Physics

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After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.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 $\frac{PV}{T}$