Class 11th

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.

13.8  (a) According to Avogadro's law, the three vessels will contain an equal number of the respective molecules. This number is equal to Avogadro's number, N = 6.023 $*{10}^{23}$

(b) The root mean square speed ( $V_{rms})$ of a gas of mass m and temperature T is given by the relation $V_{rms}=\sqrt{\frac{3kT}{m}}$ . Where k is Boltzmann constant. For the given gases, k and T are constants. Hence $V_{rms}$ depends only on the mass of the atoms

Therefore, the root mean square speed of the molecules in the three cases is not the same. Among Neon, Chlorine and Uranium hexafluoride, the mass of the neon is the smallest, so Neon will have the highest root mean square speed among the given gases.

13.7 (i) At room temperature, T = 27 $?$ = 300 K

$k_B$ is Boltzmann constant = 1.38 $*{10}^{-23}$ $m^{2}kg{s}^{-2}{K}^{-1}$

Average thermal energy = $\frac{3}{2}{k}_{B}T$ = $\frac{3*1.38\mathrm{}*{10}^{-23}*300}{2}$ = 6.21 $*{10}^{-21}$ J

Hence, the average thermal energy of a helium atom at room temperature is 6.21 $*{10}^{-21}$ J

(ii) On the surface of the Sun, T = 6000 K

Hence average thermal energy = $\frac{3}{2}{k}_{B}T$ = $\frac{3*1.38\mathrm{}*{10}^{-23}*6000}{2}$ = 1.242 $*{10}^{-19}$ J

Hence, the average thermal energy of a helium atom on the surface of the Sun is 1.242 $*{10}^{-19}$ J

(iii) Inside the core of a star, T = ${10}^7$ K

Hence average thermal energy = $\frac{3}{2}{k}_{B}T$ = $\frac{3*1.38\mathrm{}*{10}^{-23}*{10}^7\mathrm{}}{2}$ = 2.07 $*{10}^{-16}$ J

Hence, the average thermal energy of a helium atom inside the core of a star is 2.07 $*{10}^{-16}$ J