left(\text{U}_\text{mp}+\text{f} \times \text{U}_\text{mp}\right)?

View answer and explanation

Correct option: A

\frac{4 \mathrm{f}}{\mathrm{e} \sqrt{\pi}}

Option A (4f / (e * sqrt(pi))) is correct because it represents the fraction of molecules within the specified speed range, derived from the Maxwell-Boltzmann distribution.

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16 Hard

16 of 21

Application +4 | -1

As the temperature is raised from

{20^0}C\,\mathrm{to}\,{40^0}C

the average kinetic energy of neon atoms changes by a factor of which of the following?

View answer and explanation

Correct option: C

\sqrt {313/293}

Option B (

\sqrt{{313/293}}

) is correct because the average kinetic energy is directly proportional to the absolute temperature.

17 Hard

17 of 21

Application +4 | -1

What is the temperature at which the kinetic energy of 0.3 mole of helium is equal to the kinetic energy of 0.4 mole of argon at 400 K?

View answer and explanation

Correct option: C

533\mathrm{ }K

Option C (

533\mathrm{ }K

) is correct because it is calculated by equating the kinetic energies of helium and argon.

18 Hard

18 of 21

Analyzing +4 | -1

Most probable velocity

\left(u_{\mathrm{mp}}\right)

, average velocity

\left(u_{\mathrm{av}}\right)

and root mean square velocity

\left(u_{\mathrm{rms}}\right)

of a sample of argon are equal, then

View answer and explanation

Correct option: C

T_{\text {rms }} < T_{\text {av }} < T_{\text {mp }}

The correct answer is C (

T_{\text {rms }} < T_{\text {av }} < T_{\text {mp }}

) because the temperatures are related to the velocities by constants, and the given velocities are equal.

19 Hard

19 of 21

Remembering +4 | -1

The mean free path

\left( \lambda \right)

of a gas sample is given by:

View answer and explanation

Correct option: B

\lambda = \frac{1}{{\sqrt 2 \pi {\sigma ^2}N}}

Option B (

\lambda = \frac{{1}}{{\sqrt 2 \pi {{\sigma }^2}N}}

) is correct because it represents the correct formula for the mean free path of a gas molecule.

20 Hard

20 of 21

Understanding +4 | -1

Boyle’s law may be expressed as:

View answer and explanation

Correct option: B

{\left( {\partial P/\partial V} \right)_T} = - K/{V^2}

Option B (

{\left( {\partial P/\partial V} \right)_T} = - K/{V^2}

) is correct because it represents the derivative of Boyle's Law (PV = constant) with respect to volume at constant temperature.

21 Hard

21 of 21

Application +4 | -1

The rates of diffusion of gases

A\,\,\,{\text{and}}\,\,B

of molecular weights 100 and 81 respectively are in the ratio of

View answer and explanation

Correct option: A

9:10

Option A (

9:10

) is correct because the ratio of diffusion rates is inversely proportional to the square root of their molecular weights.

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Key Takeaways

✓ This set contains 21 single-correct-answer MCQs on STATES OF MATTER.
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