physics ncert solutions class 11th

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After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces, but volume remained unchanged.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

4.10

Since the motorist taking the 60° turn after every 500m, It will form a perfect hexagon of side 500m after 6 turns.

Suppose the motorist starts from point A. From A, turns 60° to reach B (first turn), then to C and so on.

1.    At  third turn, The magnitude of displacement = DG + GA = 500m + 500m = 1000m         Total path length = AB + BC + CD = 500m +500m +500m = 1500 m

2.    At sixth turn, When the motorist takes the 6th turn, he is at starting point A. Therefore,

The magnitude of displacement = 0Total path length = AB + BC + CD + DE + EF + FA = 500m + 500m +500m +500m + 500m + 500m = 3000m

3.     The motorist takes the 8th turn at point C. The magnitude of displacement:

AC= 866.03 m

4.     At sixth turn, When the motorist takes the 6th turn, he is at starting point A. Therefore,

Hence the magnitude of displacement is 866.03m at an angle 30  with AB

Total path length = 600  500 + 500 +500 = 4000 m