Kinetic Theory

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

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

13.3 (a) The dotted plot in the graph signifies the ideal behaviour of the gas, i.e. $\frac{PV}{T}$ = $?$ R, (where $?$ is the number of moles and R is the universal gas constant) is a constant quantity. It is not dependent on the pressure of the gas. The curve of the gas at temperature $T_1$ is closer to the dotted plot than the curve of the gas at temperature $T_{2.}$

(b) A real gas approaches the behaviour of an ideal gas when its temperature increases. Therefore $T_1$ > $T_2$ is true for the given plot.

(c) The value of the ratio $\frac{PV}{T}$ , where the two curves meet, is $?$ R. This is because the ideal gas equation is given as $\frac{PV}{T}$ = $?$ R. From the given data, we have

Molecular mass of oxygen = 32 g

Mass of oxygen = 1 $*{10}^{-3}$ = 1 g

R = 8.314 J/mole/K

Then $\frac{PV}{T}$ = $\frac{1}{32}*8.314$ = 0.26 J/k

Hence, the value of the ratio $\frac{PV}{T}$ , where the curves meet on the y-axis, is 0.26 J/K.

(d) If we obtain similar plots for 1.00 $*{10}^{-3}$ kg of hydrogen, then we will not get the same values of $\frac{PV}{T}$ at the point where the curves meet on the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from the oxygen (32.0 u). We have:

$\frac{PV}{T}=0.26\mathrm{}\mathrm{J}/\mathrm{K}$ .

Given, molecular mass (M) of $H_2$ = 2.02 u

$\frac{PV}{T}$ = $?$ R at constant temperature, where $?$ = $\frac{m}{M}$

M = mass of $H_2$

Then, m = $\frac{PV}{T}*\frac{M}{R}$ = $\frac{0.26*2.02}{8.31}$ = 6.3 $*{10}^{-2}$ g = 6.3 $*{10}^{-5}$ kg

Hence, 6.3 $*{10}^{-5}$ kg of $H_2$ will yield the same value of $\frac{PV}{T}$