Kinetic Theory

This is a long answer type question as classified in NCERT Exemplar

Consider the diagram

let n =number of molecules per unit volume

V_rms= rms speed of gas molecule

When block is moving with speed v_o, relative speed of molecules w.r.t front face =v+v_o

Coming head on, momentum transferred to block per collision =2m (v+v_o)

Number of collisions in time $?t$ = $\frac{1}{2}$ (v+v_o)n $?t$ A  where A is the area of cross section.

So momentum transferred in time $?t$ =m (v+v_o)²nA $?t$   this is from front surface

Similarly momentum transferred in time $?t$ = m (v-v_o)²nA $?t$ ) this is from back surface

Drag force = mnA (v+v_o)²- (v-v_o)²)

= mnA (4w_o)=4mnAvv_o

= 4 $\rho A$ vv_o

So $\rho$ =mn/V=M/V

If v = velocity along x axis

Then we can write KE= ½ mv²=1/2 K_BT

This is a long answer type question as classified in NCERT Exemplar

(a) The moon has small gravitational force and hence, the escape velocity is small .

As the moon in the proximity of the earth as seen from the sun, the moonhas the same amount of heat per unit area as that of the earth.

The air molecules have large range of speeds . even though the rms speed of the air molcules is smaller than the escape velocity on the moon, a significant number of molecules have speed grater than the escape velocity.

(b) As the molecules move higher their potential energy increases and hence kinetic energy decreases and hence temperature reduces. At greater height more volume is available and gas expands and hence some cooling takes place.

13.13 According to law of atmospheres, we have

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT] ….(i)

where $n_1$ is the number of density at height  $h_1$and $n_2$ is the number of density at height $h_2$

mg is the weight of the particle suspended in the gas column

Density of the medium = $?$

Density of the suspended particle = ${?}^{\text{'}}$

Mass of one suspended particle = m'

Mass of medium displaced = m

Volume of the suspended particle = V

According to Archimedes's principle for a particle suspended in a liquid column, the effective weight of the suspended particle is given as:

Weight of the medium displaced – weight of the suspended particle = mg – m'g

= mg- V ${?}^{\text{'}}g$ = mg – ($\frac{m}{?}$ ) ${?}^{\text{'}}g$

= mg(1- $\frac{{?}^{\text{'}}}{?}$ ) ……(ii)

Gas constant R = $k_B$ N

$k_B=\frac{R}{N}$ …(iii)

Substituting from (ii) and (iii) in (i), we get

n₂ = n₁ exp [ -mg (h₂ – h₁)/ k_BT]

= n1 exp [$\left[-mg\left(-\frac{{?}^{\text{'}}}{?}\right)(h2\mathrm{}–\mathrm{}h1)\frac{N}{RT}\right]$]

= n1 exp[$\left[-mg\left(?-{?}^{\text{'}}\right)(h2\mathrm{}–\mathrm{}h1)\frac{N}{RT?}\right]$]

13.11 Length of the narrow bore, L = 1 m = 100 cm

Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed end, $l_o$  = 15 cm

Since the bore is held vertically in air with the open end at the bottom, the mercury length that occupies the air space is 100 – (76 + 15) = 9 cm

Hence, total length of the air column = 15 + 9 = 24 cm

Let h cm of mercury flow out as a result of atmospheric pressure.

Length of the air column in the bore = 24 + h cm

Length of the mercury column = 76 – h cm

Initial pressure,  $P_1$ = 76 cm of mercury

Initial volume, $V_1$= 15 ${cm}^3$

Final pressure, $P_2$ = 76 – (76 – h) = h cm of mercury

Final volume, $V_2$ = (24 + h) ${cm}^3$

Since temperature remained constant throughout the process,

 $P_1{V}_1$=$P_2{V}_2$ or 76 $*15$ = h $*(24+h)$$$

Or, $h^2$ + 24h - 76 $*15$ =0

h = $\frac{-24\pm \sqrt{{24}^2+4*76*15}}{2}$=$\frac{-24\pm 71.67}{2}$

So h = 23.84 cm or -47.84 cm

Since height cannot be negative, so h = 23.84 cm

23.84 cm of mercury will flow out from the bore and (76-23.84) 52.16 cm of mercury will remain in the bore. The length of the air column will be 24 + 23.84 = 47.84 cm

13.10 Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2 $*1.013*{10}^5$ Pa

Temperature inside the cylinder, T = 17 $?=290K$

Radius of nitrogen molecule, r = 1.0 Å = 1 $*{10}^{-10}$ m

Diameter of nitrogen molecule, d = 2 $*{10}^{-10}$ m

Molecular mass of nitrogen molecule, M = 28 u = 28 g (assume) = 28 $*{10}^{-3}$ kg

The root means square speed of nitrogen is given by the relation

R is the universal gas constant = 8.314 J/mole/K

Hence 

The mean free path $\stackrel{}{l}$ is given by

$l ?=\frac{kT}{?2?d^{2}P}$ where k = Boltzmann constant = 1.38$*{10}^{-23}$ kg-$$$m^2{s}^{-2}{K}^{-1}$ 

$\stackrel{?}{l}=\frac{1.38\mathrm{}*{10}^{-23}*290}{?2?{(2\mathrm{}*{10}^{-10})}^2*2\mathrm{}*1.013*{10}^5}$= 1.11$*{10}^{-7}$  m

Collision frequency = $\frac{V_{rms}}{\stackrel{?}{l}}$ = $\frac{508.26}{1.11\mathrm{}*{10}^{-7}}$= 4.57 ${10}^9$ /s

Collision time, T = $\frac{d}{V_{rms}}$  =$\frac{2\mathrm{}*{10}^{-10}\mathrm{}}{508.26}$S= 3.93 $*{10}^{-13}$$\frac{\mathrm{}}{}$ s

Time taken between successive collision T' = $\frac{\stackrel{?}{l}}{V_{rms}}$ = $\frac{1.11\mathrm{}*{10}^{-7}}{508.26}$s = 2.18$*{10}^{-10}$
 s$\frac{\mathrm{T}\mathrm{\text{'}}}{T}$ =$\frac{2.18\mathrm{}*{10}^{-10}}{3.93\mathrm{}*{10}^{-13}}$ = 555.71

Hence, the time taken between successive collisions is 556 times the time taken for a collision.