# Kinetic Theory

## Class 11 NCERT Physics

### NCERT

1   Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be $3 Å$.

Diameter of an oxygen molecule,$d=3 Å.$$\\ Radius,r=\dfrac{d}{2}=\dfrac{3}{2}=1.5 Å \\ 1.5 * 10^{-8} cm$$\\$ Actual volume occupied by $1$ mole of oxygen gas at STP =$22400 cm^3$$\\ Molecular volume of oxygen gas,V=\dfrac{4}{3}\pi r^3.N$$\\$ Where, $N$ is Avogadro’s number $\\$ $=6.023*10^{23}$ molecules/mole$\\$ $\therefore V=\dfrac{4}{3}*3.14*(1.5*10^{-8})^3*6.023*10^23\\ =8.51 cm^3$$\\ Ratio of the molecular volume to the actual volume of oxygen\\ =\dfrac{8.51}{22400}=3.8*10^{-4} 2 Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard temperature and pressure (STP: 1 atmospheric pressure, 0^o C ). Show that it is 22.4 litres. ##### Solution : The ideal gas equation relating pressure (P), volume (V), and absolute temperature (T) is given as: PV = nRT Where, R is the universal gas constant =8.314 J mol^{-1}K^{-1}$$\\$ $n=$ Number of moles $=1$$\\ T = Standard temperature = 273 K$$\\$ $P =$ Standard pressure =$1$ atm =$1.013 *10^5 Nm ^{-2}$$\\ \therefore V=\dfrac{nRT}{P}\\ =\dfrac{1*8.314*273}{1.013*10^5}\\ =0.0224 m^3\\ =22.4 \text{litres}$$\\$ Hence, the molar volume of a gas at STP is $22.4$ litres.

3   Figure shows plot of $PV /T$ versus $P$ for $1.00 * 10^{-3} kg$ of oxygen gas at two different temperatures.$\\$ (a)What does the dotted plot signify?$\\$ (b)Which is true: $T_1 > T_2$ or$T_1< T_2$ ?$\\$ (c)What is the value of $PV/T$ where the curves meet on the y-axis?$\\$ (d)If we obtained similar plots for $1.00 * 10^{-3} kg$ of hydrogen, would we get the same value of $PV/T$ at the point where the curves meet on the y-axis? If not, what mass of hydrogen yields the same value of $PV/T$ (for low pressure high temperature region of the plot)? (Molecular mass of $H_2 = 2.02 u ,$ of $O_2 = 32.0 u$ , $R=8.31 J mo1^{ -1} K^{-1} . )$

##### Solution :

(a)The dotted plot in the graph signifies the ideal behaviour of the gas, i. e. , the ratio $\dfrac{PV}{T}$ is equal. $\mu R$($\mu$ is the number of moles and $R$ is the universal gas constant )is a constant quality. It is not dependent on the pressure of the gas. $\\$ (b)The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature $T_1$ is closer to the dotted plot thanthe curve of the gas at temperature $T_2$ . 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 $PV/T,$ where the two curves meet, is $\mu R$ . This is because the ideal gas equation is given as:$\\$ $PV=\mu RT\\ \dfrac{PV}{T}=\mu R$$\\ Where \\ P is the pressure \\ T is the temperature\\ V is the volume\\ \mu is the number of moles\\ R is the universal constant\\ Molecular mass of oxygen = 32.0 g$$\\$ Mass of oxygen $=1*10^{-3}kg =1g$$\\ R=8.314 J mole^{-1}K^{-1}\\ \therefore \dfrac{PV}{T}=\dfrac{1}{32}*8.314\\ =0.26 J K^{-1}$$\\$ Therefore, the value of the ratio $PV/T$, where the curves meet on the y-axis, is $0.26 J K^{-1}$$\\ (d) If we obtain similar plots for 1.00 *10^{-3} kg of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the y-axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u).$$\\$ We have:$\\$ $\therefore \dfrac{PV}{T}=0.26 JK^{-1}\\ R=8.314 J mole^{-1}K^{-1}$$\\ Molecular mass (M) of H_2=2.02 u$$\\$ $\dfrac{PV}{T}=\mu R$ at constant temperature $\\$ $\dfrac{PV}{T}=\mu R$ at constant temperature $\\$ where, $\mu =\dfrac{m}{M}$$\\ m=\text{Mass of H_2}\\ m=\dfrac{PV}{T}* \dfrac{M}{R}\\ =\dfrac{0.26* 2.02}{8.31}\\ =6.3* 10^{-2}g\\ =6.3*10^{-5}k g$$\\$ Hence ,$6.3*10^{-5}kg$ of $H_2$ will yield the same value of $\dfrac{PV}{T}$

4   An oxygen cylinder of volume $30$ litres has an initial gauge pressure of $15$ atm and a temperature of $27^o C$ . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to $11$ atm and its temperature drops to $17^o C$ . Estimate the mass of oxygen taken out of the cylinder ($R =8.31 J mol ^{-1} K^{-1}$ , molecular mass of $O_2 = 32 u$).

##### Solution :

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^oC=300K$$\\ Universal gas constant, R=8.314 J mole^{-1}K^{-1}$$\\$ Let the initial number of moles of oxygen gas in the cylinder be $n_1.$$\\ The gas equation is given as:\\ P_1V_1=n_1RT_1\\ \therefore n_1=\dfrac{P_1V_1}{RT_1}\\ =\dfrac{15.195*10^5*30*10^{-3}}{(8.314)*300}\\ =18.276 \\ But n_1=\dfrac{m_1}{M} \\ Where,\\ m_1=Initial mass of oxygen\\ M=Molecular mass of oxygen=32 g$$\\$ $\therefore m_1=n_1M=18.276 *32 \\ =584.84 g$$\\ After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.\\ Volume,V_2=30 litres =30*10^{-3}m^3$$\\$ Gauge pressure, $P_2 = 11 atm = 11 * 1.013 * 10^5 Pa$$\\ Temperature, T_2 = 17^o C = 290 K$$\\$ Let $n_2$ be the number of moles of oxygen left in the cylinder. The gas equation is given as:$\\$ $P_2V_2=n_2RT_2$

$\therefore n_2=\dfrac{P_2V_2}{RT_2}\\ =\dfrac{11.143*10^5*30*10^{-3}}{8.314*290}\\ =13.86$$\\ But,n_2=\dfrac{m_2}{M}$$\\$ Where, $\\$ $m_2$ is the mass of oxygen remaining in the cylinder$\\$ $\therefore m_2=n_2 M=13.86*32=443.52 g$$\\ The mass of oxygen taken out of the cylinder is given by the relation:\\ Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder\\ =m_1-m_2\\ =584.84 g-443.52 g\\ =141.32 g\\ =0.141 kg$$\\$ Therefore, $0.141 kg$ of oxygen is taken out of the cylinder.

5   An air bubble of volume $1.0 cm^3$ rises from the bottom of a lake $40 m$ deep at a temperature of $12^o C$ . To what volume does it grow when it reaches the surface, which is at a temperature of $35^o C$ ?

##### Solution :

Volume of the air bubble, $V_1 = 1.0 cm^3 = 1.0* 10^{-6} m^3$$\\ Bubble rises to height, d = 40 \ m$$\\$ Temperature at a depth of $40 m, T_1 = 12^o C = 285 K$$\\ Temperature at the surface of the lake, T_2= 35^o C= 308 K$$\\$ The pressure on the surface of the lake:$\\$ $P_2=1 atm =1*1.013*10^5 Pa$$\\ The pressure at the depth of 40 m:$$\\$ $P_1 = 1 atm +d \rho g$$\\ Where,\\ \rho is the density of water = 10^3 kg / m^3$$\\$ $g$ is the acceleration due to gravity $=9.8 m/s^2$$\\ \therefore P_1=1.1013*10^5+40*10^3*9.8\\ =493300 Pa$$\\$ We have:$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$$\\ Where, V _2 is the volume of the air bubble when it reaches the surface\\ V_2=\dfrac{P_1V_1T_2}{T_1P_2}\\ =\dfrac{(493300)(1.0*10^{-6})308}{285*1.013*10^5}\\ =5.263*10^{-6}m^3 or 5.263 c m^3$$\\$ Therefore, when the air bubble reaches the surface, its volume becomes $5.263 cm^3 .$