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Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Mean distance between molecules

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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.

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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. Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

so,   Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

(a) For monoatomic gases i = 3

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

(b) For rigid diatomic molecules i = 5

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET + (3N - 6) kT (vibrational) 

= (3N - 3) kT per molecule

For linear molecules, mean energy  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET+ kT (rotational) + (3 N - 5) kT (vibrational)

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Translational energy is a fraction  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEETIrodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Thus, Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

So,  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

(c) For noncollinear N- atomic molecules 

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET  for linear molecules

Thus   Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

For monoatomic gases,  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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)

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Also   Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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.  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

(b) In the process pT = const

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Thus  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

or  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

(Since a monoatomic gas has  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET and a diatomic gas has  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET [The diatomic molecule is rigid so no vibration])

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Hence  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

where  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET velocity of the vessel, N = number of molecules, each of mass m. When the vessel is stopped, internal energy becomes  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

So there is an increase in internal energy of  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET This will give rise to a rise in temperature of 

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

and finally  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEETIrodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEETIrodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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.  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

The gas must be expanded  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

or  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

or, Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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. Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Now,    Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEETIrodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

i.e. collisions decrease by a factor  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Then  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Now  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

or,  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Now   Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET


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

Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Now   Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Hence   Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

Then  Irodov Solutions: Kinetic Theory of Gases - 1 - Notes | Study Physics Class 11 - NEET

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