Class 11th

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

4.21:

(a) Velocity , $\stackrel{?}{v}$ = 10.0 ? m/s

Acceleration, $\stackrel{?}{a}$ = (8.0 ? + 2.0 ?) m s-2

We know $\stackrel{?}{a}$ = $\frac{d\stackrel{?}{v}}{dt}$ = 8.0 ? + 2.0 ?

$\stackrel{?}{dv}$ = (8.0 ? + 2.0 ?)dt

Integrating both sides we get $\stackrel{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\stackrel{?}{u}$ ,

Where, $\stackrel{?}{u}=$ velocity vector of the particle at t =0

$\stackrel{?}{v}=$ velocity vector of the particle at time t

But $\stackrel{?}{v}$ = $\frac{\underset{dr}{\to }}{dt}$

$\stackrel{?}{dr}$ = $\stackrel{?}{v}$ dt

= (8.0t ? + 2.0t ? + $\stackrel{?}{u}$ )dt

Integrating both sides with the condition at t = 0, r =0 and at t =t, r = r

$\stackrel{?}{r}=$ $\stackrel{?}{u}$ t + ½ $*$ 8.0 t2 ? + ½ $*$ 2.0 $*$ t2 ? = $\stackrel{?}{u}$ t + 4.0 t2 ? + t2 ?

Substituting the value of $\stackrel{?}{u}$ , we get

$\stackrel{?}{r}=$ ( 10.0 ?)t + 4.0 t2 ? + t2 ? . This equation can be expressed as

x ? + y ? = 4.0 t2 ? + ( 10.0t + t2) ?

Since the motion of the particle is confined to the x-y plane, on equating the coefficients of ? and ?, we get

x = 4.0 t2 and y = 10.0t + t2

t = $\sqrt{x/4}$

(a) When x = 16m, t = 2 s, y = 24m

(b) Velocity of the particle

$\stackrel{?}{v}$ (t) = 8.0t ? + 2.0t ? + $\stackrel{?}{u}$

At t = 2 s,

$\stackrel{?}{v}$ (t) = 8.0 $*$ 2 ? + 2.0 $*2$ ? + 10 $*$ ? = 16 ? + 14 ?

The magnitude of $\stackrel{?}{v}$ (t) is given by

$\left|\mathrm{}\stackrel{?}{v}\right|$ = ( 162 + 142)1/2 = 21.26 m/s

13.5 Volume of the air bubble $,$ $V_1$ = 1.0 ${cm}^3$ = 1 $*{10}^{-6}$ $m^3$

Height achieved by bubble, d = 40 m

Temperature at a depth of 40 m, $T_1$ = 12 $?$ = 285 K

Temperature at the surface of the lake, $T_2$ = 35 $?$ = 308 K

The pressure at the surface of the lake, $P_2$ = 1 atm = 1.013 $*{10}^5$ Pa

The pressure at 40m depth: $P_1$ = 1 atm + d $?g$

Where $?$ is the density of water = ${10}^3$ kg/ $m^3$

G = acceleration due to gravity = 9.8 m/ $s^2$

Hence $P_1$ = 1.013 $*{10}^5+(40*{10}^3*9.8)$ = 493300 Pa

From the relation $\frac{P_1{V}_1}{T_1}$ = $\frac{P_2{V}_2}{T_2}$ , where $V_2$ is the volume of bubble when it reaches the surface, we get, $V_2$ = $\frac{P_1{V}_1{T}_2}{T_1{P}_2}$ = $\frac{493300*1*{10}^{-6}*308}{285*1.013\mathrm{}*{10}^5}$ $=$ 5.263 $*$ ${10}^{-6}{m}^3=5.263{cm}^3$

13.4 Volume of oxygen, $V_1$ = 30 litres = 30 $*{10}^{-3}{m}^3$

Gauge pressure, $P_1$ = 15 atm = 15 $*1.013*{10}^5$ Pa

Temperature, $T_1$ = 27 $?$ = 300 K

Universal gas constant, R = 8.314 J/mole/K

Let the initial number of moles of oxygen gas cylinder be $n_1$

From the gas equation, we get $P_1{V}_1=n_1\mathrm{R}{T}_1$

$n_1=\frac{P_1{V}_1}{\mathrm{R}{T}_1}$ = $\frac{15\mathrm{}*1.013*{10}^5*30*{10}^{-3}}{8.314*300}$ = 18.276

But $n_1$ = $\frac{m_1}{M}$ , where $m_1$ = initial mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_1$ = $n_1$ M = 18.276 $*32=584.84$ g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.

Hence, Volume, $V_2$ = 30 litres = 30 $*{10}^{-3}{m}^3$

Gauge pressure, $P_2$ = 11 atm = 11 $*1.013*{10}^5$ Pa

Temperature, $T_2$ = 17 $?$ = 290 K

Let the final number of moles of oxygen left in the cylinder be $n_2$

From the gas equation, we get $P_2{V}_2=n_2\mathrm{R}{T}_2$

$n_2=\frac{P_2{V}_2}{\mathrm{R}2}$ = $\frac{11\mathrm{}*1.013*{10}^5*30*{10}^{-3}}{8.314*290}$ = 13.864

But $n_2$ = $\frac{m_2}{M}$ , where $m_2$ = remaining mass of oxygen

M = Molecular mass of oxygen = 32 g

$m_2$ = $n_2$ M = 13.864 $*32=443.65$ g

So the mass of oxygen taken out = $m_1-m_2$ = 141.19 gm = 0.141 kg