Kinetic Theory Class 11 Physics Important Questions

Very Short Answer Questions [1 Mark Questions]

Ques. What does gas constant R signify and what is its value?

Ans. The universal gas constant represented as R signifies the work done by (or on) a gas per mole per kelvin. Its value is 8.31 J mol^-1 K

Ques. What is the nature of the curve obtained when:

1. Pressure versus reciprocal of volume is plotted for an ideal gas at constant temperature.

Ans. It is a straight line.

2. Volume of an ideal gas is plotted against absolute temperature at a constant pressure.

Ans. It is a straight line.

Ques. The graph shows the variation of the product of PV with the pressure of the constant mass of three gases A, B and C. If all the changes are at a constant temperature, then which of the three gases is an ideal gas? Why?

Ans. A is an ideal gas because the product PV is constant at constant temperature for an ideal gas.

Ques. According to Charle’s law what is the minimum possible temperature?

Ans. – 273.15°C.

Ques. Find the ratio of initial and final pressures if the masses of all the molecules of a gas are halved and their speeds are doubled.

Ans. 1:2

Ques. Water solidifies into ice at 273 K. What happens to the K.E. of water molecules?

Ans. It is partly transformed into the binding energy of ice.

Ques. Name three gas laws that can be obtained from the gas equation.

Ans. Boyle’s law, Charle’s law ,Gay Lussac’s law.

Ques. What is the average velocity of the molecules of a gas in equilibrium?

Ans. Zero.

Ques. A vessel is filled with a mixture of two different gases. Will the mean kinetic energies per molecule of both gases be equal and why?

Ans. Yes, it will be. This is because the mean K.E. per molecule i.e. 32 kT depends only upon the temperature.

Ques. Four molecules of a gas are having speeds, v₁, v₂, v₃ and v₄.

1. What is their average speed?

Ans. V_av = v₁+v₂+v₃+v₄₄

2. What is the r.m.s. speed?

Ans. V_rms = v₂₁+v₂₂+v₂₃+v₂₄₄−−−−−−−−−√

Ques. If the density of a gas is doubled keeping all other factors unchanged, what will be the effect of the same on the pressure of the gas?

Ans. It will be doubled.

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Short Answer Questions [2 Marks Questions]

Ques. Why is cooling caused by evaporation?

Ans. Evaporation occurs in consequence of faster molecules escaping from the surface of the liquid. The liquid is thus left with molecules having lower speeds. The decrease in the mean speed of molecules results in lowering the temperature and hence cooling is caused.

Ques. Equal masses of O₂ and He gases are supplied equal amounts of heat. Which gas would undergo a greater rise in temperature and why?

Ans. Helium is a monatomic gas while O₂ is a diatomic gas. In the case of helium, the supplied heat has to work in increasing only the translational K.E. of the gas molecules. On the other hand, in the case of oxygen, the supplied heat has to increase the translation, vibrational and rotational K.E. of gas molecules which means all three of them. Thus, helium would undergo a greater rise in temperature.

Ques. State law of equipartition of energy.

Ans. It states that in equilibrium, the total energy of the system is divided equally in all possible energy modes with each mode i.e. degree of freedom having an average energy equal to 12 k_BT.

Ques. A glass of water is stirred and then allowed to stand until the water stops moving. What has happened to the K.E. of the moving water?

Ans. The K.E. of moving water is dissipated into internal energy. The temperature of water thus increases.

Ques. Why does the pressure of a gas increase when it is heated up?

Ans. This is due to the two reasons:

1. The gas molecules move faster than before after getting heating up and so strike the container walls more often.
2. Each impact yields a greater momentum to the walls of the container.

Ques. R.m.s. velocities of gas molecules are comparable to those of a single bullet, yet a gas takes several seconds to diffuse through a room. Explain why?

Ans. Gas molecules collide with one another at a very high frequency. Hence, a molecule moves along a random and long path to go from one point to another. Therefore, gas takes several seconds to go from one corner of the room to the other.

Ques. When a gas is heated, its molecules move apart. Does this increase the P.E. or K.E. of the molecules? Explain.

Ans. It increases the K.E. of the molecules. Because of heating, the temperature increases and hence the average velocity of the molecules and their collision with the wall also increases which increases the K.E.

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Long Answer Questions [3 Marks Questions]

Ques. There are n molecules of a gas in a container. If the number of molecules is increased to 2n, what will be:

1. the pressure of the gas.
2. the total energy of the gas.
3. r.m.s. speed of the gas molecules.

Ans.

1. We know that

P = 13mnC².

where n = no. of molecules per unit volume.

Thus when no. of molecules is increased from n to 2n, no. of molecules per unit volume (n) will increase from n 2n

nV to 2nV, hence pressure will become double.

2. The K.E. of a gas molecule is,

12mC² = 32kT

If the no. of molecules is increased from n to 2n. There is no effect on the average K.E. of a gas molecule, but the total energy is doubled.

100. r.m.s speed of gas is C_rms = 3Pρ−−−√=3Pmn−−−√

When n is increased from n to 2n. both n and P become double and the ratio Pn remains unchanged. So there will be no effect of increasing the number of molecules from n to 2n on r.m.s. speed of gas molecules.

Ques. Two bodies of specific heats S₁ and S₂ having the same heat capacities are combined to form a single composite body. What is the specific heat of the composite body?

Ans. Let m₁ and m₂ be the masses of two bodies having heat capacities S₁ and S₂ respectively.

∴ (m₁ + m₂)S = m₁S₁ + m₂S₂ = m₁S₁ + m₁S₁ = 2m₁S₁

S = 2m₁S₁m₁+m₂.

Also, m₂S₂ = m₁S₁

or

m₂ = m₁ S₁ S₂

∴ S = 2 m₁ S₁ m₁+m₁ S₁ S₂=2 S₁ S₂ S₁+S₂

Ques. Tell the degree of freedom of:

1. Monoatomic gas moles.

Ans. A monatomic gas possesses 3 translational degrees of freedom for each molecule.

2. Diatomic gas moles.

Ans. A diatomic gas molecule has 5 degrees of freedom including 3 translational and 2 rotational degrees of freedom.

100. Polyatomic gas moles.

Ans. The polyatomic gas molecule has 6 degrees of freedom (3 translational and 3 rotational).

Ques. Explain why it is not possible to increase the temperature of gas while keeping its volume and pressure constant?

Ans. It is not possible to increase the temperature of a gas keeping volume and pressure constant can be explained as follows:

According to the Kinetic Theory of gases,

T ∝ PV

Now, because T is directly proportional to the product of P and V. If P and V are constant, then T is also constant.

Ques. Calculate the value of the universal gas constant (R).

Ans. We know that R is given by

R = PVT

Now one mole of all gases at S.T.P. occupy 22.4 litrês.

P = 0.76 m of Hg

= 0.76 × 13.6 × 10³ × 9.8

= 1.013 × 10⁵ Nm^-2

V = 22.4 litre

= 22.4 × 10 3m³

T = 273 K .

n = 1

∴ R = 1.013×105×22.4273 × 10^-3

= 8.31 J mol^-1 k^-1

Ques. On reducing the volume of the gas at a unvarying temperature, the pressure of the gas increases. Explain this phenomenon on the basis of the kinetic theory of gases.

Ans. On reducing the volume, the amount of space for the given number of molecules of the gas decreases i.e. no. of molecules per unit volume increases. As a result, more molecules collide with the walls of the vessel when considered per second. Hence a larger momentum is transferred to the walls each second. This is the reason that the pressure of gas increases.

Ques. Why is temperature less than absolute zero not possible?

Ans. According to the kinetic interpretation of temperature, absolute temperature means the kinetic energy of molecules stops. As heat is taken out, the temperature falls down and hence velocity decreases. At absolute zero, the velocity of the molecules becomes zero i.e. kinetic energy becomes zero. So, there is no more decrease of K.E. possible, hence temperature cannot fall further.

Ques. What are the different ways of increasing the number of molecular collisions per unit time against the walls of the vessel containing a gas?

Ans. The number of collisions per unit time .an be increased in the following ways:

1. By increasing the temperature of the gas.
2. By increasing the number of molecules.
3. By decreasing the volume of the gas.

Ques. Distinguish between the terms evaporation, boiling and vaporisation.

Ans. Evaporation: It is defined as the process of conversion of the liquid into a vapour state at all temperatures and it occurs only at the surface of the liquid.

Boiling: It is the process of rapid conversion of the liquid into the vapour state at a definite or fixed temperature and it occurs throughout the liquid.

Vaporisation: It is the general term for the conversion of liquid into the vapour state. It includes both evaporation and boiling.

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Very Long Answer Questions [5 Marks Questions]

Ques. Derive gas laws from the kinetic theory of gases.

Ans.

1. Boyle’s law: It states that P ∝ 1⁄V, If T = constant.

Derivation:

We know from the kinetic theory of gases that

Here R = constant

If T = constant, then PV = constant

or

P = 1⁄V

2. Charles’ law: It states that for a given mass of a gas, the volume of the gas is directly proportional to the absolute temperature of the gas if pressure is constant

1. e. V ∝ T.

Derivation:

We know that

PV = 1⁄3MC²= 1⁄3mNC²

Were N = Avogadro’s number

Also, we know that mean K..E. of a molecule is

If P = constant, then V ∝ T.

Hence proved.

100. Avogadro’s Law: It states that equal volumes of all gases contain an equal number of molecules if Temperature and Pressure are the same.

Derivation:

Consider two gases A and B having n₁ and n₂ as the no. of molecules, C₁ and C₂ are the r.m.s. velocities of these molecules respectively.

According to the kinetic theory of gases,

Also, we know that

Hence proved.

500. Graham’s law of diffusion of gases: It states that the rate of diffusion of a gas is inversely proportional to the square root of the density of the gas.

Derivation:

We know that

Also, we know that r.m.s. velocity is directly proportional to the rate of diffusion (r) of the gas, i.e.

Ques. Answer of the following

1. What is an ideal perfect gas?
2. What is the Mean free path?
3. State the law of equipartition of energy?

Ans.

1. A gas that obey all the laws like the Boye’s Law, Charles Law, Gay Lussac’s Law, Avogadro’s Law, etc is called an ideal gas. In an ideal gas, the size of the molecules of gas is zero and there is no force of attraction or repulsion between the molecules of either same or different material.
2. Mean free path is defined as the mean distance a molecule can travel between two successive collisions. It is represented by λ and its unit is metres (m).
3. Law of Equipartition of Energy states that if a dynamic system is said to be in thermal equilibrium, the energy of the system is equally distributed amongst the various degrees of freedom and the energy associated with each degree of freedom per molecule is 1/2 k_bT.

Ques. Define and derive an expression for the mean free path.

Ans. It is defined as the average distance travelled by a gas molecule between two successive collisions. It is denoted by X.

Derivation of Expression – Let us assume that only one molecule is in motion and all other molecules are at rest. ,

Let d = diameter of each molecule.

l = distance travelled by the moving molecule.

The moving molecule will collide with all those molecules whose centres lie inside a volume πd2l.

Let n = no. of molecules per unit volume in the gas.

Now λ = distance travelled no. of collisions =lnπd2l

or

λ = 1nπd2 …(1)

In this derivation, we have assumed that all but one molecules are at rest. But this assumption is not correct. All the molecules are in random motion. So the chances of a collision by a molecule are greater.

Thus taking it into account, the mean free path can be shown to be 2–√ times less than that in equation (1),

∴ λ = 12√nπd2

which is the required expression.