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`color{blue}{star}` KINETIC THEORY OF AN IDEAL GAS
`color{blue}{star}` PRESSURE OF AN IDEAL GAS


`color{blue} ✍️` Kinetic theory of gases is based on the molecular picture of matter. A given amount of gas is a collection of a large number of molecules (typically of the order of Avogadro’s number) that are in incessant random motion.

`color{blue} ✍️` At ordinary pressure and temperature, the average distance between molecules is a factor of 10 or more than the typical size of a molecule `(2 Å)`.

`color{blue} ✍️`Thus the interaction between the molecules is negligible and we can assume that they move freely in straight lines according to Newton’s first law.

`color{blue} ✍️`However, occasionally, they come close to each other, experience intermolecular forces and their velocities change. These interactions are called collisions.

`color{blue} ✍️`The molecules collide incessantly against each other or with the walls and change their velocities. The collisions are considered to be elastic. We can derive an expression for the pressure of a gas based on the kinetic theory.

`color{blue} ✍️`We begin with the idea that molecules of a gas are in incessant random motion, colliding against one another and with the walls of the container.

`color{blue} ✍️`All collisions between molecules among themselves or between molecules and the walls are elastic. This implies that total kinetic energy is conserved. The total momentum is conserved as usual.

Pressure of an Ideal Gas

`color{blue} ✍️`Consider a gas enclosed in a cube of side l. Take the axes to be parallel to the sides of the cube, as shown in Fig. 13.4.

`color{blue} ✍️`A molecule with velocity `(v_x, v_y, v_z )` hits the planar wall parallel to `yz-` plane of area `A (= l^2)`. Since the collision is elastic, the molecule rebounds with the same velocity; its y and z components of velocity do not change in the collision but the x-component reverses sign.

`color{blue} ✍️`That is, the velocity after collision is `(-v_x, v_y, v_z )` . The change in momentum of the molecule is : `–mv_x – (mv_x) = – 2mv_x` . By the principle of conservation of momentum, the momentum imparted to the wall in the collision `= 2mv_x` .

`color{blue} ✍️`To calculate the force (and pressure) on the wall, we need to calculate momentum imparted to the wall per unit time. In a small time interval `Δt`, a molecule with x-component of velocity `v_x` will hit the wall if it is within the distance `v_x Δt` from the wall.

`color{blue} ✍️`That is, all molecules within the volume `Av_x Δt` only can hit the wall in time `Δt`. But, on the average, half of these are moving towards the wall and the other half away from the wall.

`color{blue} ✍️`Thus the number of molecules with velocity `(v_x, v_y, v_z )` hitting the wall in time `Δt` is `½A v_x Δt` n where n is the number of molecules per unit volume. The total momentum transferred to the wall by these molecules in time `Δt` is :

`color{blue} { Q = (2mv_x) ( 1/2 n Av_x Delta t)}`


`color{blue} ✍️`The force on the wall is the rate of momentum transfer Q/Δt and pressure is force per unit area :

`color{blue} { P = Q // (A Delta t ) = n m_x^2}`


`color{blue} ✍️`Actually, all molecules in a gas do not have the same velocity; there is a distribution in velocities. The above equation therefore, stands for pressure due to the group of molecules with speed `vx` in the x-direction and n stands for the number density of that group of molecules. The total pressure is obtained by summing over the contribution due to all groups:

`color{blue} { P = n m bar(v_x^2)}`


where `bar(v_x^2)` is the average of `v_x ^2` . Now the gas is isotropic, i.e. there is no preferred direction of velocity of the molecules in the vessel.

`color{blue} ✍️`Therefore, by symmetry,

`color{purple} {bar(v_x^2) = bar(v_y^2) = bar(v_z^2)}`

`color{blue} { = (1/3)} color{blue} { [bar(v_x^2) + bar(v_z^2)] = (1/3) bar(v^2)}`


`color{blue} ✍️`where `v` is the speed and `bar(v^2)` denotes the mean of the squared speed. Thus

`color{blue} {P = (1//3) nm bar(v^2)} `


`color{blue} ✍️`Some remarks on this derivation. First, though we choose the container to be a cube, the shape of the vessel really is immaterial. For a vessel of arbitrary shape, we can always choose a small infinitesimal (planar) area and carry through the steps above.

`color{blue} ✍️`Notice that both A and `Δt ` do not appear in the final result. By Pascal’s law, given in Ch. 10, pressure in one portion of the gas in equilibrium is the same as anywhere else. Second, we have ignored any collisions in the derivation.

`color{blue} ✍️`Though this assumption is difficult to justify rigorously, we can qualitatively see that it will not lead to erroneous results. The number of molecules hitting the wall in time `Δt` was found to be `½ n Av_x Δt`. Now the collisions are random and the gas is in a steady state.

`color{blue} ✍️`Thus, if a molecule with velocity `(v_x, v_y, v_z )` acquires a different velocity due to collision with some molecule, there will always be some other molecule with a different initial velocity which after a collision acquires the velocity `(v_x, v_y, v_z )`.

`color{blue} ✍️` If this were not so, the distribution of velocities would not remain steady. In any case we are finding `bar(v_x^2)` . Thus, on the whole, molecular collisions (if they are not too frequent and the time spent in a collision is negligible compared to time between collisions) will not affect the calculation above.

Kinetic Interpretation of Temperature

`color{blue} ✍️`Equation (13.14) can be written as

`color{blue} { PV = (1//3) n Vm bar (v^2)}` ...........(13.15a)

`color{blue} {PV = (2/3) N x ½ m bar(v^2)}` ...................(13.15b)

`color{blue} ✍️`where `N = nV` is the number of molecules in the sample.

`color{blue} ✍️`The quantity in the bracket is the average translational kinetic energy of the molecules in the gas. Since the internal energy `E` of an ideal gas is purely kinetic

`color{blue} {E = N × (1/2) m bar(v^2)}` ................(13.16)

`color{blue} ✍️`Equation (13.15) then gives :

`color{blue} {PV = (2/3) E}` ........... (13.17)

`color{blue} ✍️`We are now ready for a kinetic interpretation of temperature. Combining Eq. (13.17) with the ideal gas Eq. (13.3), we get

`color{blue} {E = (3/2) kB NT}` ................ (13.18)

or ` color{purple} {E//N = 1/2 m bar(v^2) = (3/2) k_B T}` ....................(13.19)

`color{blue} ✍️`i.e., the average kinetic energy of a molecule is proportional to the absolute temperature of the gas; it is independent of pressure, volume or the nature of the ideal gas.

`color{blue} ✍️`This is a fundamental result relating temperature, a macroscopic measurable parameter of a gas (a thermodynamic variable as it is called) to a molecular quantity, namely the average kinetic energy of a molecule. The two domains are connected by the Boltzmann constant.

`color{blue} ✍️`We note in passing that Eq. (13.18) tells us that internal energy of an ideal gas depends only on temperature, not on pressure or volume. With this interpretation of temperature, kinetic theory of an ideal gas is completely consistent with the ideal gas equation and the various gas laws based on it.

`color{blue} ✍️`For a mixture of non-reactive ideal gases, the total pressure gets contribution from each gasin the mixture. Equation (13.14) becomes

`color{blue} {P = (1/3) [ n_1m_1bar(v_1^2) + n_2m_2 bar(v_2^2) +.....]}` .........................(13.20)

`color{blue} ✍️`In equilibrium, the average kinetic energy of the molecules of different gases will be equal.

`color{blue} ✍️`That is,

`color{purple} {1/2 m_1 bar(v_1^2) = 1/2 m_2 bar(v_2^2) = (3/2)k_BT}`

`color{blue} ✍️`so that

`color{blue} {P = ( n_1 + n_2 + .... ) k_BT}`


`color{blue} ✍️`which is Dalton’s law of partial pressures.

`color{blue} ✍️`From Eq. (13.19), we can get an idea of the typical speed of molecules in a gas. At a temperature `T = 300 K`, the mean square speed of a molecule in nitrogen gas is :

`color{purple} { m = M_(N_2)/N_A = 28/(6.02 xx 10^26) = 4.65 xx 10^-26 kg }`

`color{purple} { bar(v^2) = 3 k_B T//m = (516)^2 m^2 s^-2}`

`color{blue} ✍️`The square root of `bar(v^2)` is known as root mean square `(r m s)` speed and is denoted by `v_(rms)`,

`color{blue} ✍️`( We can also write `bar(v^2)` as `< v^2 >`.)

`color{purple} {v_(rms) = 516 m s^-1}`

`color{blue} ✍️`The speed is of the order of the speed of sound in air. It follows from Eq. (13.19) that at the same temperature, lighter molecules have greater rms speed.
Q 3280501417

A flask contains argon and chlorine in the ratio of 2:1 by mass. The temperature of the mixture is `27 °C`. Obtain the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed `v_(rms)` of the molecules of the two gases. Atomic mass of argon = 39.9 u; Molecular mass of chlorine = 70.9 u.
Class 11 Chapter 13 Example 5

The important point to remember is that the average kinetic energy (per molecule) of any (ideal) gas (be it monatomic like argon, diatomic like chlorine or polyatomic) is always equal to `(3/2) k_BT`. It depends only on temperature, and is independent of the nature of the gas.

(i) Since argon and chlorine both have the same temperature in the flask, the ratio of average kinetic energy (per molecule) of the two gases is 1:1.

(ii) Now `1/2 m v_(rms)^ 2 =` average kinetic energy per molecule `= (3/2) ) k_BT` where m is the mass of a molecule of the gas. Therefore,

`((V_(rms)^2)_(Ar))/((V_(rms)^2)_(Cl) ) = ((m)_(Cl))/((m)_(Ar)) = ((M)_(Cl))/((M)_(Ar)) = 70.9/39.9 = 1..77`

where M denotes the molecular mass of the gas. (For argon, a molecule is just an atom of argon.)

Taking square root of both sides,

`(V_(rms))_(Ar)/(V_(rms))_(Cl) = 1.33`

You should note that the composition of the mixture by mass is quite irrelevant to the above calculation. Any other proportion by mass of argon and chlorine would give the same answers to (i) and (ii), provided the temperature remains unaltered.
Q 3210501419

Uranium has two isotopes of masses 235 and 238 units. If both are present in Uranium hexafluoride gas which would have the larger average speed ? If atomic mass of fluorine is 19 units, estimate the percentage difference in speeds at any temperature.
Class 11 Chapter 13 Example 6

At a fixed temperature the average energy `= 1/2 m < v^2 >` is constant. So smaller the mass of the molecule, faster will be the speed. The ratio of speeds is inversely proportional to the square root of the ratio of the masses. The masses are 349 and 352 units. So

`v_349 / v_352 = ( 352/ 349)^(1/2) = 1.0044` .

Hence difference `(DeltaV)/V = 0.44 %`

[`text()^235U` is the isotope needed for nuclear fission. To separate it from the more abundant isotope `text()^238U`, the mixture is surrounded by a porous cylinder. The porous cylinder must be thick and narrow, so that the molecule wanders through individually, colliding with the walls of the long pore. The faster molecule will leak out more than the slower one and so there is more of the lighter molecule (enrichment) outside the porous cylinder (Fig. 13.5). The method is not very efficient and has to be repeated several times for sufficient enrichment.].
Q 3210601510

(a) When a molecule (or an elastic ball) hits a ( massive) wall, it rebounds with the same speed. When a ball hits a massive bat held firmly, the same thing happens. However, when the bat is moving towards the ball, the ball rebounds with a different speed. Does the ball move faster or slower? (Ch.6 will refresh your memory on elastic collisions.)

(b) When gas in a cylinder is compressed by pushing in a piston, its temperature rises. Guess at an explanation of this in terms of kinetic theory using (a) above

(c) What happens when a compressed gas pushes a piston out and expands. What would you observe ?

(d) Sachin Tendulkar uses a heavy cricket bat while playing. Does it help him in anyway ?
Class 11 Chapter 13 Example 7

(a) Let the speed of the ball be u relative to the wicket behind the bat. If the bat is moving towards the ball with a speed V relative to the
wicket, then the relative speed of the ball to bat is `V + u ` towards the bat. When the ball rebounds (after hitting the massive bat) its speed, relative to bat, is V + u moving away from the bat. So relative to the wicket the speed of the rebounding ball is ` V + (V + u) = 2V + u`, moving away from the wicket. So the ball speeds up after the collision with the bat. The rebound speed will be less than u if the bat is not massive. For a molecule this would imply an increase in temperature.

You should be able to answer (b) (c) and (d) based on the answer to (a).

(Hint: Note the correspondence, piston`->` bat, cylinder `->` wicket, molecule `->` ball.)