**NCERT Solutions Class 11 Physics Chapter 13 Kinetic Theory** – Here are all the NCERT solutions for Class 11 Physics Chapter 13. This solution contains questions, answers, images, explanations of the complete chapter 13 titled Of Kinetic Theory taught in Class 11. If you are a student of Class 11 who is using NCERT Textbook to study Physics, then you must come across chapter 13 Kinetic Theory After you have studied the lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory in one place.

## NCERT Solutions Class 11 Physics Chapter 13 Kinetic Theory

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Class | 11 |

Subject | Physics |

Book | Physics Part I |

Chapter Number | 13 |

Chapter Name |
Kinetic Theory |

### NCERT Solutions Class 11 Physics chapter 13 Kinetic Theory

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### Question & Answer

Q.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 3Å.

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

Q.2:Molar volume is the volume occupied by 1 mol of any (ideal) gas at standard temperature and pressure (STP : 1 atmospheric pressure, O \(^{\circ} \mathrm{C}\)). Show that it is 22.4 litres.

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

Q.3:\(\begin{array}{l}{\text { Figure } 13.8 \text { shows plot of } P V / T \text { versus For } 1.00 \times 10^{-3} \mathrm{kg} \text { of oxygen gas at two different }} \\ {\text { temperatures. }}\end{array}\)(a) does the dotted plot signify? (b) Which is true. Tl > T2 or Tl

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: \(\begin{array}{l}{\frac{P Y}{T}=0.26 \mathrm{JK}^{-1}} \\ {R=8.314 \mathrm{Jmole}^{-1} \mathrm{K}^{-1}} \\ {\text { Molecular mass }(M) \text { of } \mathrm{H}_{2}=2.02 \mathrm{u}} \\ {\frac{P V}{T}=\mu R \text { at constant temperature }} \\ {\text { Where, } \mu=\frac{m}{M}} \\ {m=\text { Mass of } \mathrm{H}_{2}}\end{array}\) \(\begin{array}{l}{\quad m=\frac{P V}{T} \times \frac{M}{R}} \\ {=\frac{0.26 \times 2.02}{8.31}} \\ {=6.3 \times 10^{-2} \mathrm{g}=6.3 \times 10^{-5} \mathrm{kg}} \\ {\text { Hence, } 6.3 \times 10^{-5} \mathrm{kg} \text { of } \mathrm{H}_{2} \text { will yield the same value of } P V / T}\end{array}\)

Q.4:An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27 oc. After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17 oc. Estimate the mass of oxygen taken out of the cylinder (R = 8.31 J \({mole}^{-1}\) \(K^{-1}\), molecular mass of 02 = 32 u).

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 \(\begin{array}{l}{\text { Volume, } V_{2}=30 \text { litres }=30 \times 10^{-3} \mathrm{m}^{3}} \\ {\text { Gauge pressure, } P_{2}=11 \text { atm }=11 \times 1.013 \times 10^{5} \mathrm{Pa}} \\ {\text { Temperature, } T_{2}=17^{\circ} \mathrm{C}=290 \mathrm{K}}\end{array}\) \(\begin{array}{l}{P_{2} V_{2}=n_{2} R T_{2}} \\ {\therefore n_{2}=\frac{P_{2} V_{2}}{R T_{2}}} \\ {=\frac{11.143 \times 10^{5} \times 30 \times 10^{-3}}{8.314 \times 290}=13.86}\end{array}\) \(\begin{array}{l}{n_{2}=\frac{m_{2}}{M}} \\ {\text { Where, }} \\ {m_{2} \text { is the mass of oxygen remaining in the cylinder }}\end{array}\) \(\begin{array}{l}{\therefore m_{2}=r_{2} M=13.86 \times 32=453.1 \mathrm{g}} \\ {\text { The mass of oxygen taken out of the cylinder is given by the relation: }} \\ {\text { Initial mass of oxygen in the cylinder - Final mass of oxygen in the cylinder }} \\ {=m_{1}-m_{2}} \\ {=584.84 \mathrm{g}-453.1 \mathrm{g}} \\ {=131.74 \mathrm{g}} \\ {=0.131 \mathrm{kg}} \\ {\text { Therefore, } 0.131 \mathrm{kg} \text { of oxygen is taken out of the cylinder. }}\end{array}\)

Q.5:An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C ?

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.

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