NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

Ans : \(\begin{array}{l}{\text { Diameter of an oxygen molecule, } d=3 A} \\ {=\frac{d}{2}=\frac{3}{2}=1.5 \hat{A}=1.5 \times 10^{-8} \mathrm{cm}}\end{array}\)
\(\begin{array}{l}{\text { Actual volume occupied by 1 mole of oxygen gas at STP = } 22400 \mathrm{cm}^{3}} \\ {V=\frac{4}{3} \pi r^{3} \cdot N}\end{array}\)
\(\begin{array}{l}{\text { Where, } N \text { is Avogadro's number }=6.023 \times 10^{23} \text { molecules/mole }} \\ {\therefore V=\frac{4}{3} \times 3.14 \times\left(1.5 \times 10^{-8}\right)^{3} \times 6.023 \times 10^{23}=8.51 \mathrm{cm}^{3}}\end{array}\)
\(\begin{array}{l}{\text { Ratio of the molecular volume to the actual volume of oxygen }=\frac{8.51}{22400}} \\ {=3.8 \times 10^{-4}}\end{array}\) 

Ans : \(\begin{array}{l}{\text { The ideal gas equation relating pressure (P), volume }(V), \text { and absolute temperature (T) }} \\ {\text { is given as: }} \\ {P V=n R T}\end{array}\)
\(\begin{array}{l}{\text { Where, }} \\ {R \text { is the universal gas constant }=8.314 \mathrm{Jmol}^{-1} \mathrm{K}^{-1}} \\ {n=\text { Number of moles }=1} \\ {T=\text { Standard temperature }=273 \mathrm{K}} \\ {P=\text { Standard pressure }=1 \mathrm{atm}=1.013 \times 10^{5} \mathrm{Nm}^{-2}}\end{array}\)
\(\begin{array}{l}{\therefore V=\frac{n R T}{P}} \\ {=\frac{1 \times 8.314 \times 273}{1.013 \times 10^{5}}} \\ {=0.0224 \mathrm{m}^{3}} \\ {=22.4 \text { litres }}\end{array}\) 

Ans : \(\begin{array}{l}{\text { (a) The dotted plot in the graph signifies the ideal behaviour of the gas, l.e., the ratio }} \\ {\frac{P V}{T} \text { is equal. } \mu R(\mu \text { is the number of moles and } R \text { is the universal gas constant) is a }} \\ {\text { constant quality. It is not dependent on the pressure of the gas. }}\end{array}\)


(b) The dotted plot in the given graph represents an ideal gas. The curve of the gas at 
temperature Tl is closer to the dotted plot than the 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. 


\(\begin{array}{l}{\text { (c) The value of the ratio } P V / T, \text { where the two curves meet, is } \mu R \text { . This is because the }} \\ {\text { ideal gas equation is given as: }}\end{array}\)
\(\begin{array}{l}{P V=\mu R T} \\ {\frac{P V}{T}=\mu R} \\ {\text { Where, }}\end{array}\)
P is the pressure 
T is the temperature 
V is the volume 
p is the number of moles 
R is the universal constant 
Molecular mass Of oxygen = 32.0 g 
\(\begin{array}{l}{\text { Mass of oxygen }=1 \times 10^{-3} \mathrm{kg}=1 \mathrm{g}} \\ {R=8.314 \mathrm{mole}^{-1} \mathrm{K}^{-1}} \\ {\quad \frac{P V}{T}=\frac{1}{32} \times 8.314}\end{array}\)
\(\begin{array}{l}{=0.26 \mathrm{JK}^{-1}} \\ {\text { Therefore, the value of the ratio } \mathrm{PV} / T, \text { where the curves meet on the } y \text { -axis, is }} \\ {0.26 \mathrm{JK}^{-1}}\end{array}\)


(d) If We obtain similar plots for \(1.00 \times 10^{-3}\) kg Of hydrogen, then we will not get the same value of PV/T at the point where the curves meet the V axis. This is because the molecular mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u) We have: 
\(\begin{array}{l}{\frac{P Y}{T}=0.26 \mathrm{JK}^{-1}} \\ {R=8.314 \mathrm{Jmole}^{-1} \mathrm{K}^{-1}} \\ {\text { Molecular mass }(M) \text { of } \mathrm{H}_{2}=2.02 \mathrm{u}} \\ {\frac{P V}{T}=\mu R \text { at constant temperature }} \\ {\text { Where, } \mu=\frac{m}{M}} \\ {m=\text { Mass of } \mathrm{H}_{2}}\end{array}\)
\(\begin{array}{l}{\quad m=\frac{P V}{T} \times \frac{M}{R}} \\ {=\frac{0.26 \times 2.02}{8.31}} \\ {=6.3 \times 10^{-2} \mathrm{g}=6.3 \times 10^{-5} \mathrm{kg}} \\ {\text { Hence, } 6.3 \times 10^{-5} \mathrm{kg} \text { of } \mathrm{H}_{2} \text { will yield the same value of } P V / T}\end{array}\) 

Ans : \(\begin{array}{l}{\text { Volume of oxygen, } V_{1}=30 \text { litres }=30 \times 10^{-3} \mathrm{m}^{3}} \\ {\text { Gauge pressure, } P_{1}=15 \mathrm{atm}=15 \times 1.013 \times 10^{5} \mathrm{Pa}}\end{array}\)
\(\begin{array}{l}{\text { Temperature, } T_{1}=27^{\circ} \mathrm{C}=300 \mathrm{K}} \\ {\text { Universal gas constant, } R=8.314 \mathrm{Jmole}^{-1} \mathrm{K}^{-1}} \\ {\text { Let the initial number of moles of oxygen gas in the cylinder be } n_{1}} \\ {\text { The gas equation is given as: }} \\ {P_{1} V_{1}=n_{1} R T_{1}}\end{array}\)
\(\begin{array}{l}{\therefore n_{1}=\frac{P V_{1}}{R T_{1}}} \\ {=\frac{15.195 \times 10^{5} \times 30 \times 10^{-3}}{(8.314) \times 300}=18.276}\end{array}\)
\(\begin{array}{l}{\text { Where, }} \\ {m_{1}=\text { Initial mass of oxygen }} \\ {M=\text { Molecular mass of oxygen }=32 \mathrm{g}} \\ {\therefore m_{1}=n_{1} M=18.276 \times 32=584.84 \mathrm{g}}\end{array}\)
After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces 
\(\begin{array}{l}{\text { Volume, } V_{2}=30 \text { litres }=30 \times 10^{-3} \mathrm{m}^{3}} \\ {\text { Gauge pressure, } P_{2}=11 \text { atm }=11 \times 1.013 \times 10^{5} \mathrm{Pa}} \\ {\text { Temperature, } T_{2}=17^{\circ} \mathrm{C}=290 \mathrm{K}}\end{array}\)
\(\begin{array}{l}{P_{2} V_{2}=n_{2} R T_{2}} \\ {\therefore n_{2}=\frac{P_{2} V_{2}}{R T_{2}}} \\ {=\frac{11.143 \times 10^{5} \times 30 \times 10^{-3}}{8.314 \times 290}=13.86}\end{array}\)
\(\begin{array}{l}{n_{2}=\frac{m_{2}}{M}} \\ {\text { Where, }} \\ {m_{2} \text { is the mass of oxygen remaining in the cylinder }}\end{array}\)
\(\begin{array}{l}{\therefore m_{2}=r_{2} M=13.86 \times 32=453.1 \mathrm{g}} \\ {\text { The mass of oxygen taken out of the cylinder is given by the relation: }} \\ {\text { Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder }} \\ {=m_{1}-m_{2}} \\ {=584.84 \mathrm{g}-453.1 \mathrm{g}} \\ {=131.74 \mathrm{g}} \\ {=0.131 \mathrm{kg}} \\ {\text { Therefore, } 0.131 \mathrm{kg} \text { of oxygen is taken out of the cylinder. }}\end{array}\) 

Ans : \(\begin{array}{l}{\text { Volume of the air bubble, } V_{1}=1.0 \mathrm{cm}^{3}=1.0 \times 10^{-6} \mathrm{m}^{3}} \\ {\text { Bubble rises to height, } d=40 \mathrm{m}} \\ {\text { Temperature at a depth of } 40 \mathrm{m}, T_{1}=12^{\circ} \mathrm{C}=285 \mathrm{K}} \\ {\text { Temperature at the surface of the lake, } T_{2}=35^{\circ} \mathrm{C}=308 \mathrm{K}}\end{array}\)
\(\begin{array}{l}{\text { The pressure on the surface of the lake: }} \\ {P_{2}=1 \text { atm }=1 \times 1.013 \times 10^{5} \mathrm{Pa}} \\ {\text { The pressure at the depth of } 40 \mathrm{m} \text { : }} \\ {P_{1}=1 \mathrm{atm}+\mathrm{d} \rho \mathrm{g}} \\ {\text { Where, }} \\ {\rho \text { is the density of water }=10^{3} \mathrm{kg} / \mathrm{m}^{3}} \\ {\therefore P_{1} \text { is the acceleration due to gravity }=9.8 \mathrm{m} / \mathrm{s}^{2}} \\ {\therefore P_{1}=1.013 \times 10^{5}+40 \times 10^{3} \times 9.8=493300 \mathrm{Pa}}\end{array}\)
\(\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}\)
Where, \(V_{2}\) is the volume Of the air bubble when it reaches the surface 
\(\begin{array}{l}{V_{2}=\frac{P_{1} V_{1} T_{2}}{T_{1} P_{2}}} \\ {=\frac{(493300)\left(1.0 \times 10^{-6}\right) 308}{285 \times 1.013 \times 10^{5}}} \\ {=5.263 \times 10^{-6} \mathrm{m}^{3} \text { or } 5.263 \mathrm{cm}^{3}}\end{array}\)
Therefore, where, the air bubble reaches the surface, its Volume becomes 5.263 crn³.