Kinetic Theory Class 11 Physics Important Questions

Students can read the important questions given below for Kinetic Theory Class 11 Physics. All Kinetic Theory Class 11 Notes and questions with solutions have been prepared based on the latest syllabus and examination guidelines issued by CBSE, NCERT and KVS. You should read all notes provided by us and Class 11 Physics Important Questions provided for all chapters to get better marks in examinations. Physics Question Bank Class 11 is available on our website for free download in PDF.

Important Questions of Kinetic Theory Class 11

Very Short Answer Type Questions :

Question. On reducing the volume of a gas at constant temperature, the pressure of the gas increases. Explain it on the basis of kinetic theory.
Answer : On reducing the volume, the number of molecules per unit volume increases. Hence, a large number of molecules collide with the walls of the vessel per second and a larger  momentum is transferred to the wall per second. This increases the pressure of the gas.

Question. Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 Å and 2 Å. The gases may be considered under identical conditions of  temperature, pressure and volume. 
Answer : As, λ ∝ 1/d²
Given, d₁ = 1Å and d₂ = 2Å ⇒ λ₁ : λ₂ = 4 : 1

Question. When a gas is heated, its temperature increases. Explain it on the basis of kinetic theory of gases.
Answer : When a gas is heated, the root mean square velocity of its molecules increases. As V_rms α√ T so temperature of the gas increases.

Question. On which factors does the average kinetic energy of gas molecules depend?
Answer : The average K.E. of a gas molecule depends only on the absolute temperature of the gas and is directly proportional to it.

Question. Name two factors on which the degrees of freedom of a gas depend?
Answer : (i) Atomicity of the gas molecule.
(ii) Shape of the molecule.
(iii) Temperature of gas.

Question. Following fig. shows the variation of the product PV with respect to the pressure (P) of given masses of three gases, A,B,C. The temperature is kept constant. State with proper arguments which of these gases is ideal.
Answer : Gas ‘C’ is ideal because PV is constant for it. That is gas ‘C’ obeys Boyle’s law at all pressures.

Question. What type of motion is associated with the molecules of a gas?
Answer : Brownian motion.

Question. The ratio of vapour densities of two gases at the same temperature is 8:9. Compare the rms. velocity of their molecules
Answer :

Question. What is the mean translational kinetic energy of a perfect gas molecule at temperature T?
Answer : A perfect gas molecule has only translational K.E.
E = 3/2 kBT

Question. Why do the gases at low temperature and high pressure, show large deviations from ideal behaviour?
Answer : At low temperature and high pressure , the intermolecular attractions become appreciable. So, the volume occupied by the gas molecules cannot be neglected in comparison to the volume of the gas. Hence the real gases show large from ideal gas behaviour.

Question. Define absolute zero, according to kinetic interpretation of temperature?
Answer : Absolute zero is the temperature at which all molecular motion ceases.

Question. Cooking gas containers are kept in a lorry moving with uniform speed. What will be the effect on temperature of the gas molecules?
Answer : As the lorry is moving with a uniform speed, there will be no change in the translational motion or K.E. of the gas molecules. Hence the temperature of the gas will remain same.

Question. The volume of a given mass of a gas at 27°C and 1 atm is 100 cm³. What will be its volume at 327°C?
Answer :

Question. What would be the effect on the rms velocity of gas molecules if the temperature of the gas is increased by a factor of 4?
Answer : ∴ v_rms ∝ √T . If temperature is increased 4 times then, v_rms is doubled.

Question. What is the order of mean free path (λ) of the gas molecules?
Answer : λ is of the order of a micron.
1 micron = 10^–6 m

Question. A gas mixture consists of molecules of types A, B and C with masses m_A > m_B >m_C . Rank the  three types of molecules in decreasing order of their rms speeds.
Answer : 

Question. Two gases are at temperatures of 300 K and 350 K respectively. What is the ratio of the average kinetic energies of their molecules ?
Answer : 

Question. Find the energy per mole per degree of freedom of an ideal gas.
Answer : According to the law of equipartition of energy, the energy per mole per degree of freedom of an ideal gas

Question. Name the two factors on which the degree of freedom of gas depend.
Answer : The degree of freedom of gas depend on
(i) N = number of particles in the system
(ii) k = number of independent relations between the particles.

Question. Plot pressure (P) versus volume (V) graphs for a fixed mass of a gas are drawn at two  different temperatures.
Answer : 

Short Answer Type Questions :

Question. When a gas filled in a closed vessel is heated through 1⁰C, its pressure increases by 0.4 %. What is the initial temperature of the gas?
Answer : P` = P = 0.4/100. P, T` = T + 1

Question. A box contains equal number of molecules of hydrogen and oxygen. If there is a fine hole in the box, which gas will leak rapidly? Why?
Answer :

Hence hydrogen gas will leak more rapidly because of its smaller molecular mass.

Question. What will be the ratio of the root mean square speeds of the molecules of an ideal gas at 270K and 30K?
Answer : 

Question. Should the specific heat of monoatomic gas be less than, equal to or greater than that of a diatomic gas at room temperature? Justify your answer.
Answer : Specific heat of a gas at constant volume is equal to f/2R.
For monoatomic gases f = 3 so Cv = 3/2 R.
For diatomic gases f = 5 so Cv = 5/2 R.
Hence the specific heat for monoatomic gas is less than that for a diatomic gas.

Question. Write the relation between the pressure and kinetic energy per unit volume of a gas. Water solidifies into ice at 273 K. What happens to the K.E. of water molecules?
Answer : P = 2/3 E. The K.E. of water molecules gas partly converted into the binding energy of the ice.

Question. The absolute temperature of a gas is increased 4 times its original value. What will be the change in r.m.s. velocity of its molecules?
Answer : V_rms α√T
V’_rms α√4T
V’_rms / V_rms = 2
V’_rms = 2V_rms
Change in rms velocity of molecules = V’_rms – V_rms
= V_rms

Question. The velocities of three molecules are 3V,4V and 5V.Determine the root mean square velocity.
Answer :

Question. A gas in a closed vessel is at the pressure Po. If the masses of all the molecules be made half and their speeds be made double, then find the resultant pressure?
Answer :

Question. Write the equation of state for 16g of O₂.
Answer : No. of moles in 32g of O₂ = 1
No. of moles in 16g of O₂ = 1/9 x 16 = ½
As pv = nRT and n = 1/2
So, PV= 1/2 RT

Question. A mixture of Helium and Hydrogen gas is filled in a vessel at 30 degree Celsius.
Compare the root mean square velocities of the molecules of these gases at this temperature. (atomic weight of Hydrogen is 4)
Answer :

Question. Figure shows plot of PV/T versus P for 1.00 × 10^–3 kg of oxygen gas at two different  temperatures.

(a) Which is true: T₁ > T₂ or T₁ < T₂?
(b) What is the value of PV/T where the curves meet on the y-axis?
Answer : (a) Curve at temperature T₁ is more close to the dotted plot than the curve at the temperature T₂. Since the behaviour of a real gas approaches the perfect gas behaviour, as the  temperature is increased therefore T₁ > T₂ .
(b) If the amount of gas under consideration is 1 mole, then the value of PV/T, where the curves meet PV/Taxis will be R (= 8.31 J mole^–1 K^–1). For oxygen, molecular mass is 32.0 g i.e., 32.0 × 10^–3 kg. Since mass of oxygen gas under considerations is 1.0 × 10^–3 kg, the value of PV/T, where the curves meet the PV/T-axis is given by

Question. (a) What do you understand by specific heat capacity of water (assume water behaves as a solid)?
(b) If one mole of ideal monatomic gas
(ϒ = 5/3) is mixed with one mole of diatomic gas
(ϒ = 7/5). What is the value of g for the mixtures?
(here, ϒ represents the ratio of specific heat at constant pressure to that at constant volume)
Answer : (a) Consider water molecule as a solid made up of 3 atoms (2 hydrogen and 1 oxygen) and each atom if free to vibrate in three dimensions about its mean position.
In 3-dimension,
[the average energy per atom of water molecule = 3k_BT ]
∴ Total energy of one molecule of water is

Question. The velocities of ten particles of a sample of gas in m s^–1 are 8, 1, 4, 4, 4, 4, 7, 7, 6, 5. Find the
(i) average speed
(ii) rms speed
(iii) most probable speed.
Answer : (i) Average speed,

Question. Two vessels of the same volume are filled with the same gas at the same temperature. If the pressure of the gas in these vessels be in the ratio 1 : 2 then, find
(i) the ratio of the rms speeds of the molecules.
(ii) the ratio of the number of molecules.
Answer : 

Question. Calculate specific heat of water using law of equipartition of energy.
Answer : Consider water molecule as a solid made up of 3 atoms (2 hydrogen and 1 oxygen) and each atom if free to vibrate in three dimensions about its mean position. 
In 3-dimension, the average energy per atom of water molecule = 3k_BT

Question. For a rigid diatomic molecule, universal gas constant R = nC_P where C_P is the molar specific heat at constant pressure and n is a number. Find the value of n.
Answer : For a rigid diatomic molecule

Question. State the law of equipartition of energy. 
Answer : In equilibrium the total energy is equally distributed in all possible energy modes, with each mode having an average energy equal to (1/2) kT. This is known as law of equipartition energy.
For mono-atomic of a single molecule is
E = 3/2 k_BT
For diatomic, KE of a single molecule is
E = 5/2 kT

Question. A flask contains argon and chlorine in the ratio of 2 : 1 by mass. The temperature of the  mixture is 27°C. Obtain the ratio of average kinetic energy per molecule and root mean  square speed v_rms of the molecules of the two gases. Given : atomic mass of argon = 39.9 u and molecular mass of chlorine = 70.9 u.
Answer : Average kinetic energy per molecule of any ideal gas

Question. Isothermal curves for a given mass of gas are shown at two different temperatures T₁ and T₂. State whether T₁ > T₂ or T₂ > T₁. Justify your

Answer : Since, T = PV / nR, ∴ T ∝ P V
                                                                                    [∴ R = constant]
If area under PV is greater for the curve at T₂ than for the curve T₁. Hence, T₂ > T₁.

Question. (a) Write the SI unit of R.
(b) Write the values of R in different units, where R is universal gas constant.
Answer : (a) SI unit of R is J mol^–1 K^–1
(b) R = 8.31×10⁷ ergs mol^–1 K^–1
= 8.31 J mol^–1 K^–1 = 2 cal mol^–1 K^–1

Question. Two monatomic ideal gases A and B occupying the same volume V, are at the same temperature T and pressure P. If they are mixed, the resultant mixture has volume V and temperature T. Calculate the pressure of the mixture.
Answer : As number of moles is conserved

Question. If C_P and C_V denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then what is the value of C_P – C_V ?
Answer : Molar heat capacity
                                         = Molar mass × specific heat capacity
So, the molar heat capacities of nitrogen at constant pressure and constant volume will be 28C_P and 28C_V respectively.
∴ 28C_P – 28C_V = R or C_p – C_v = R/28

Question. A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?
Answer : When the tube is held horizontally, the mercury thread of length 76 cm traps a length of air = 15 cm. A length of 9 cm of the tube will be left at the open end, figure (a). The pressure of air enclosed in tube will be atmospheric pressure. Let area of cross-section of the tube be 1 sq. cm.

∴ P₁ = 76 cm and V₁ = 15 cm³
When the tube is held vertically, 15 cm air gets another 9 cm of air filled in the right hand side (in the horizontal position) and let h cm of mercury flows out to balance the atmospheric pressure, figure (b) Then the heights of air column and mercury column are (24 + h) cm and (76 – h) cm respectively. 
The pressure of air = 76 – (76 – h) = h cm of mercury
∴ V2 = (24 + h) cm³ and P² = h cm.
If we assume that temperature remains constant, then

Since h cannot be negative (because more mercury cannot flow into the tube), therefore h = 23.8 cm. Thus, in the vertical position of the tube, 23.8 cm of mercury flows out.

Question. (a) How much volume does one mole of a gas occupy at NTP?
(b) What are the essential characteristics of an ideal gas?
(c) What are the conditions under which real gases behave nearly as a perfect gas?
Answer : (a) 22.4 litre of volume is occupied by one mole of gas at NTP.
(b) A gas which obeys the ideal gas equation, PV = nRT, at all temperature and pressures is called an ideal gas. No such a gas does not really exists. An ideal gas is a simple theoretical model of a gas.
Important characteristics of an ideal gas are
(i) The size of the molecule is negligibly small.
(ii) There is no force of attraction or repulsion amongst its molecules.
(c) At high temperature and low pressure the real gas behaves as an ideal gas.

Question. An insulated container containing monoatomic gas of molar mass m is moving with a velocity v_o . If the container is suddenly stopped, find the change in temperature.
Answer : 

Question. Explain why
(a) there is no atmosphere on moon.
(b) there is fall in temperature with altitude.
Answer : (a) Gravitational force of the moon is roughly 1/6^th of the gravitational force of earth, so the escape velocity of air molecules on the moon is smaller than that on earth. As the moon is in the proximity of the Earth as seen from the sun so solar irradiance on the moon surface is same as that of the earth.
Escape velocity of the moon is 2.38 km s^–1. The rms speed of the constituents of air (O₂, N₂, CO₂ and water vapour) lies between 0.4 km s^–1 to 0.8 km s^–1, which is smaller than 2.38 km s^–1. Inspite of this a significant number of molecules have speeds greater than 2.38 km s^–1 and hence escape easily. Now rest of molecules arrange the speed distribution for the equilibrium temperature. Again a  significant number of molecules get their speeds greater than 2.38 km s^–1 and escape. Due to this continuous process, the moon has lost its atmosphere over a long period of time.
(b) We know
E = U + K = constant
As the molecules move higher their potential energy (U) increase and due to this its kinetic energy (K) decrease and hence fall in temperature. At greater height more volume is available for gas to expand and hence some cooling takes place i.e. a fall in temperature.

Question. A gas mixture consists of molecules of types A, B and C with masses m_A > m_B >m_C . Rank the three types of molecules in decreasing order of (a) average K.E., (b) rms speeds.
Answer : 

Question. Estimate the average thermal energy of a helium atom at
(i) room temperature (27°C), 
(ii) the temperature on the surface of the Sun (6000 K),
(iii) the temperature of 10 million kelvin (the typical core temperature in case of a star). 
Answer : The average kinetic energy of the gas at a temperature T

Question. Calculate the internal energy of 1g of oxygen gas at STP.
Answer : 

Question. From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7 cm³ s^–1. The diffusion of another gas under the same conditions is measured to have an average rate of 7.2 cm³ s^–1. Identify the gas.
Answer : According to Graham’s law of diffusion,

Question. Ten small planes are flying at a speed of 150 km h^–1 in total darkness in an air space that is 20 × 20 × 1.5 km³ in volume. You are in one of the planes, flying at random within this space with no way of knowing where the other planes are. On the average about how long a time will elapse between near collision with your plane. 
Assume for this rough computation that a safety region around the plane can be approximated by a sphere of radius 10 m.
Answer : Here, v = 150 km h^–1

Question. Calculate the number of degrees of freedom of molecules of hydrogen in 1 cm³ of hydrogen gas at NTP.
Answer : 

Question. What do you understand by mean speed, root mean square speed and most probable speed of a
gas. The velocities of ten particles in m/s are 0, 2, 3, 4, 4, 4, 5, 5, 6, 9. Calculate
(i) average speed
(ii) rms speed.
Answer : 

Question. Hydrogen is at temperature T and helium is at temperature 2T. The internal energy of both gases is the same. What is the ratio of number of moles of hydrogen and helium gases?
Answer : Internal energy of a gas, U = nC_VT
where n is the number of the moles of a gas and CV is the molar specific heat at constant volume and T is the temperature of the gas.

Question. A closed vessel contains a mixture of two diatomic gases A and B. Molar mass of A is 16 times that of B and mass of gas A, contained in the vessel is 2 times that of B. Find the ratio  of 
(i) average kinetic energy per molecule of gas A is equal to that of gas B.
(ii) root mean square value of translational velocity of gas B to that of A.
(iii) pressure exerted by gas B to that exerted by gas A.
Answer : 

Question. A balloon has 5 g mole of helium at 7°C. Calculate,
(a) the number of atoms of helium in the balloon,
(b) the total internal energy of the system.
Answer : 

Long Answer Type Questions :

Question. Using the law of equipartition of energy, determine the values of CP, CV and g for
(a) monatomic,
(b) diatomic and
(c) triatomic gases.
Answer : (a) Monoatomic gases : For monoatomic gases, degree of freedom, f = 3

Question. Consider a rectangular block of wood moving with a velocity v0 in a gas at temperature T and  mass density ρ. Assume the velocity is along x-axis and the area of cross-section of the block perpendicular to v₀ is A. Show that the drag force on the block is 4pAv₀ √kT/m where m is the mass of the gas molecule.
Answer : 

Speed of wooden block = v_o
Cross-sectional area of block = A
Temperature of the gas = T
Let n be the number of molecules per unit volume.
rms speed of gas molecules = v_rms.
Relative speed of the molecule with respect to front face of the block = v + v_o
In head on collision, momentum transferred to block per collision = 2m(v + v_o)

Question. What is meant by mean free path of a gas molecule? Derive expression for it. On which  factors does the mean free path depends? 
Answer : The mean free path of a gas molecule is defined as the average distance travelled by a molecule between two successive collisions.
Let us consider the mean free path, we make use of the following assumptions:
(i) Each molecule of the gas is a sphere of diameter d.
(ii) All molecules of the gas except the molecule A under consideration are at rest.
As shown figure, suppose the molecule A has average speed v . It will collide with all those molecules whose centres lie within a distance d from its path. In time Δt, it will collide with all those molecules in the cylinder of volume pd 2v Δt. Let n be the number of molecules per unit volume.

Question. Derive an expression for pressure of a gas in a container. Using it, relate K.E. with pressure.
Answer : Consider a cylinder with piston of cross section area A and length L. The cylinder contains ideal gas and we need to find the pressure exerted by gas molecule on the wall of the cylinder. According to assumptions, the collisions are perfectly elastic hence KE is conserved.
Consider the motion of particle only in one direction say (x-axis), velocities in y and z directions remain unaltered, only x-component changes by reversing itself from v_x to –v_x. Now, change in momentum of a particle along x-axis, Δp_x = 2mv_x.
If length of cylinder is d, the time taken to hit the wall and bouncing back to the position be Δt.