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JEE • PHYSICS • KINETIC THEORY OF GASES

Kinetic theory of gases 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 151-180 of 290 questions

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151 Medium
151 of 290
Application +4 | -1
Calculate the r.m.s. velocity of air molecules at S.T.P. Given density of air at S.T.P. is 1.296 kg m−31.296 \mathrm{~kg} \mathrm{~m}^{-3}.
View answer and explanation
Correct option: D
482.24 ms−1482.24 \mathrm{~ms}^{-1}
Option D (482.24 ms−1482.24 \mathrm{~ms}^{-1}) is correct because it is the result of applying the formula for root mean square velocity using the given pressure and density.
152 Medium
152 of 290
Application +4 | -1
When heat Q is supplied to a diatomic gas of rigid molecules, at constant volume, its temperature increases by ΔT\Delta T. The heat required to produce the same change in temperature, at a constant pressure is
View answer and explanation
Correct option: D
75Q\frac{7}{5}Q
The correct answer is 75Q\frac{7}{5}Q because the ratio of heat required at constant pressure to that at constant volume is equal to the adiabatic index for a diatomic gas.
153 Medium
153 of 290
Understanding +4 | -1
A diatomic molecule has how many degrees of freedom
View answer and explanation
Correct option: C
5
Option C (5) is correct because a diatomic molecule has 3 translational and 2 rotational degrees of freedom.
154 Medium
154 of 290
Application +4 | -1
A vessel contains 14 g14 \mathrm{~g} of nitrogen gas at a temperature of 27∘C27^{\circ} \mathrm{C}. The amount of heat to be transferred to the gas to double the r.m.s speed of its molecules will be:

Take R=8.32 J mol−1 K−1R=8.32 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}.
View answer and explanation
Correct option: C
9360 J
Option C (9360 J) is correct because it is the amount of heat required to double the RMS speed, calculated using the specific heat at constant volume.
155 Easy
155 of 290
Understanding +4 | -1
The temperature of the gas is a measure of
View answer and explanation
Correct option: A
The average translational kinetic energy of the
gas molecules
Option A (The average translational kinetic energy of the gas molecules) is correct because temperature is directly proportional to the average translational kinetic energy of the gas molecules.
156 Medium
156 of 290
Application +4 | -1
How many times a diatomic gas should be expanded adiabatically so as to reduce the root mean square velocity to half
View answer and explanation
Correct option: B
32
Option B (32) is correct because the volume should be increased by 32 times to reduce the root mean square velocity to half.
157 Medium
157 of 290
Application +4 | -1
Find out the temperature of oxygen at which vrmsv_{r m s} of oxygen is equal to vrmsv_{r m s} of hydrogen at 27∘C27^{\circ} \mathrm{C}.
View answer and explanation
Correct option: A
4800 K4800 \mathrm{~K}
Option A (4800 K4800 \text{{ K}}) is correct because it is calculated by equating the RMS velocities of oxygen and hydrogen and solving for the temperature of oxygen.
158 Medium
158 of 290
Application +4 | -1
4.0 g4.0 \mathrm{~g} of helium occupies 22400 cm322400 \mathrm{~cm}^3 at STP. The specific heat capacity of helium at constant pressure is 5.0 calK−1 mol−15.0\, \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}. Calculate the speed of sound in helium at STP.
View answer and explanation
Correct option: C
962.28 m/s
Option C (962.28 m/s) is correct because it is the result of correctly applying the formula for the speed of sound in a gas, considering the adiabatic index and STP conditions.
159 Medium
159 of 290
Understanding +4 | -1
What will be the average value of energy along one degree of freedom for an ideal gas in thermal equilibrium at a temperature TT ?(kB{k_B} is Boltzmann constant)
View answer and explanation
Correct option: A
12kBT\frac{1}{2}{k_B}T
Option A (12kBT\frac{1}{2}{k_B}T) is correct because the average energy per degree of freedom for an ideal gas is 12kBT\frac{1}{2}k_BT.
160 Medium
160 of 290
Understanding +4 | -1
For an ideal gas of diatomic molecules
View answer and explanation
Correct option: D
Cp=72R{C_p} = \frac{7}{2}R
Option D (Cp=72R{C_p} = \frac{7}{2}R) is correct because for a diatomic molecule, the degrees of freedom are 5, and therefore, Cp=(f/2+1)R=(5/2+1)R=7/2RC_p = (f/2 + 1)R = (5/2 + 1)R = 7/2 R.
161 Medium
161 of 290
Remembering +4 | -1
The correct relation between the degrees of freedom ff and the ratio of specific heat γ\gamma is
View answer and explanation
Correct option: A
f=2γ−1f = \frac{2}{{\gamma - 1}}
Option A (f=2γ−1f = \frac{2}{{\gamma - 1}}) is correct because it represents the correct relationship between degrees of freedom (ff) and the ratio of specific heats (γ\gamma).
162 Medium
162 of 290
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 has 5 degrees of freedom (3 translational and 2 rotational) at normal temperatures.
163 Medium
163 of 290
Application +4 | -1
At a pressure of 24×105  dyne/cm2,24 \times {10^5}\,\,\mathrm{dyne/c}{\mathrm{m}^\mathrm{2}}, the volume of O2{O_2} is 10 litre and mass is 20 gm. The rms velocity is [in m/sec]
View answer and explanation
Correct option: C
600
Option C (600) is correct because it accurately calculates the RMS velocity using the given pressure, volume, and mass.
164 Medium
164 of 290
Understanding +4 | -1
Vessel AA is filled with hydrogen while vessel B,B, whose volume is twice that of A,A, is filled with the same mass of oxygen at the same temperature. The ratio of the mean kinetic energies of hydrogen and oxygen is
View answer and explanation
Correct option: D
1  :11\;:1
Option D (1:1) is correct because the mean kinetic energy of a gas depends only on its temperature, and both gases are at the same temperature.
165 Medium
165 of 290
Application +4 | -1
The rms velocity of the molecules in a sample of helium is 5/75 / 7 in that of the molecules in a sample of hydrogen. If the temperature of hydrogen sample is 0∘C0^{\circ} {C}. The temperature of the helium sample is nearly
View answer and explanation
Correct option: A
0∘C0^{\circ} {C}
Option A (0∘C0^{\circ} {C}) is correct because the rms velocity depends on the square root of the temperature, and the ratio of velocities is used to find the temperature of helium.
166 Medium
166 of 290
Understanding +4 | -1
In a mixture of gases at a fixed temperature
View answer and explanation
Correct option: A
Lighter molecule has higher average speed
Option A is correct because at a fixed temperature, lighter molecules have a higher average speed due to their lower mass.
167 Medium
167 of 290
Analyzing +4 | -1
Assertion (A) : If a gas container is placed in a moving train, the temperature of gas will increase.

Reason (R) : Kinetic energy of gas molecules will increase.
View answer and explanation
Correct option: D
If Assertion is false but the Reason is true.
Option D is correct because the assertion about gas temperature increase in a moving train is false, while the reason about increased kinetic energy of gas molecules is true.
168 Medium
168 of 290
Application +4 | -1
Determine the absolute gas temperature at which the root mean square speed of helium molecules exceeds their most probable speed by 200 m/s200 \mathrm{~m} / \mathrm{s}.
View answer and explanation
Correct option: C
190.2 K190.2 \mathrm{~K}
Option C (190.2 K190.2 \mathrm{~K}) is correct because the temperature is calculated by equating the difference between RMS and most probable speeds to 200 m/s.
169 Easy
169 of 290
Analyzing +4 | -1
Assertion: The ratio of specific heat gas at constant pressure and specific heat at constant volume for a diatomic gas is more than that for a monatomic gas.
Reason: The molecules of a monatomic gas have more degrees of freedom than those of a diatomic gas.
View answer and explanation
Correct option: D
If both assertion and reason are false
Option D is correct. Both the assertion and reason are false. The ratio of specific heats is lower for diatomic gases, and monatomic gases have fewer degrees of freedom.
170 Medium
170 of 290
Application +4 | -1
The average degree of freedom per molecule for a gas is 6 . The gas performs 25 J25 \mathrm{~J} of work when it expands at constant pressure. The heat absorbed by the gas is
View answer and explanation
Correct option: B
100 J100 \mathrm{~J}
Option B (100 J100 \mathrm{~J}) is correct because it is calculated using the relation between work done and heat absorbed at constant pressure, considering the degrees of freedom.
171 Medium
171 of 290
Understanding +4 | -1
A tyre kept outside in sunlight bursts off after sometime because of
View answer and explanation
Correct option: A
increase in pressure
Option A (increase in pressure) is correct because, according to Gay-Lussac's law, as the temperature increases, the pressure inside the tire increases, leading to it bursting.
172 Medium
172 of 290
Application +4 | -1
Consider a sample of oxygen at 300 K300 \mathrm{~K}. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth.
View answer and explanation
Correct option: D
8 hrs
Option D (8 hrs) is correct because it is the approximate time calculated using the average speed of oxygen molecules and the Earth's diameter.
173 Medium
173 of 290
Understanding +4 | -1
The root mean square velocity of a gas molecule of mass mm at a given temperature is proportional to
View answer and explanation
Correct option: D
1m\frac{1}{{\sqrt m }}
Option D (1m\frac{1}{{\sqrt m }}) is correct because the root mean square velocity is inversely proportional to the square root of the mass of the gas molecule.
174 Medium
174 of 290
Application +4 | -1
Two containers A and B are such that,VA=0.10 m3,PA=1.40MPa,VB=0.15 m3,PB=0.7MPaV_A=0.10 \mathrm{~m}^3, P_A=1.40 \mathrm{MPa}, V_B=0.15 \mathrm{~m}^3, P_B=0.7 \mathrm{MPa}

After opening the tap, final pressure in the containers is (Assume temperature of A and B are same)
View answer and explanation
Correct option: B
0.98 MPa0.98 \mathrm{~MPa}
Option B (0.98 MPa0.98 \mathrm{~MPa}) is correct because it is calculated using the conservation of moles and the ideal gas law, considering the combined volume.
175 Medium
175 of 290
Application +4 | -1
The amount of heat required to raise the temperature of 1 mole of a monoatomic gas from 20∘C20^{\circ} \mathrm{C} to 30∘C30^{\circ} \mathrm{C} at constant volume is HH. Then, the amount of heat required to raise the temperature of 2 mol2 \mathrm{~mol} of a diatomic gas from 20∘C20^{\circ} \mathrm{C} to 25∘C25^{\circ} \mathrm{C} at constant pressure is
View answer and explanation
Correct option: D
73H\frac{7}{3} H
Option D (73H\frac{7}{3} H) is correct because it correctly relates the heat required for monoatomic and diatomic gases at constant volume and pressure, respectively.
176 Medium
176 of 290
Understanding +4 | -1
In case of hydrogen and oxygen at NTP, which of the following is the same of both ?
View answer and explanation
Correct option: B
Average KE per molecule
Option B (Average KE per molecule) is correct because at the same temperature, the average kinetic energy per molecule is the same for all gases.
177 Medium
177 of 290
Understanding +4 | -1
The average kinetic energy of a gas molecule can be determined by knowing
View answer and explanation
Correct option: C
The temperature of the gas only
Option C (The temperature of the gas only) is correct because the average kinetic energy of a gas molecule is directly proportional to the absolute temperature of the gas, as described by the equipartition theorem.
178 Medium
178 of 290
Application +4 | -1
A mixture of 2 moles of helium gas (atomic mass = 4  amu4\;amu), and 1 mole of argon gas (atomic mass = 40amu40amu) is kept at 300K300K in a container. The ratio of the rmsrms speeds [Vrms(helium)Vrms(argon)]\left[ {\frac{{{V_{rms}}\left( {\mathrm{helium}} \right)}}{{{V_{rms}}\left( {\mathrm{argon}} \right)}}} \right] is
View answer and explanation
Correct option: D
3.16
Option D (3.16) is correct because it is calculated directly from the ratio of the square roots of the molecular masses.
179 Medium
179 of 290
Understanding +4 | -1
Consider a gas with density ρ\rho and cˉ{\bar c} as the root mean square velocity of its molecules contained in a volume. If the system moves as whole with velocity v,  v,\; then the pressure exerted by the gas is
View answer and explanation
Correct option: A
13ρcˉ2\frac{1}{3}\rho {{\bar c}^2}
Option A is correct because the pressure exerted by the gas depends on the root mean square velocity of its molecules and is independent of the system's overall velocity.
180 Medium
180 of 290
Application +4 | -1
The kinetic energy of emitted electron is EE when the light incident on the metal has wavelength λ\lambda. To double the kinetic energy, the incident light must have wavelength :
View answer and explanation
Correct option: B
hcλEλ+hc\frac{h c \lambda}{E \lambda+h c}
Option B (hcλEλ+hc\frac{h c \lambda}{E \lambda+h c}) is correct because it is derived from the photoelectric effect equation, relating kinetic energy and wavelength.
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