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JEE • CHEMISTRY • STATES OF MATTER

States of matter practice · Mixed difficulty

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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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Showing 121-150 of 262 questions

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121 Medium
121 of 262
Analyzing +4 | -1
S-I: Helium shows only positive deviation from ideal behavior at room temperature.

S-II: Helium is an inert gas.
View answer and explanation
Correct option: B
Both statements are correct and statement-II is not correct explanation of statement-I
The correct answer is B because both statements are correct, but the inertness of helium doesn't explain its positive deviation from ideal behavior, which is due to its small size and weak intermolecular forces.
122 Medium
122 of 262
Application +4 | -1
If pressure of a gas contained in a closed vessel is increased by 0.4%0.4\% when heated by 1∘C1^\circ C its initial temperature must be:
View answer and explanation
Correct option: A
250  K250\,\,\mathrm{K}
The correct answer is A (250 K) because the initial temperature is calculated using the relationship between pressure and temperature for a gas at constant volume.
123 Easy
123 of 262
Remembering +4 | -1
Gay-Lussac’s law of gaseous volumes is derived from:
View answer and explanation
Correct option: C
Experimental observations
Option C (Experimental observations) is correct because Gay-Lussac's law was formulated based on experimental data and observations of gas behavior under constant temperature and pressure.
124 Medium
124 of 262
Application +4 | -1
If the absolute temperature of a gas is doubled and the pressure is reduced to one half, the volume of the gas will
View answer and explanation
Correct option: C
Increase four fold
Option C (Increase four fold) is correct because using the ideal gas law, when temperature doubles and pressure is halved, the volume increases by a factor of four.
125 Medium
125 of 262
Application +4 | -1
What is the difference in the density of dry air at 1 atm and 25∘C25^{\circ} \mathrm{C} and moist air with 50%50 \% relative humidity under the same condition? The vapour pressure of water at 25∘C25^{\circ} \mathrm{C} is 23.7 torr and dry air has 75.5%   N275.5 \%\,\, {~N}_2 and 24.5%  O224.5 \%\,\, {O}_2.
View answer and explanation
Correct option: B
0.007  kg  m−30.007\, \mathrm{~kg}\, \mathrm{~m}^{-3}
Option B (0.007  kg  m−30.007\, \mathrm{~kg}\, \mathrm{~m}^{-3}) is correct because it represents the difference in density between dry air and moist air calculated considering the partial pressure of water vapor.
126 Medium
126 of 262
Understanding +4 | -1
Assertion : Gases combine in simple ratio of their volume but, not always.

Reason : Gases deviate from ideal behaviour.
View answer and explanation
Correct option: A
If both assertion and reason are true and the reason is the correct explanation of the assertion.
Option A (If both assertion and reason are true and the reason is the correct explanation of the assertion) is correct because gases combine in simple ratios when they behave ideally, and deviations from ideal behavior explain why this isn't always the case.
127 Medium
127 of 262
Understanding +4 | -1
At low pressure, the Vanderwaal's equation is expressed as
View answer and explanation
Correct option: D
Z=pVmRT=1−aRTVmZ=\frac{p V_m}{R T}=1-\frac{a}{R T V_m}
Option D (Z=pVmRT=1−aRTVmZ=\frac{p V_m}{R T}=1-\frac{a}{R T V_m}) is correct because it represents the van der Waals equation at low pressure, where the volume is large and the 'b' term can be neglected.
128 Medium
128 of 262
Remembering +4 | -1
The inter molecular interaction that is dependent on the inverse cube of distance between the molecules is
View answer and explanation
Correct option: D
Dipole-dipole interaction
Option D (Dipole-dipole interaction) is correct because the dipole-dipole interaction energy is inversely proportional to the cube of the distance between the molecules.
129 Medium
129 of 262
Application +4 | -1
A perfect gas of a given mass is heated first in a small vessel and then in a large vessel, such that their volume remains unchanged. The P−TP - T curves are:
View answer and explanation
Correct option: D
Linear with different slopes
The correct answer is Linear with different slopes because the P−TP-T relationship is linear, and the slope depends on the volume.
130 Medium
130 of 262
Understanding +4 | -1
Following properties will decrease with increase in temperature except
View answer and explanation
Correct option: D
Vapour pressure
Option D (Vapour pressure) is correct because vapor pressure increases with increasing temperature due to increased kinetic energy of molecules.
131 Easy
131 of 262
Remembering +4 | -1
The units of the Vander Waal’s constant ‘b’ are
View answer and explanation
Correct option: C
L.mol−1{\text{L}}{\text{.mo}}{{\text{l}}^{ - 1}}
Option C (L.mol−1{\text{L}}{\text{.mo}}{{\text{l}}^{ - 1}}) is correct because the van der Waals constant 'b' represents the volume per mole of gas molecules.
132 Easy
132 of 262
Remembering +4 | -1
In van der Waal's equation of state of the gas law, the constant ' bb ' is a measure of
View answer and explanation
Correct option: A
Volume occupied by the molecules
Option A (Volume occupied by the molecules) is correct because the van der Waals constant 'b' accounts for the finite volume of gas molecules.
133 Hard
133 of 262
Application +4 | -1
At what temperature do the average speed of CH4(g)\mathrm{CH}_{4(\mathrm{g})} molecule equal the average speed of O2\mathrm{O}_2 molecule at 300 K?300 \mathrm{~K}?
View answer and explanation
Correct option: A
150 K150 \mathrm{~K}
Option A (150 K150 \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.
134 Medium
134 of 262
Understanding +4 | -1
If the assumption that there is no force of attraction between the molecules of a gas is correct, what will be the consequences?
View answer and explanation
Correct option: B
The gases will never liquefy when cooled and compressed
Option B is correct because if there is no attraction between gas molecules, they will not come close enough to liquefy, even when cooled and compressed.
135 Medium
135 of 262
Application +4 | -1
X mLX \mathrm{~mL} of H2\mathrm{H}_2 gas effuse through a hole in a container in 5 s5 \mathrm{~s}. The time taken for the effusion of the same volume of the gas specified below under ideal conditions is
View answer and explanation
Correct option: B
20 s,O220 \mathrm{~s}, \mathrm{O}_2
Option B (20 s,O220 \mathrm{~s}, \mathrm{O}_2) is correct because the effusion rate is inversely proportional to the square root of the molar mass.
136 Easy
136 of 262
Understanding +4 | -1
At normal conditions, Dalton's law of partial pressure is not applicable to which gaseous mixture
View answer and explanation
Correct option: A
H2\mathrm{H}_2 and Cl2\mathrm{Cl}_2 mixture
Option A (H2\mathrm{H}_2 and Cl2\mathrm{Cl}_2 mixture) is correct because hydrogen and chlorine react to form HCl under normal conditions, violating the assumption of non-reactivity in Dalton's law.
137 Medium
137 of 262
Application +4 | -1
A gas of unknown identity effuses at the rate of 83.3  mLs−183.3\,\,\mathrm{mL}{\mathrm{s}^{ - 1}} in an effusion apparatus in which carbon dioxide effuses at the rate of 102  mLs−1102\,\,\mathrm{mL}{\mathrm{s}^{ - 1}}. Calculate molar mass of the unknown gas
View answer and explanation
Correct option: B
65.97  g mol−165.97\,\,\mathrm{g}\,\mathrm{mo}{\mathrm{l}^{ - 1}}
Option B (65.97  g mol−165.97\,\,\mathrm{g}\,\mathrm{mo}{\mathrm{l}^{ - 1}}) is correct because it is calculated using Graham's law of effusion, relating effusion rates to molar masses.
138 Medium
138 of 262
Application +4 | -1
Equal masses of CH4C{H_4} and O2{O_2} are mixed in a container at 25∘ C25^\circ \,C. Fraction of total pressure exerted by O2{O_2} is
View answer and explanation
Correct option: B
13\frac{1}{3}
Option B (13\frac{1}{3}) is correct because the partial pressure of a gas is proportional to its mole fraction in the mixture.
139 Medium
139 of 262
Application +4 | -1
In two separate bulbs containing ideal gases AA and BB respectively, the density of gas AA is twice that of gas BB while molecular weight of gas AA is half that of gas BB at the same temperature, pressure ratio PA/PB{P_A}/{P_B} will be:
View answer and explanation
Correct option: C
4
Option C (4) is correct because the pressure ratio is directly proportional to the ratio of densities and inversely proportional to the ratio of molecular weights, resulting in a ratio of 4.
140 Medium
140 of 262
Application +4 | -1
4.0g4.0\mathrm{g} of argon has pressure PP and temperature T  KT\,\,\mathrm{K} in a vessel. On keeping the vessel at 50∘50^\circ higher temperature, 0.8g0.8\mathrm{g} of argon was given out to maintain the pressure PP. The original temperature was:
View answer and explanation
Correct option: C
200  K200\,\,\mathrm{K}
Option C (200  K200\;\mathrm{K}) is correct because it is derived from applying the ideal gas law under the given conditions of constant pressure and volume.
141 Medium
141 of 262
Analyzing +4 | -1
Assertion : Compressibility factor (Z)(Z) for non-ideal gases can be greater than 1 .

Reason : Non-ideal gases always exert higher pressure than expected.
View answer and explanation
Correct option: C
If Assertion is correct but Reason is incorrect.
The correct answer is C because the assertion is correct (Z can be > 1) but the reason is incorrect (non-ideal gases can exert lower pressure than expected).
142 Medium
142 of 262
Understanding +4 | -1
Correct order of van der Waals constant b of given gases is
View answer and explanation
Correct option: A
CO2>N2>O2>H2>He\mathrm{CO}_2>\mathrm{N}_2>\mathrm{O}_2>\mathrm{H}_2>\mathrm{He}
Option D (All of these) is correct because all the given expressions are valid representations of Kf, the cryoscopic constant.
143 Medium
143 of 262
Understanding +4 | -1
Dipole-Dipole interaction energy b/w polar molecule in solid depends on the radius of the molecule (r)\left( r \right) and it is directly proportional to
View answer and explanation
Correct option: D
1r3\frac{1}{{{r^3}}}
Option D (1r3\frac{1}{{{r^3}}}) is correct because the dipole-dipole interaction energy in solids is inversely proportional to the cube of the distance (radius) between the molecules.
144 Medium
144 of 262
Application +4 | -1
Under what conditions will a pure sample of an ideal gas not only exhibit a pressure of 1 atm but also a concentration of 1mol  litre−11\mathrm{mol}\,\,\mathrm{litr}{\mathrm{e}^{ - 1}} ? (  R=0.082(\;R = 0.082 litre atm mol−1deg⁡−1)\mathrm{mo}{\mathrm{l}^{ - 1}}{\deg ^{ - 1}})
View answer and explanation
Correct option: C
When T=12KT = 12K
Option C (When T=12KT = 12K) is correct because it satisfies the ideal gas law equation with the given conditions.
145 Medium
145 of 262
Application +4 | -1
A closed cylinder having volume 20 L20 \mathrm{~L} contains 56 g N256 \mathrm{~g} {~N}_2 and 16 g He16 \mathrm{~g} {~He} gases. If pressure exerted by the gaseous mixture is 8.2 atm8.2 \mathrm{~atm} then the temperature of the gas molecules will be
View answer and explanation
Correct option: B
333.3 K333.3 \mathrm{~K}
Option B (333.3 K) is correct because it is calculated using the ideal gas law with the given pressure, volume, and number of moles.
146 Medium
146 of 262
Understanding +4 | -1
If pressure of a fixed quantity of a gas is increased 4 times keeping the temperature constant, the r.m.s. velocity will be
View answer and explanation
Correct option: C
Same
Option C (Same) is correct because the root mean square (rms) velocity depends only on temperature and molar mass, and since the temperature is constant, the rms velocity remains the same.
147 Easy
147 of 262
Application +4 | -1
A sample of air is saturated with benzene(vapour pressure =100 mmHg=100 \mathrm{~mmHg} at 298 K298 \mathrm{~K} ), 750 mmHg750 \mathrm{~mmHg} pressure. If it is isothermally compressed to one third of its initial volume, the final pressure of the system is
View answer and explanation
Correct option: A
2050
Option A (2050) is correct because the final pressure is the sum of the partial pressure of benzene and the compressed air pressure.
148 Easy
148 of 262
Application +4 | -1
At STP 16 mL of O2{O_2} diffused through a porous partition in ‘t’ seconds. What volume of CO2C{O_2} will diffuse in the same time and under the same conditions.
View answer and explanation
Correct option: A
13.65 mL
Option A (13.65 mL) is correct because it is calculated using Graham's law of diffusion, relating diffusion rate to molar mass.
149 Medium
149 of 262
Understanding +4 | -1
Which of the assumption of the kinetic molecular theory best explains the observation that a balloon collapses when exposed to liquid nitrogen (which is much colder than a cold winter day)?
View answer and explanation
Correct option: B
The velocity of gas molecules is proportional to their kelvin temperature
Option B is correct because the kinetic molecular theory states that the velocity of gas molecules is proportional to their Kelvin temperature, explaining why a balloon collapses when exposed to liquid nitrogen.
150 Easy
150 of 262
Understanding +4 | -1
A: Vapour pressure of the liquid increases with increase in temperature.
R: At elevated temperature kinetic energy of the liquid molecules increases.
View answer and explanation
Correct option: A
If both A and R are correct and R is the correct explanation of A
Option A (If both A and R are correct and R is the correct explanation of A) is correct because the kinetic energy increase at higher temperatures allows more liquid molecules to overcome intermolecular forces and enter the vapor phase, thus increasing vapor pressure.
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