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NEET • PHYSICS • KINETIC THEORY OF GASES

Kinetic theory of gases practice · Mixed difficulty

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61 Medium
61 of 115
Remembering +4 | -1
The relation between the gas pressure PP and average kinetic energy per unit volume EE is
View answer and explanation
Correct option: D
P=23EP = \frac{2}{3}E
The correct answer is D because the relation between gas pressure and average kinetic energy per unit volume is P = (2/3)E, derived from kinetic theory of gases.
62 Medium
62 of 115
Understanding +4 | -1
Relation between the ratio of specific heats (γ)\left( \gamma \right) of gas and degree of freedom 'f' will be
View answer and explanation
Correct option: C
f=2/(γ−1)f = 2/\left( {\gamma - 1} \right)
Option C (f=2/(γ−1)f = 2/(\gamma - 1)) is correct because it represents the correct relationship between degrees of freedom and the ratio of specific heats.
63 Medium
63 of 115
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 (13ρcˉ2\frac{1}{3}\rho {\bar c}^2) is correct because the pressure exerted by a gas depends on the root mean square velocity of its molecules, not the bulk velocity of the system.
64 Medium
64 of 115
Application +4 | -1
A monoatomic gas is kept at room temperature 300K. Calculate the average kinetic energy of gas molecule (Use k=1.38×10−23MKSk = 1.38 \times {10^{ - 23}}MKS units)
View answer and explanation
Correct option: C
0.039eV0.039eV
Option C (0.039eV0.039eV) is correct because it is the average kinetic energy of a monoatomic gas molecule at 300K, calculated using the formula KE=32kTKE = \frac{3}{2}kT.
65 Medium
65 of 115
Analyzing +4 | -1
Read the given statements and decide which is/are correct on the basis of kinetic theory of gases
(I) Energy of one molecule at absolute temperature is zero
(II) r.m.s.r.m.s.speeds of different gases are same at same temperature
(III) For one gram of all ideal gas kinetic energy is same at same temperature
(IV) For one mole of all ideal gases mean kinetic energy is same at same temperature
View answer and explanation
Correct option: D
None of these
Option D (None of these) is correct because all the given statements are incorrect based on the kinetic theory of gases.
66 Medium
66 of 115
Analyzing +4 | -1
Given below are two statements :

Statement I : The temperature of a gas is −73∘C-73^{\circ} \mathrm{C}. When the gas is heated to 527∘C527^{\circ} \mathrm{C}, the root mean square speed of the molecules is doubled.

Statement II : The product of pressure and volume of an ideal gas will be equal to translational kinetic energy of the molecules.

In the light of the above statements, choose the correct answer from the options given below:
View answer and explanation
Correct option: D
Statement I is true but statement II is false
Option D is correct because Statement I is true (RMS speed doubles when temperature quadruples in Kelvin), but Statement II is false (PV = 2/3 * Translational KE).
67 Easy
67 of 115
Analyzing +4 | -1
Assertion: A gas has a unique value of specific heat.
Reason: Specific heat is defined as the amount of heat required to raise the temperature of unit mass of the substance through unit degree.
View answer and explanation
Correct option: D
If reason is true but assertion is false
The correct answer is D because the reason is true, but the assertion is false. A gas can have different specific heats under different conditions (e.g., constant pressure or constant volume).
68 Easy
68 of 115
Remembering +4 | -1
A monoatomic gas molecule has
View answer and explanation
Correct option: A
Three degrees of freedom
Option A (Three degrees of freedom) is correct because a monoatomic gas molecule can only move in three independent directions (x, y, and z).
69 Easy
69 of 115
Application +4 | -1
A gas consisting of rigid diatomic molecules is expanded adiabatically. How many time has the gas to be expanded to reduce the root mean square velocity of the molecules η=1.50\eta=1.50 times.
View answer and explanation
Correct option: A
7.6
The correct answer is 7.6 because it involves adiabatic expansion and the relationship between temperature and volume for diatomic gases.
70 Medium
70 of 115
Understanding +4 | -1
Water is used to cool radiators of engines, because
View answer and explanation
Correct option: D
It has high specific heat
Option D (It has high specific heat) is correct because water's high specific heat allows it to absorb a large amount of heat with a small temperature change.
71 Medium
71 of 115
Understanding +4 | -1
Molecules of a gas behave like
View answer and explanation
Correct option: C
Perfectly elastic rigid sphere
Option C (Perfectly elastic rigid sphere) is correct because the kinetic theory of gases assumes that gas molecules undergo perfectly elastic collisions and are rigid.
72 Easy
72 of 115
Application +4 | -1
The root mean square velocity of the molecules of a gas is 1260 m/s1260 \mathrm{~m} / \mathrm{s}. The most probable speed of the molecules is
View answer and explanation
Correct option: A
1029 m s−11029 \mathrm{~m} \mathrm{~s}^{-1}
Option A (1029 m s−11029 \mathrm{~m} \mathrm{~s}^{-1}) is correct because the most probable speed is related to the RMS speed by a constant factor.
73 Medium
73 of 115
Application +4 | -1
The Molecules of a given mass of a gas have r.m.s. Velocity of 200ms−1 at 270C200ms_{}^{ - 1}\,at\,{27^0}C and 1.0×105Nm−21.0 \times {10^5}N{m^{ - 2}} pressure. When the temperature and pressure of the gas are respectively, 1270C  and  0.05×105Nm−2{127^0}C{\mkern 1mu} {\mkern 1mu} and{\mkern 1mu} {\mkern 1mu} 0.05 \times {10^5}N{m^{ - 2}} , The r.m.s velocity of its molecule in   ms−1\;ms_{}^{ - 1} is
View answer and explanation
Correct option: B
4003\frac{{400}}{{\sqrt 3 }}
Option B (4003\frac{{400}}{{\sqrt 3 }}) is correct because it is the calculated rms velocity based on the given temperature and pressure changes.
74 Medium
74 of 115
Application +4 | -1
The speeds of 5 molecules of a gas (in arbitary units) are as follows : 2,3,4,5,6. The r.m.s speed for these molecules is
View answer and explanation
Correct option: D
4:244:24
Option D (4:244:24) is correct because it is the calculated root mean square speed for the given set of molecular speeds.
75 Easy
75 of 115
Analyzing +4 | -1
Assertion: Mean free path of a gas molecules varies inversely as density of the gas.
Reason: Mean free path varies inversely as pressure of the gas.
View answer and explanation
Correct option: A
If both Assertion and Reason are true and Reason is the correct explanation of Assertion
Option A is correct because both the assertion and the reason are true, and the reason correctly explains the assertion. Mean free path is inversely proportional to both density and pressure.
76 Medium
76 of 115
Application +4 | -1
For a gas molecule with a 6 degrees of freedom the law of equipartition of energy gives the following relation between the molar specific heat (Cv)\left( {Cv} \right) and gas constant (R)\left( R \right)
View answer and explanation
Correct option: D
Cv=3R{C_v} = 3R
Option D (Cv=3R{C_v} = 3R) is correct because, according to the equipartition theorem, each degree of freedom contributes R/2 to the molar specific heat at constant volume, and with 6 degrees of freedom, Cv=6∗(R/2)=3RC_v = 6 * (R/2) = 3R.
77 Medium
77 of 115
Analyzing +4 | -1
Same gas is filled in two vessels of the same volume at the same temperature. If the ratio of the number of molecules is 1:41: 4, then
A. The r.m.s. velocity of gas molecules in two vessels will be the same.
B. The ratio of pressure in these vessels will be 1:41 : 4.
C. The ratio of pressure will be 1:11 : 1.
D. The r.m.s. velocity of gas molecules in two vessels will be in the ratio of 1:41 : 4.
Choose the correct answer from the options given
below
View answer and explanation
Correct option: C
A and B only
Option A (A and B only) is correct because the r.m.s. velocity depends only on temperature, and the pressure ratio is directly proportional to the number of molecules.
78 Medium
78 of 115
Remembering +4 | -1
The translational kinetic energy of gas molecule for one mole of the gas is equal to
View answer and explanation
Correct option: A
32RT\frac{3}{2}RT
Option A (32RT\frac{3}{2}RT) is correct because the translational kinetic energy of one mole of gas is given by this formula.
79 Medium
79 of 115
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 is false (temperature doesn't necessarily increase), but the reason is true (kinetic energy of the gas molecules will increase due to the train's motion).
80 Easy
80 of 115
Application +4 | -1
One mole of an ideal gas undergoes a process: P=Po1+(Vo/V)2P=\frac{P_o}{1+\left(V_o / V\right)^2}. Here PoP_o and VoV_o are constants. Change in temperature of the gas when volume is changed from V=VoV=V_o to V=2VoV=2 V_o is
View answer and explanation
Correct option: B
11PoVo10R\frac{11 P_o V_o}{10 R}
Option B (11PoVo10R\frac{11 P_o V_o}{10 R}) is correct because it's calculated by finding the initial and final temperatures using the given process equation and then finding the difference.
81 Medium
81 of 115
Analyzing +4 | -1
Assertion: An ideal gas has infinitely many molar specific heats.
Reason: Specific heat is amount of heat needed to raise the temperature of 1 mole of gas by 1 K1 \mathrm{~K}.
View answer and explanation
Correct option: B
If both Assertion and reason are true and the Reason is not a correct explanation of the Assertion
Option B is correct because while both the assertion and reason are true, the reason doesn't explain why an ideal gas has infinitely many molar specific heats.
82 Medium
82 of 115
Analyzing +4 | -1
Let V\mathrm{V} denote the root mean square speed of the molecules in an ideal diatomic gas at absolute temperature T\mathrm{T}. The mass of a molecule is ' mm '. Neglecting vibrational energy terms, which is false?
View answer and explanation
Correct option: C
The average rotational kinetic energy of a molecule is mv2/4\mathrm{mv}^2 / 4
The statement that is false is C (The average rotational kinetic energy of a molecule is mv2/4\mathrm{mv}^2 / 4) because for a diatomic molecule, the average rotational kinetic energy is kT\mathrm{kT}, which is equivalent to mv2/6\mathrm{mv}^2 / 6.
83 Medium
83 of 115
Application +4 | -1
A sample of gas is 00C{0^0}C. To what temperature it must be raised in order to double the r.m.s. speed
of the molecule
View answer and explanation
Correct option: B
8190C{819^0}C
Option B (8190C{819^0}C) is correct because the RMS speed is proportional to the square root of the absolute temperature, so doubling the RMS speed requires quadrupling the absolute temperature.
84 Medium
84 of 115
Analyzing +4 | -1
The specific heats, Cp  and  CV{C_p}\mathrm{ \,\,and }\,\,{C_V} of a gas of diatomic molecules, A are given (in units of Jmol−1K−1\mathrm{J mo}{\mathrm{l}^{ - 1}}{K^{ - 1}}) by 29 and 22, respectively. Another gas of diatomic molecules B, has the corresponding values 30 and 21. If they are treated as ideal gases, then
View answer and explanation
Correct option: A
A has a vibrational mode but B has none
The correct answer is A (A has a vibrational mode but B has none) because the experimental value of Cp for gas A is greater than the theoretical value without vibration, indicating the presence of a vibrational mode.
85 Hard
85 of 115
Application +4 | -1
The temperature at which the ‘rms’ speed of an atom. In an argon gas cylinder is equal to the ‘rms’ speed of a helium gas atom at −20∘ C- 20^\circ \,C is (Atomic mass) of argon and that of helium = 4.04)
View answer and explanation
Correct option: A
2530 K
Option A (2530 K) is correct because it's the temperature at which the RMS speed of Argon equals the RMS speed of Helium at -20°C, calculated using the RMS speed formula.
86 Medium
86 of 115
Understanding +4 | -1
If the rms velocity of a gas is vv ,then
View answer and explanation
Correct option: B
v2/T=constant{v^2}/T = \mathrm{constant}
Option B (v2/T=constant{v^2}/T = \mathrm{constant}) is correct because the rms velocity squared is directly proportional to the temperature.
87 Hard
87 of 115
Application +4 | -1
An electron tube was sealed off during manufacture at a pressure of 1.2×10−7mm1.2 \times {10^{ - 7}}mm of mercury of 27∘C.27^\circ C. If volume is 1000 cm3,1000\,c{m^3},. The number of molecules that remains in the tube is
View answer and explanation
Correct option: C
3.86×10113.86 \times {10^{11}}
Option C (3.86×10113.86 \times {10^{11}}) is correct because it represents the number of molecules remaining in the electron tube after it was sealed off, calculated using the ideal gas law.
88 Medium
88 of 115
Application +4 | -1
The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the root mean square velocity of the gas molecule is V. at 480 K it becomes
View answer and explanation
Correct option: B
3 V
Option A (4 V) is incorrect. The correct answer is 2V. The rms velocity is proportional to the square root of the temperature. If the temperature is increased by a factor of 4, the rms velocity increases by a factor of 2.
89 Hard
89 of 115
Understanding +4 | -1
Assertion: Vibrational degree of freedom of a di-atomic gas molecule appears at very high temperature
Reason: Di-atomic gas has two vibrational degree of freedom in one direction.
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 (If both Assertion and Reason are true, but Reason is not the correct explanation of Assertion) is correct because both statements are true, but the reason doesn't explain why vibrational degrees of freedom appear at high temperatures.
90 Easy
90 of 115
Application +4 | -1
0.014 kg0.014 \mathrm{~kg} of nitrogen is enclosed in a vessel at a temperature of 27∘C27^{\circ} \mathrm{C}. How much heat has to be transferred to the gas to double the r.m.s. velocity of its molecules?
View answer and explanation
Correct option: A
9315
The correct answer is 9315 because it involves calculating the heat required to double the RMS velocity of nitrogen gas, considering its diatomic nature.
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