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Q. 83. Calculate the most probable, the mean, and the root mean square velocities of a molecule of a gas whose density under standard atmospheric pressure is equal to p = 1.00 g/1. 

Solution. 83. 

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 84. Find the fraction of gas molecules whose velocities differ by less than δη = 1.00% from the value of
 (a) the most probable velocity;
 (b) the root mean square velocity. 

Solution. 84. (a) The formula is

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 85. Determine the gas temperature at which
 (a) the root mean square velocity of hydrogen molecules exceeds their most probable velocity by Δv = 400 m/s;
 (b) the velocity distribution function F (v) for the oxygen molecules will have the maximum value at the velocity v = 420 m/s. 

Solution. 85.  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 86. In the case of gaseous nitrogen find:
 (a) the temperature at which the velocities of the molecules v1 = 300 m/s and v2 = 600 m/s are associated with equal values of the Maxwell distribution function F (v);
 (b) the velocity of the molecules v at which the value of the Maxwell distribution function F (v) for the temperature T0  will be the same as that for the temperature η times higher

Solution. 86. (a) We have,

Irodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEE

So  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

(b)    Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Thus   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Thus   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 87. At what temperature of a nitrogen and oxygen mixture do the most probable velocities of nitrogen and oxygen molecules differ by Δv = 30 m/s? 

Solution. 87.  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 88. The temperature of a hydrogen and helium mixture is T = 300 K. At what value of the molecular velocity v will the Maxwell distribution function F (v) yield the same magnitude for both gases? 

Solution. 88.    

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE  Putting the values we get v = 1.60 km/s


Q. 89. At what temperature of a gas will the number of molecules, whose velocities fall within the given interval from v to v + dv, be the greatest? The mass of each molecule is equal to m. 

Solution. 89.

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

For a given range v to v + dv (i.e. given v and dv ) this is maximum when

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

or,    Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 90. Find the fraction of molecules whose velocity projections on the x axis fall within the interval from vx to vx + dvx,  while the moduli of perpendicular velocity components fall within the interval from v to v + dv. The mass of each molecule is m, and the temperature is T.

Solution. 90.  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Thus   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 91. Using the Maxwell distribution function, calculate the mean velocity projection Irodov Solutions: Kinetic Theory of Gases - 2 - JEE and the mean value of the modulus of this projection Irodov Solutions: Kinetic Theory of Gases - 2 - JEE if the mass of each molecule is equal to m and the gas temperature is T.

Solution. 91. Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 92. From the Maxwell distribution function find Irodov Solutions: Kinetic Theory of Gases - 2 - JEE the mean value of the squared vx projection of the molecular velocity in a gas at a temperature T. The mass of each molecule is equal to m. 

Solution. 92.   

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 93. Making use of the Maxwell distribution function, calculate the number v of gas molecules reaching a unit area of a wall per unit time, if the concentration of molecules is equal to n, the temperature to T, and the mass of each molecule is m. 

Solution. 93. Here vdA = No. of molecules hitting an area dA of the wall per second

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

or,   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 94. Using the Maxwell distribution function, determine the pressure exerted by gas on a wall, if the gas temperature is T and the concentration of molecules is n. 

Solution. 94.    Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

be the number of molecules per unit volume with x component of velocity in the range  vx to vx + dvx

Then    Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE
Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 95. Making use of the Maxwell distribution function, find Irodov Solutions: Kinetic Theory of Gases - 2 - JEE, the mean value of the reciprocal of the velocity of molecules in an ideal gas at a temperature T, if the mass of each molecule is equal to m. Compare the value obtained with the reciprocal of the mean velocity.

Solution. 95. 

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 96. A gas consists of molecules of mass m and is at a temperature T. Making use of the Maxwell velocity distribution function, find the corresponding distribution of the molecules over the kinetic energies ε. Determine the most probable value of the kinetic energy εp. Does εp correspond to the most probable velocity? 

Solution. 96. Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

or,  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Now,   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

or,  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

i.e.   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

The most probable kinetic energy is given from

Irodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEE

The corresponding velocity is   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 97. What fraction of monatomic molecules of a gas in a thermal equilibrium possesses kinetic energies differing from the mean value by δη = 1.0 % and less? 

Solution. 97. The mean kinetic energy is

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Thus

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 98. What fraction of molecules in a gas at a temperature T has the kinetic energy of translational motion exceeding ε0 if ε0  ≫ ≫kT? 

Solution. 98.  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

(In evaluating the integral, we have taken out  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE since the integral is dominated by the lower limit.)


Q. 99. The velocity distribution of molecules in a beam coming out of a hole in a vessel is described by the function F (v)= A V3e-mv2/2hT, where T is the temperature of the gas in the vessel. Find the most probable values of 
 (a) the velocity of the molecules in the beam; compare the result obtained with the most probable velocity of the molecules in the vessel;
 (b) the kinetic energy of the molecules in the beam. 

Solution. 99. Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

For the most probable value of the velocity

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

So,    Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

This should becom pared with the value  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE for the Maxwellian distribution.

(b) In terms of eneigy, Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

From this the probable eneigy comes out as follows : F' (ε) = 0 implies

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 100. An ideal gas consisting of molecules of mass m with concentration n has a temperature T. Using the Maxwell distribution function, find the number of molecules reaching a unit area of a wall at the angles between θ and θ + dθ to its normal per unit time. 

Solution. 100. The number of molecules reaching a unit area of wall at angle between θ and θ + dθ to its normal per unit time is

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 101. From the conditions of the foregoing problem find the number of molecules reaching a unit area of a wall with the velocities in the interval from v to v + dv per unit time.

Solution. 101. Similarly the number of molecules reaching the wall per unit area of the wall with velocities in the interval v to v + dv oer unit time is

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 102. Find the force exerted on a particle by a uniform field if the concentrations of these particles at two levels separated by the distance Δh = 3.0 cm (along the field) differ by η = 2.0 times. The temperature of the system is equal to T = 280 K. 

Solution. 102. If the force exerted is F then the law of variation of concentration with height reads

Irodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 103. When examining the suspended gamboge droplets under a microscope, their average numbers in the layers separated by the distance h = 40 urrn were found to differ by η = 2.0 times. The environmental temperature is equal to T = 290 K. The diameter of the droplets is d = 0.40 μm, and their density exceeds that of the surrounding fluid by Δp = 0.20 g/cm3. Find Avogadro's number from these data. 

Solution. 103. Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 104. Suppose that η0  is the ratio of the molecular concentration of hydrogen to that of nitrogen at the Earth's surface, while η  is the corresponding ratio at the height h = 3000 m. Find the ratio η/η0  at the temperature T = 280 K, assuming that the temperature and the free fall acceleration are independent of the height. 

Solution. 104. Irodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 105. A tall vertical vessel contains a gas composed of two kinds of molecules of masses m1 and m2, with m2 > m1. The concentrations of these molecules at the bottom of the vessel are equal to n1 and n2 respectively, with n2 > n1. Assuming the temperature T and the free-fall acceleration g to be independent of the height, find the height at which the concentrations of these kinds of molecules are equal. 

Solution. 105. Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

They are equal at a height h where  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 106. A very tall vertical cylinder contains carbon dioxide at a certain temperature T. Assuming the gravitational field to be uniform, find how the gas pressure on the bottom of the vessel will change when the gas temperature increases η times. 

Solution. 106. At a temperature T the concentration n (z) varies with height according to

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

This means that the cylinder contains  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

particles per unit area of the base. Clearly this cannot change. Thus n0 kT = p0 = pressure at the bottom of the cylinder must not change with change of temperature. 


Q. 107. A very tall vertical cylinder contains a gas at a temperature T'. Assuming the gravitational field to be uniform, find the mean value of the potential energy of the gas molecules. Does this value depend on whether the gas consists of one kind of molecules or of several kinds? 

Solution. 107.

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

When there are many kinds of molecules, this formula holds for each kind and the average energy

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

where fi α fractional concentration of each kind at the ground level.


Q. 108. A horizontal tube of length l = 100 cm closed from both ends is displaced lengthwise with a constant acceleration w. The tube contains argon at a temperature T = 330 K. At what value of w will the argon concentrations at the tube's ends differ by η = 1.0%? 

Solution. 108. The constant acceleration is equivalent to a pseudo force wherein a concentration gradient is set up. Then

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

or  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 109. Find the mass of a mole of colloid particles if during their centrifuging with an angular velocity ω about a vertical axis the concentration of the particles at the distance r2 from the rotation axis is η times greater than that at the distance r1 ( in the same horizontal plane). The densities of the particles and the solvent are equal to p and to P0 respectively. 

Solution. 109. In a centrifuge rotating with angular velocity co about an axis, there is a centrifugal acceleration ω2 r where r is the radial distance from the axis. In a fluid if there are suspended colloidal particles they experience an additional force. If m is the mass of each particle then its volume  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE and the excess force on this particle is  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE outward corresponding to a potential energy  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

This gives rise to a concentration variation 

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Thus  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

where   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE m is the molecular weight

Thus  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE


Q. 110. A horizontal tube with closed ends is rotated with a constant angular velocity ω about a vertical axis passing through one of its ends. The tube contains carbon dioxide at a temperature T = 300 K. The length of the tube is l = 100 cm. Find the value ω at which the ratio of molecular concentrations at the opposite ends of the tube is equal to η = 2.0. 

Solution. 110. The potential energy associated with each molecule is :  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

and there is a concentration variation  

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Thus   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Using  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

we get ω = 280 radians per second.


Q. 111. The potential energy of gas molecules in a certain central field depends on the distance r from the field's centre as U (r) = ar2, where a is a positive constant. The gas temperature is T, the concentration of molecules at the centre of the field is n0. Find:
 (a) the number of molecules located at the distances between r and r + dr from the centre of the field;
 (b) the most probable distance separating the molecules from the centre of the field;
 (c) the fraction of molecules located in the spherical layer between r and r + dr;
 (d) how many times the concentration of molecules in the centre of the field will change if the temperature decreases η times.

Solution. 111.  Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

(a) The number of molecules located at the distance between r and r + dr is

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

(b)  Irodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEE

(c) The fraction of molecules lying between r and r + dr is 

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

Thus   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

(d)     Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

So   Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

When T decreases η times (n0) = n0 will increase η3/2 times


Q. 112.  From the conditions of the foregoing problem find:
 (a) the number of molecules whose potential energy lies within the interval from U to U + dU;
 (b) the most probable value of the potential energy of a molecule; compare this value with the potential energy of a molecule located at its most probable distance from the centre of the field. 

Solution. 112.  Irodov Solutions: Kinetic Theory of Gases - 2 - JEEIrodov Solutions: Kinetic Theory of Gases - 2 - JEE

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

The most probable value of U is given by

Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

From Q.111 (b), the potential energy at the most probable distance is kT.

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FAQs on Irodov Solutions: Kinetic Theory of Gases - 2 - JEE

1. What is the kinetic theory of gases?
Ans. The kinetic theory of gases is a scientific model that explains the behavior of gases based on the motion of their particles. It states that gases are composed of tiny particles, such as molecules or atoms, that are in constant random motion. This motion is responsible for the various properties of gases, such as pressure, temperature, and volume.
2. How does the kinetic theory of gases explain pressure?
Ans. According to the kinetic theory of gases, the pressure exerted by a gas is a result of the constant collisions between the gas particles and the walls of the container. When gas particles collide with the container walls, they exert a force, which collectively leads to the overall pressure. The more frequent and energetic the collisions, the higher the pressure.
3. Can the kinetic theory of gases explain the relationship between temperature and volume?
Ans. Yes, the kinetic theory of gases can explain the relationship between temperature and volume, known as Charles's Law. According to this law, when the temperature of a gas is increased, its volume also increases if the pressure remains constant. This is because an increase in temperature leads to an increase in the average kinetic energy of the gas particles, causing them to move faster and occupy a larger space.
4. How does the kinetic theory of gases explain the concept of absolute zero?
Ans. The kinetic theory of gases states that at absolute zero temperature (-273.15°C or 0 Kelvin), all molecular motion ceases. This means that the gas particles have zero kinetic energy and do not move. Therefore, the concept of absolute zero is explained by the complete absence of molecular motion, as predicted by the kinetic theory of gases.
5. Can the kinetic theory of gases explain the diffusion of gases?
Ans. Yes, the kinetic theory of gases provides an explanation for the diffusion of gases. Diffusion refers to the gradual mixing of gases due to the random motion of their particles. According to the kinetic theory, gas particles move randomly and collide with each other. These collisions cause the gas particles to spread out and mix with other particles, leading to diffusion. The rate of diffusion depends on factors such as the size of the particles and the temperature.
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