Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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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. 

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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.  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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,

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

So  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b)    Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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.  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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.    

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev  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.

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,    Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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.  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


Q. 91. Using the Maxwell distribution function, calculate the mean velocity projection Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev and the mean value of the modulus of this projection Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev if the mass of each molecule is equal to m and the gas temperature is T.

Solution. 91. Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


Q. 92. From the Maxwell distribution function find Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev 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.   

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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.    Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

Then    Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev
Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


Q. 95. Making use of the Maxwell distribution function, find Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev, 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. 

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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. Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Now,   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

or,  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

i.e.   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

The most probable kinetic energy is given from

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

The corresponding velocity is   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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.  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

(In evaluating the integral, we have taken out  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev 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. Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

For the most probable value of the velocity

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

So,    Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

This should becom pared with the value  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev for the Maxwellian distribution.

(b) In terms of eneigy, Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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. Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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. Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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. Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

They are equal at a height h where  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

This means that the cylinder contains  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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.

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

or  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev and the excess force on this particle is  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev outward corresponding to a potential energy  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

This gives rise to a concentration variation 

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

where   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev m is the molecular weight

Thus  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev


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 :  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

and there is a concentration variation  

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Using  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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.  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

(b)  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Thus   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

(d)     Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

So   Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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.  Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRevKinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

The most probable value of U is given by

Kinetic Theory Of Gases (Part - 2) - Heat, Irodov JEE Notes | EduRev

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

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