Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

I. E. Irodov Solutions for Physics Class 11 & Class 12

: Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

The document Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev is a part of the Course I. E. Irodov Solutions for Physics Class 11 & Class 12.

Q. 62. Modern vacuum pumps permit the pressures down to p = 4.10-15  atm to be reached at room temperatures. Assuming that the gas exhausted is nitrogen, find the number of its molecules per 1 cm3  and the mean distance between them at this pressure. 

Solution. 62. From the formula p - n k T

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Mean distance between molecules

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 63. A vessel of volume V = 5.0 l contains m = 1.4 g of nitrogen at a temperature T = 1800 K. Find the gas pressure, taking into account that η = 30% of molecules are disassociated into atoms at this temperature.

Solution. 63. A fter dissociation each N2 molecule becomes two Adatoms and so contributes, 2 x 3 degrees of freedom. Thus the number of moles becomes

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Here M is the molecular weight in grams of N2.


Q. 64. Under standard conditions the density of the helium and nitrogen mixture equals p = 0.60 g/l. Find the concentration of helium atoms in the given mixture. 

Solution. 64. Let n1 = number density of He atoms, n2 = number density of N2 molecules 

Then p = n1 m1 + n2 m2

where m1 = mass of He atom, m2 = mass of N2 molecule also p = (n1 + n2) kT

From these two equations we get

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 65. A parallel beam of nitrogen molecules moving with velocity v = 400 m/s impinges on a wall at an angle θ = 30° to its normal. The concentration of molecules in the beam n = 0.9.1019  cm-3. Find the pressure exerted by the beam on the wall assuming the molecules to scatter in accordance with the perfectly elastic collision law. 

Solution. 65.

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 66. How many degrees of freedom have the gas molecules, if under standard conditions the gas density is p = 1.3 mg/cm3 and the velocity of sound propagation in it is v = 330 m/s. 

Solution. 66. From the formula

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

If i = number of degrees of freedom of the gas then 

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 67. Determine the ratio of the sonic velocity v in a gas to the root mean square velocity of molecules of this gas, if the molecules are
 (a) monatomic; (b) rigid diatomic. 

Solution. 67. Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

so,   Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

(a) For monoatomic gases i = 3

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

(b) For rigid diatomic molecules i = 5

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 68. A gas consisting of N-atomic molecules has the temperature T at which all degrees of freedom (translational, rotational, and vibrational) are excited. Find the mean energy of molecules in such a gas. What fraction of this energy corresponds to that of translational motion? 

Solution. 68. For a general noncollinear, nonplanar molecule 

mean energy  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev + (3N - 6) kT (vibrational) 

= (3N - 3) kT per molecule

For linear molecules, mean energy  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev+ kT (rotational) + (3 N - 5) kT (vibrational)

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Translational energy is a fraction  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 69. Suppose a gas is heated up to a temperature at which all degrees of freedom (translational, rotational, and vibrational) of its molecules are excited. Find the molar heat capacity of such a gas in the isochoric process, as well as the adiabatic exponent γ, if the gas consists of
 (a) diatomic;
 (b) linear N-atomic;
 (c) network N-atomic molecules. 

Solution. 69. (a) A diatomic molecule has 2 translational, 2 rotational and one vibrational degrees of freedom. The corresponding energy per mole is

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Thus, Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

(b) For linear N - atomic molecules energy per mole

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

So,  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

(c) For noncollinear N- atomic molecules 

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 70. An ideal gas consisting of N-atomic molecules is expanded isobarically. Assuming that all degrees of freedom (translational, rotational, and vibrational) of the molecules are excited, find what fraction of heat transferred to the gas in this process is spent to perform the work of expansion. How high is this fraction in the case of a monatomic gas? 

Solution. 70. In the isobaric process, work done is

A = pdv = RdT per mole.

On the other hand heat transferred Q = CpdT

Now Cp = (3N - 2) R for non-collinear molecules and  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev  for linear molecules

Thus   Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

For monoatomic gases,  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 71. Find the molar mass and the number of degrees of freedom of molecules in a gas if its heat capacities are known: cv = 0.65 J/(g•K) and cp  = 0.91 J/(g•K).

Solution. 71. Given specific heats cp, cv (per unit mass)

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Also   Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 72. Find the number of degrees of freedom of molecules in a gas whose molar heat capacity
 (a) at constant pressure is equal to Cp  = 29 J/(mol.K);
 (b) is equal to C = 29 J/(mol•K) in the process pT = const. 

Solution. 72.  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

(b) In the process pT = const

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Thus  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

or  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Hence i = 3 (monoatomic)


Q. 73. Find the adiabatic exponent γ for a mixture consisting of v1 moles of a monatomic gas and v2 moles of gas of rigid diatomic molecules. 

Solution. 73.  Obviously

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

(Since a monoatomic gas has  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev and a diatomic gas has  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev [The diatomic molecule is rigid so no vibration])

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Hence  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 74. A thermally insulated vessel with gaseous nitrogen at a temperature t = 27°C moves with velocity v = 100 m/s. How much (in per cent) and in what way will the gas pressure change on a sudden stoppage of the vessel?

Solution. 74. The internal energy of the molecules are

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

where  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev velocity of the vessel, N = number of molecules, each of mass m. When the vessel is stopped, internal energy becomes  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

So there is an increase in internal energy of  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev This will give rise to a rise in temperature of 

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

there being no flow of heat This change of temperature will lead to an excess pressure

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

and finally  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

where M = molecular weight of N2 , i = number of degrees of freedom of N2


Q. 75. Calculate at the temperature t = 17°C:
 (a) the root mean square velocity and the mean kinetic energy of an oxygen molecule in the process of translational motion;
 (b) the root mean square velocity of a water droplet of diameter d = 0.10 μm suspended in the air. 

Solution. 75. (a) From the equipartition theorem

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

(b) In equilibrium the mean kinetic energy of the droplet will be equal to that of a molecule.

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 76. A gas consisting of rigid diatomic molecules is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of the molecules η = 1.50 times? 

Solution. 76.  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

The gas must be expanded  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 77. The mass m = 15 g of nitrogen is enclosed in a vessel at a temperature T = 300 K. What amount of heat has to be transferred to the gas to increase the root mean square velocity of its molecules η = 2.0 times?

Solution. 77. Here  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

m = mass of the gas, M = molecular weight If vrms increases η times, the temperature will have increased η2 times. This will require (neglecting expansion of the vessels) a heat flow of amount

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 78. The temperature of a gas consisting of rigid diatomic molecules is T = 300 K. Calculate the angular root mean square velocity of a rotating molecule if its moment of inertia is equal to I = 2.1.10-39  g• cm2.

Solution. 78. The root mean square angular velocity is given by

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

or  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 79. A gas consisting of rigid diatomic molecules was initially under standard conditions. Then the gas was compressed adiabatically η = 5.0 times. Find the mean kinetic energy of a rotating molecule in the final state.

Solution. 79. Under compression, the temperature will rise

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

or, Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 80. How will the rate of collisions of rigid diatomic molecules against the vessel's wall change, if the gas is expanded adiabatically η times? 

Solution. 80. Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Now,    Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

(When the gas is expanded η times, n decreases by a factor η). Also

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

i.e. collisions decrease by a factor  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 81. The volume of gas consisting of rigid diatomic molecules was increased η = 2.0 times in a polytropic process with the molar heat capacity C = R. How many times will the rate of collisions of molecules against a vessel's wall be reduced as a result of this process?

Solution. 81. In a polytropic process pVn = constant., where n is called the polytropic index. For this process

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Then  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Now  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

or,  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Now   Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev


Q. 82. A gas consisting of rigid diatomic molecules was expanded in a polytropic process so that the rate of collisions of the molecules against the vessel's wall did not change. Find the molar heat capacity of the gas in this process. 

Solution. 82. If a is the polytropic index then

Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Now   Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Hence   Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

Then  Kinetic Theory Of Gases (Part - 1) - Heat, Irodov JEE Notes | EduRev

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