Skip to main content
Back to KINETIC THEORY OF GASES

Your Progress

0 / 115 Answered
0
Correct
0
Incorrect
0%
Accuracy
NEET • PHYSICS • KINETIC THEORY OF GASES

Kinetic theory of gases practice · Mixed difficulty

Try a question, then open “View answer” to check it. Sign in to submit answers and save progress.

Practice format and chapter quiz

The practice player uses single-correct-answer MCQs. Chapter quizzes help you review a topic; they are not full-length NEET mock papers.

Take a chapter quiz →
Showing 31-60 of 115 questions
31 Easy
31 of 115
Understanding +4 | -1
Assertion: Absolute zero is the temperature corresponding to zero energy.
Reason: The temperature at which no molecular temperature.
View answer and explanation
Correct option: D
If both assertion and reason are false
The correct answer is D because absolute zero corresponds to minimal, but not zero, energy, and molecular motion doesn't cease entirely at any temperature.
32 Hard
32 of 115
Application +4 | -1
A balloon carries a total load of 185 kg at normal pressure and temperature of 27∘ C \mathrm{2}{\mathrm{7}^ \circ }\,\mathrm{C}\,.
What load will the balloon carry on rising to a height at which the barometric pressure is 45 cm of Hg and the temperature is −7∘ C- {\mathrm{7}^ \circ }\,\mathrm{C} ?
[Assuming, the volume constant.]
View answer and explanation
Correct option: D
123.54 kg
Option D (123.54 kg) is correct because it represents the load the balloon can carry at the new pressure and temperature, calculated using the combined gas law and considering the constant volume.
33 Hard
33 of 115
Application +4 | -1
NN moles of a polyatomic gas (f=6)(f=6) must be mixed with two moles of a monoatomic gas so that the mixture behaves as a diatomic gas. The value of NN is :
View answer and explanation
Correct option: B
4
The correct answer is 4 because it satisfies the condition that the mixture behaves as a diatomic gas, meaning it has 5 degrees of freedom.
34 Medium
34 of 115
Analyzing +4 | -1
Two identical containers A\mathrm{A} and B\mathrm{B} have frictionless pistons. They contain the same volume of a ideal gas at the same temperature. The mass of the gas in A\mathrm{A} is mA\mathrm{m}_{\mathrm{A}} and that in B\mathrm{B} is mB\mathrm{mB}. The gas in each cylinder is now allowed to expand isothermally to double the initial volume. The changes in the pressure in A and B are found to be Δp\Delta p and 1.5Δp1.5 \Delta p respectively.
View answer and explanation
Correct option: C
3 mAA=2mB3 \mathrm{~mA}_{\mathrm{A}}=2 \mathrm{mB}
Option C (3 mA=2mB3 \mathrm{~mA}=2 \mathrm{mB}) is correct because it relates the masses based on the changes in pressure during isothermal expansion using the ideal gas law.
35 Medium
35 of 115
Analyzing +4 | -1
Assertion: Vrms{V}_{\mathrm{rms}} and Vmean{V}_{\mathrm{mean}} of gaseous molecules is nearly of the order of velocity of sound.
Reason: The sound travels in air because of vibrational molecular motion.
View answer and explanation
Correct option: B
If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion
Option B is correct because both the assertion and reason are true, but the reason is not the correct explanation of the assertion. RMS and mean speeds are of the order of sound speed, but sound travels due to pressure variations, not vibrational motion.
36 Medium
36 of 115
Understanding +4 | -1
In the absence of intermolecular forces of attraction, the observed pressure pp will be
View answer and explanation
Correct option: C
>p> p
Option C (>p> p) is correct because intermolecular forces of attraction reduce the pressure exerted by the gas, so their absence would result in a higher observed pressure.
37 Easy
37 of 115
Remembering +4 | -1
The critical volume of a gas obeying van der Waals' equation is
View answer and explanation
Correct option: C
3b3 b
The correct answer is C (3b) because the critical volume in the van der Waals equation is defined as three times the van der Waals parameter 'b'.
38 Medium
38 of 115
Application +4 | -1
The root mean square speed of molecules of a gas is 1260 m/s. The average speed of the gas molecules is
View answer and explanation
Correct option: A
1161 m/s
Option A (1161 m/s) is correct because the average speed is related to the RMS speed by a constant factor derived from the Maxwell-Boltzmann distribution.
39 Medium
39 of 115
Application +4 | -1
If number of molecules of H2{H_2} are double than that of O2,{O_2}, then ratio of kinetic energy of hydrogen and that of oxygen at 300  K300\;K is
View answer and explanation
Correct option: A
1  :11\;:1
Option A (1:11:1) is correct because the average kinetic energy depends only on temperature, which is the same for both gases.
40 Easy
40 of 115
Application +4 | -1
If the rms velocity of the molecules of a gas in a container be doubled than pressure of gas will
View answer and explanation
Correct option: A
Becomes 4 times of its previous value
Option A (Becomes 4 times of its previous value) is correct because pressure is proportional to the square of the RMS velocity.
41 Easy
41 of 115
Understanding +4 | -1
The root mean square velocity, vrmsv_{\text {rms}}, the average velocity, vavv_{{av}} and the most probable velocity, vmpv_{{mp}} of the molecules of the gas are in the order:
View answer and explanation
Correct option: B
vrms>vav>vmpv_{\mathrm{rms}} > v_{{av}} > v_{{mp}}
Option B (vrms>vav>vmpv_{\mathrm{rms}} > v_{{av}} > v_{{mp}}) is correct because the root mean square velocity is greater than the average velocity, which is greater than the most probable velocity.
42 Medium
42 of 115
Application +4 | -1
The value closest to the thermal velocity of a Helium atom at room temperature (300K) in ms−1m{s^{ - 1}} is [kB=1.4×10−23J/K;mHe=7×10−27kg]\left[ {{k_B} = 1.4 \times {{10}^{ - 23}}J/K;{m_{He}} = 7 \times {{10}^{ - 27}}kg} \right]
View answer and explanation
Correct option: D
1.3×1031.3 \times {10^3}
Option D (1.3×1031.3 \times {10^3}) is correct because it is the closest value to the calculated thermal velocity of a Helium atom at room temperature.
43 Medium
43 of 115
Understanding +4 | -1
At constant pressure, which of the following is true?
View answer and explanation
Correct option: D
v∝1ρv \propto \frac{1}{{\sqrt \rho }}
Option D (v∝1ρv \propto \frac{1}{{\sqrt \rho }}) is correct because at constant pressure, the volume is inversely proportional to the square root of the density.
44 Medium
44 of 115
Application +4 | -1
Two gases AA and BB having same pressure pp, volume V  V\; and absolute temperature TT are mixed. If the mixture has the volume and temperature as VV and TT respectively, then the pressure of the mixture is
View answer and explanation
Correct option: A
2p2p
Option A (2p2p) is correct because when two gases with the same pressure, volume, and temperature are mixed, the total number of moles doubles, resulting in double the pressure.
45 Medium
45 of 115
Remembering +4 | -1
The rate of diffusion is
View answer and explanation
Correct option: D
Faster in gases than in liquids and solids
The correct answer is D (Faster in gases than in liquids and solids) because diffusion is fastest in gases due to higher kinetic energy and larger intermolecular spaces.
46 Medium
46 of 115
Remembering +4 | -1
The specific heat of an ideal gas is
View answer and explanation
Correct option: D
Independent of TT
Option D (Independent of TT) is correct because the specific heat of an ideal gas is a constant value that does not depend on temperature.
47 Easy
47 of 115
Application +4 | -1
The root mean square velocity of the gas molecules is 300 m/s300 \mathrm{~m} / \mathrm{s}. What will be the root mean square speed of the molecules if the atomic weight is doubled and absolute temperature is halved?
View answer and explanation
Correct option: B
150 m/s150 \mathrm{~m} / \mathrm{s}
Option B (150 m/s150 \mathrm{~m} / \mathrm{s}) is correct because the rms velocity is proportional to TM\sqrt{\frac{T}{M}}, so doubling M and halving T reduces the velocity by a factor of 2.
48 Easy
48 of 115
Understanding +4 | -1
A gas is taken through a number of thermodynamic states. What happens to its specific heat?
View answer and explanation
Correct option: D
It can have any value depending upon process of heat absorbed or evolved
Option D (It can have any value depending upon process of heat absorbed or evolved) is correct because specific heat depends on the process.
49 Easy
49 of 115
Understanding +4 | -1
If the molecular weight of two gases are M1{M_1} and M2{M_2}, then at a temperature the ratio of root mean square velocity v1{v_1} and v2{v_2} will be
View answer and explanation
Correct option: B
M2M1\sqrt {\frac{{{M_2}}}{{{M_1}}}}
Option B (M2M1\sqrt {\frac{{{M_2}}}{{{M_1}}}}) is correct because the RMS velocity is inversely proportional to the square root of the molecular weight.
50 Medium
50 of 115
Analyzing +4 | -1
To raise the temperature of a certain mass of gas by 50oC5{\mathrm{0}^\mathrm{o}}\mathrm{C} at a constant pressure, 160 cal of heat is required. When the same mass of gas is cooled by 100oC10{\mathrm{0}^\mathrm{o}}\mathrm{C} at constant volume, 240 cal of heat is released. How many degrees of freedom does each molecule of this gas have (assume gas to be ideal)?
View answer and explanation
Correct option: C
6
Option C (6) is correct because the gas has 6 degrees of freedom, calculated from the given heat capacities at constant pressure and volume.
51 Easy
51 of 115
Understanding +4 | -1
The root mean square speeds of molecules of ideal gases at the same temperature are:
View answer and explanation
Correct option: B
Inversely proportional to the square root of the molecular weight
Option B (Inversely proportional to the square root of the molecular weight) is correct because the RMS speed is inversely proportional to the square root of the molar mass at the same temperature.
52 Medium
52 of 115
Application +4 | -1
An ideal gas has molecules with 5 degrees of freedom. The ratio of specific heats at constant pressure (CP)\left( {{C_P}} \right) and at constant volume (CV)\left( {{C_V}} \right) is:
View answer and explanation
Correct option: D
7/57/5
Option D (7/57/5) is correct because the ratio of specific heats is calculated using the formula γ=1+2f\gamma = 1 + \frac{2}{f}, where f is the degrees of freedom.
53 Easy
53 of 115
Application +4 | -1
The root mean square speed of smoke particles of mass 5×10−17 kg5 \times 10^{-17} \mathrm{~kg} in their Brownian motion in air at NTP is approximately. [Given k=1.38×10−23JK−1k=1.38 \times 10^{-23} \mathrm{JK}^{-1} ]
View answer and explanation
Correct option: C
15 mm s−115 \mathrm{~mm} \mathrm{~s}{ }^{-1}
Option C (15 mm s−115 \mathrm{~mm} \mathrm{~s}{ }^{-1}) is correct because it's calculated using the formula for root mean square speed, vrms=3kTmv_{rms} = \sqrt{\frac{3kT}{m}}, with appropriate substitutions.
54 Easy
54 of 115
Application +4 | -1
Find the number of molecules in 1 cm31 \mathrm{~cm}^3 of an ideal gas at 0∘C0^{\circ} \mathrm{C} and at a pressure of 10−5 mm10^{-5} \mathrm{~mm} of mercury.
View answer and explanation
Correct option: D
3.538×10113.538 \times 10^{11}
Option D (3.538×10113.538 \times 10^{11}) is correct because it's calculated using the ideal gas law to find the number of molecules per unit volume.
55 Medium
55 of 115
Application +4 | -1
The diameter of oxygen atom is 3A∘3\mathop A\limits^ \circ. The fraction of molecular volume to the actual volume occupied by oxygen at STP is
View answer and explanation
Correct option: B
8×10−48 \times {10^{ - 4}}
Option B (8×10−48 \times {10^{ - 4}}) is correct because it represents the ratio of the volume occupied by oxygen molecules to the total volume at STP, considering the diameter of the oxygen atom.
56 Medium
56 of 115
Understanding +4 | -1
The absolute temperature of a gas is determined by
View answer and explanation
Correct option: D
The mean square velocity of the molecules
Option D (The mean square velocity of the molecules) is correct because the absolute temperature of a gas is directly related to the average kinetic energy, which is proportional to the mean square velocity of the molecules.
57 Medium
57 of 115
Application +4 | -1
If the mean free path of atoms is doubled then the pressure of gas will become
View answer and explanation
Correct option: B
P2\frac{P}{2}
Option B (P2\frac{P}{2}) is correct because the pressure is inversely proportional to the mean free path.
58 Medium
58 of 115
Understanding +4 | -1
At O K which of the following properties of a gas will be zero
View answer and explanation
Correct option: A
Kinetic energy
Option A (Kinetic energy) is correct because at 0 K, all molecular motion ceases, resulting in zero kinetic energy.
59 Easy
59 of 115
Understanding +4 | -1
The parameter that remains the same for molecules
of all gases at a given temperature is :
View answer and explanation
Correct option: C
kinetic energy
The correct answer is C because at a given temperature, the average kinetic energy of molecules is the same for all gases, according to the kinetic theory of gases.
60 Medium
60 of 115
Understanding +4 | -1
Supposing the distance between the atoms of a diatomic gas to be constant, its specific heat at constant volume per mole (gram mole) is
View answer and explanation
Correct option: A
52R\frac{5}{2}R
Option A (52R\frac{5}{2}R) is correct because a diatomic gas with constant distance between atoms has 3 translational and 2 rotational degrees of freedom, leading to Cv=52RC_v = \frac{5}{2}R.
Revision notes and key takeaways

Review KINETIC THEORY OF GASES in your textbook or class notes, then use this set to check your understanding.

Showing 31-60 of 115 questions
0 /115
0✓ 0✗
0% Accuracy
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.