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NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

Edited By Vishal kumar | Updated on Sep 06, 2023 01:18 PM IST

NCERT Solutions Class 11 Physics Chapter 13 – Free PDF Access

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory are essential resources for achieving high marks in school exams and competitive exams like JEE and NEET. On this Careers360 page, you'll find the comprehensive kinetic theory of gases class 11 solutions covering questions from 13.1 to 13.10 (exercise questions) and 13.11 to 13.14 (additional exercise questions).

The solution of chapter 13 Physics Class 11 is prepared by our highly experienced Subject Matter which gives all the answers to NCERT Solution. By using of NCERT solutions for Class 11, specifically, the kinetic theory of gases class 11 NCERT solutions, can aid students in practising questions and grasping important syllabus concepts before their exams. These solutions are available for free and can be accessed below.

Maxwell, Boltzmann, and others developed the kinetic theory of gas in the 19th century. Before going to solutions of NCERT for Class 11 Physics Chapter 13 Kinetic Theory one must be familiar with ideal gas equations and gas laws. The Kinetic Theory interprets the pressure and temperature at the molecular level. Kinetic Theory is consistent with gas laws and it correctly explains specific heat and relates measurable properties of gases such as viscosity, diffusion, etc. The CBSE NCERT solutions for Class 11 Physics chapter 13 KTG gives an idea about how to apply the formulas studied in the chapter in numerical problems. class 11 physics ch 13 ncert solutions explain all the exercises and additional exercise questions.

** This chapter has been renumbered as Chapter 12 in accordance with the CBSE Syllabus 2023–24.

Free download kinetic theory of gases class 11 solutions PDF for CBSE exam.

NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory

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NCERT Solutions for Class 11 Physics Chapter 13 Kinetic Theory - Exercise Questions

Q13.1 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.

Answer:

The diameter of an oxygen molecule, d = 3 Å.

The actual volume of a mole of oxygen molecules V actual is

\\V_{actual}=N_{A}\frac{4}{3}\pi \left ( \frac{d}{2} \right )^{3}\\ V_{actual}=6.023\times 10^{23}\times \frac{4}{3}\pi \times \left ( \frac{3\times 10^{-10}}{2} \right )^{3}\\ V_{actual}=8.51\times 10^{-6}m^{3}\\ V_{actual}=8.51\times 10^{-3}litres

The volume occupied by a mole of oxygen gas at STP is V molar = 22.4 litres

\\\frac{V_{actual}}{V_{molar}}=\frac{8.51\times 10^{-3}}{22.4}\\ \frac{V_{actual}}{V_{molar}}=3.8\times 10^{-4}

Q13.3 Figure 13.8 shows plot of PV/T versus P for 1.00\times 10^{-3} kg of oxygen gas at two different temperatures.

1650449178452

(a) What does the dotted plot signify?
(b) Which is true: T_{1}> T_{2}\: \: orT_{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\times 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\mu , of O_{2}=32.0\mu ,
R=8.31J mol^{-1}K^{-1} .)

Answer:

(a) The dotted plot corresponds to the ideal gas behaviour.

(b) We know the behaviour of a real gas tends close to that of ideal gas as its temperature increases and since the plot corresponding to temperature T 1 is closer to the horizontal line that the one corresponding to T 2 we conclude T 1 is greater than T 2 .

(c) As per the ideal gas equation

\frac{PV}{T}=nR

The molar mass of oxygen = 32 g

n=\frac{1}{32}

R = 8.314

\\nR=\frac{1}{32}\times 8.314\\ nR=0.256JK^{-1}

(d) If we obtained similar plots for 1.00\times 10^{-3} kg of hydrogen we would not get the same
value of PV/T at the point where the curves meet on the y-axis as 1 g of Hydrogen would contain more moles than 1 g of Oxygen because of having smaller molar mass.

Molar Mass of Hydrogen M = 2 g

mass of hydrogen

m=\frac{PV}{T} \frac{M}{R}={0.256}\times \frac{2}{8.314}=5.48\times10^{-5}Kg

Q13.4 An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a temperature of 27^{0}C . After some oxygen is withdrawn from the cylinder, the gauge pressure drops to 11 atm and its temperature drops to 17^{0}C . Estimate the mass of
oxygen taken out of the cylinder ( R=8.31Jmol^{-1}K^{-1} , molecular mass of O_{2}=32\mu ).

Answer:

Initial volume, V 1 = Volume of Cylinder = 30 l

Initial Pressure P 1 = 15 atm

Initial Temperature T 1 = 27 o C = 300 K

The initial number of moles n 1 inside the cylinder is

\\n_{1}=\frac{P_{1}V_{1}}{RT_{1}}\\ n_{1}=\frac{15\times 1.013\times 10^{5}\times 30\times 10^{-3}}{8.314\times 300}\\ n_{1}=18.28

Final volume, V 2 = Volume of Cylinder = 30 l

Final Pressure P 2 = 11 atm

Final Temperature T 2 = 17 o C = 290 K

Final number of moles n 2 inside the cylinder is

\\n_{2}=\frac{P_{2}V_{2}}{RT_{2}}\\ n_{2}=\frac{11\times 1.013\times 10^{5}\times 30\times 10^{-3}}{8.314\times 290}\\ n_{2}=13.86

Moles of oxygen taken out of the cylinder = n 2 -n 1 = 18.28 - 13.86 = 4.42

Mass of oxygen taken out of the cylinder m is

\\m=4.42\times 32\\ m=141.44g

Q13.5 An air bubble of volume 1.0 cm^{3} rises from the bottom of a lake 40m deep at a temperature of 12^{0}C . To what volume does it grow when it reaches the surface, which is at a temperature of 35^{0}C ?

Answer:

Initial Volume of the bubble, V 1 = 1.0 cm 3

Initial temperature, T 1 = 12 o C = 273 + 12 = 285 K

The density of water is \rho _{w}=10^{3}\ kg\ m^{-3}

Initial Pressure is P 1

Depth of the bottom of the lake = 40 m

\\P_{1}=Atmospheric\ Pressure+Pressure\ due\ to\ water\\ P_{1}=P_{atm}+\rho _{w}gh\\ P_{1}=1.013\times 10^{5}+10^{3}\times 9.8\times 40\\ P_{1}=4.93\times 10^{5}Pa

Final Temperature, T 2 = 35 o C = 35 + 273 = 308 K

Final Pressure = Atmospheric Pressure =1.013\times 10^{5}Pa

Let the final volume be V 2

As the number of moles inside the bubble remains constant we have

\\\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}\\ V_{2}=\frac{P_{1}T_{2}V_{1}}{P_{2}T_{1}}\\ V_{2}=\frac{4.93\times 10^{5}\times 308\times 1}{1.013\times 10^{5}\times 285}\\ V_{2}=5.26\ cm^{3}

Q13.6 Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0m^{3} at a temperature of 27^{0}C and 1atm pressure.

Answer:

The volume of the room, V = 25.0 m 3

Temperature of the room, T = 27 o C = 300 K

The pressure inside the room, P = 1 atm

Let the number of moles of air molecules inside the room be n

\\n=\frac{PV}{RT}\\ n=\frac{1.013\times 10^{5}\times 25}{8.314\times 300}\\ n=1015.35

Avogadro's Number, N_{A}=6.022\times 10^{23}

Number of molecules inside the room is N

\\N=nN_{A}\\ N=1015.35\times 6.022\times 10^{23}\\ N=6.114\times 10^{26}

Q13.8 Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second contains chlorine (diatomic), and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal number of respective molecules? Is the root mean square speed of molecules the same in the three cases? If not, in which case is v rms the largest?

Answer:

As per Avogadro's Hypothesis under similar conditions of temperature and pressure equal volumes of gases contain equal number of molecules. Since the volume of the vessels are the same and all vessels are kept at the same conditions of pressure and temperature they would contain equal number of molecules.

Root mean square velocity is given as

v_{rms}=\sqrt{\frac{3kT}{m}}

As we can see v rms is inversely proportional to the square root of the molar mass the root mean square velocity will be maximum in case of Neon as its molar mass is the least.

Q13.9 At what temperature is the root mean square speed of an atom in an argon gas cylinder equal to the rms speed of a helium gas atom at -20^{0}C ? (atomic mass of Ar=39.9\mu , of He=4.0\mu ).

Answer:

As we know root mean square velocity is given as v_{rms}=\sqrt{\frac{3RT}{M}}

Let at temperature T the root mean square speed of an atom in an argon cylinder equal to the rms speed of a helium gas atom at -20^{0}C

\\\sqrt{\frac{3R\times T}{39.9}}=\sqrt{\frac{3R\times 253}{4}}\\ T=2523.7\ K

Q13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17^{0}C . Take the radius of a nitrogen molecule to be roughly 1.0A . Compare the collision time with the time the
molecule moves freely between two successive collisions (Molecular mass of N_{2}=28.0\mu ).

Answer:

Pressure, P = 2atm

Temperature, T = 17 o C

The radius of the Nitrogen molecule , r=1\ \AA

The molecular mass of N 2 = 28 u

The molar mass of N 2 = 28 g

From ideal gas equation

\\PV=nRT\\ \frac{n}{V}=\frac{P}{RT}\\

The above tells us about the number of moles per unit volume, the number of molecules per unit volume would be given as

\\n'=\frac{N_{A}n}{V}=\frac{6.022\times 10^{23}\times 2\times 1.013\times 10^{5}}{8.314\times (17+273)}\\ n'=5.06\times 10^{25}

The mean free path \lambda is given as

\\\lambda =\frac{1}{\sqrt{2}\pi n'd^{2}}\\ \lambda =\frac{1}{\sqrt{2}\times \pi \times 5.06\times 10^{25}\times (2\times 1\times 10^{-10})^{2}}\\ \lambda =1.11\times 10^{-7}\ m

The root mean square velocity v rms is given as

\\v_{rms}=\sqrt{\frac{3RT}{M}}\\ v_{rms}=\sqrt{\frac{3\times 8.314\times 290}{28\times 10^{-3}}}\\ v_{rms}=508.26\ m\ s^{-1}

The time between collisions T is given as

\\T=\frac{1}{Collision\ Frequency}\\ T=\frac{1}{\nu }\\ T=\frac{\lambda }{v_{rms}}\\ T=\frac{1.11\times 10^{-7}}{508.26}\\ T=2.18\times 10^{-10}s

Collision time T' is equal average time taken by a molecule to travel a distance equal to its diameter

\\T'=\frac{d}{v_{rms}}\\ T'=\frac{2\times 1\times 10^{-10}}{508.26}\\ T'=3.935\times 10^{-13}s

The ratio of the average time between collisions to the collision time is

\\\frac{T}{T'}=\frac{2.18\times 10^{-10}}{3.935\times 10^{-13}}\\ \frac{T}{T'}=554

Thus we can see time between collisions is much larger than the collision time.


NCERT solutions for class 11 physics chapter 13 kinetic theory - Additional Exercise Questions

Q13.11 A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm long mercury thread, which traps a 15 cm column of air. What happens if the tube is held vertically with the open end at the bottom?

Answer:

Initially, the pressure of the 15 cm long air column is equal to the atmospheric pressure, P 1 = 1 atm = 76 cm of Mercury

Let the crossectional area of the tube be x cm 2

The initial volume of the air column, V 1 = 15x cm 3

Let's assume once the tube is held vertical y cm of Mercury flows out of it.

The pressure of the air column after y cm of Mercury has flown out of the column P 2 = 76 - (76 - y) cm of Mercury = y cm of mercury

Final volume of air column V 2 = (24 + y)x cm 3

Since the temperature of the air column does not change

\\P_{1}V_{1}=P_{2}V_{2}\\ 76\times 15x=y\times (24+y)x\\ 1140=y^{2}+24y\\ y^{2}+24y-1140=0

Solving the above quadratic equation we get y = 23.8 cm or y = -47.8 cm

Since a negative amount of mercury cannot flow out of the column, y cannot be negative. Therefore y = 23.8 cm.

Length of the air column = y + 24 = 47.8 cm.

Therefore once the tube is held vertically, 23.8 cm of Mercury flows out of it and the length of the air column becomes 47.8 cm

Q 13.12 From a certain apparatus, the diffusion rate of hydrogen has an average value of 28.7cm^{3}s^{-1} . The diffusion of another gas under the same conditions is measured to have an average rate of 7.2cm^{3}s^{-1} . Identify the gas.

Answer:

As per Graham's Law of diffusion if two gases of Molar Mass M 1 and M 2 diffuse with rates R 1 and R 2 respectively their diffusion rates are related by the following equation

\frac{R_{1}}{R_{2}}=\sqrt{\frac{M_{2}}{M_{1}}}

In the given question

R 1 = 28.7 cm 3 s -1

R 2 = 7.2 cm 3 s -1

M 1 = 2 g

\\M_{2}=M_{1}\left ( \frac{R_{1}}{R_{2}} \right )^{2}\\ M_{2}=2\times \left ( \frac{28.7}{7.2} \right )^{2}\\ M_{2}=31.78g

The above Molar Mass is close to 32, therefore, the gas is Oxygen.

Q13.13 A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called law of atmospheres n_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ]

where n2, n1 refer to number density at heights h2 and h1 respectively. Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column:

n_{2}=n_{1}exp\left [ -mg N_{A}(\rho -\rho ^{'})(h_{2}-h_{1})/ (\rho RT)\right ]
where \rho is the density of the suspended particle, and \rho ^{'} , that of the surrounding medium. [ N_{A} s Avogadro’s number, and R the universal gas constant.] [Hint: Use Archimedes principle to find the apparent weight of the suspended particle.]

Answer:

n_{2}=n_{1}exp\left [ -mg(h_{2}-h_{1}) /k_{b}T\right ] (i)

Let the suspended particles be spherical and have radius r

The gravitational force acting on the suspended particles would be

F_{G}=\frac{4}{3}\pi r^{3}\rho g

The buoyant force acting on them would be

F_{B}=\frac{4}{3}\pi r^{3}\rho' g

The net force acting on the particles become

\\F_{net}=F_{G}-F_{B}\\ F_{net}=\frac{4}{3}\pi r^{3}\rho g-\frac{4}{3}\pi r^{3}\rho' g\\F_{net}=\frac{4}{3}\pi r^{3}g(\rho -\rho ')

Replacing mg in equation (i) with the above equation we get

\\n_{2}=n_{1}exp\left [ -\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1}) /k_{b}T\right ]\\ n_{2}=n_{1}exp\left [ \frac{-\frac{4}{3}\pi r^{3}g(\rho -\rho ')(h_{2}-h_{1})}{\frac{RT}{N_{A}}} \right ]\\ n_{2}=n_{1}exp\left [ \frac{-mgN_{A}(\rho -\rho ')(h_{2}-h_{1})}{RT\rho '} \right ]

The above is the equation to be derived

Q13.14 Given below are densities of some solids and liquids. Give rough estimates of the size of their atoms :

Substance
Atomic Mass (u)
Density (10 3 Kg m 3 )
Carbon (diamond)
12.01
2.22
Gold
197
19.32
Nitrogen (liquid)
14.01
1
Lithium
6.94
0.53
Fluorine
19
1.14

Answer:

Let one mole of a substance of atomic radius r and density \rho have molar mass M

Let us assume the atoms to be spherical

Avogadro's number is N_{A}=6.022\times 10^{23}

\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3M}{4N_{A}\pi \rho } \right )^{\frac{1}{3}}

For Carbon

\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 12.01}{4\times 6.022\times 10^{23}\times \pi\times 2.22\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.29\AA

For gold

\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 197.00}{4\times 6.022\times 10^{23}\times \pi\times 19.32\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.59\AA

For Nitrogen

\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 14.01}{4\times 6.022\times 10^{23}\times \pi\times 1.00\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.77\AA

For Lithium

\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times 6.94}{4\times 6.022\times 10^{23}\times \pi\times 0.53\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.73\AA

For Fluorine

\\N_{A}\frac{4}{3}\pi r^{3}\rho =M\\ r=\left ( \frac{3\times19.00}{4\times 6.022\times 10^{23}\times \pi\times 1.14\times 10^{3} } \right )^{\frac{1}{3}}\\ r=1.88\AA

Class 11 physics ch 13 ncert solutions are essential for comprehending how tiny particles in matter behave. This chapter helps students understand how gases work and their thermal properties. It also connects to other chapters in thermodynamics and has practical applications. By studying this chapter, students can do well in their exams and gain a solid understanding of how matter works on a fundamental level.

NCERT Solutions for Class 11 Physics Chapter Wise

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kinetic theory of gases class 11 physics NCERT solutions: Important Formulas and Diagrams

  • The ideal gas equation is given by

PV=\mu RT=K_BNT

where \mu is the number of moles

N is the number of molecules

R and k B are universal constants.

  • The pressure of an ideal gas

P=\frac{1}{3}nm\bar v^2

where n is the number density, m is the mass of the molecule and \bar v^2 is the mean of the squared speed.

  • The RMS velocity of the gas molecule

v_{rms}=\sqrt{\bar v^2}=\sqrt{\frac{3K_BT}{m}} Where T is the temperature.

  • The translational kinetic energy is

\frac{3K_BNT}{2}

  • The mean free path

l=\frac{1}{\sqrt{2}n\pi d^2}

where n is the number density and d is the diameter of the molecule

Use of NCERT Solutions For Class 11 Physics Chapter 13 Kinetic Theory

  • To familiarise yourself with the concepts studied in the chapter, it is important to study the CBSE NCERT solutions for Class 11 Physics Chapter 13 Kinetic Theory.
  • All the equations studied in the chapter are utilized in the solutions of NCERT Class 11 Physics Chapter 13 Kinetic Theory.
  • The NCERT solutions for Class 11 are important for class exams and competitive exams. For exams like NEET and JEE Mains, one question can be expected from the chapter and the questions can be solved easily if you are familiar with the concepts and formulas of the chapter.

Key Features of Class 11 physics ch 13 NCERT solutions 13 Kinetic Theory

  1. Comprehensive Coverage: These kinetic theory of gases exercise solutions cover all the important topics and concepts presented in the chapter, ensuring a thorough understanding of kinetic theory.

  2. Exercise and Additional Exercise Solutions: Detailed kinetic theory of gases class 11 physics ncert solutions for exercise questions (13.1 to 13.10) and additional exercise questions (13.11 to 13.14) are provided, facilitating practice and self-assessment.

  3. Clarity and Simplicity: The kinetic theory of gases class 11 questions and answers are explained in clear and simple language, making complex concepts more accessible to students.

  4. Problem-Solving Skills: By practising with these kinetic theory class 11 solutions, students develop strong problem-solving skills, which are essential in physics and other subjects.

  5. Exam Preparation: These kinetic theory of gases exercise solutions are designed to help students prepare effectively for their exams, including board exams and competitive exams like JEE and NEET.

Dropped Topics –

13.6.5 - Specific Heat Capacity of Water
Exercises - (13.11–13.14)

NCERT solutions for class 11 Subject wise

Also Check NCERT Books and NCERT Syllabus here

Subject wise NCERT Exemplar solutions

We trust that the kinetic theory of gases class 11 questions and answers prove to be beneficial for your studies. If you have any questions or require further clarification regarding these solutions, please feel free to leave a comment below, and we will promptly address your inquiries.

Frequently Asked Question (FAQs)

1. What is the weightage of the Class 11 Physics chapter Kinetic Theory for JEE Main and NEET

For both JEE Main and NEET one question can be expected from the chapter Kinetic Theory. To practice more problems on Kinetic Theory refer to NCERT book questions, NCERT exemplar questions and previous year papers of NEET and JEE Main.

2. What are the topics covered in Kinetic Theory Class 11 Physics NCERT book?

The following are the topics covered in Class 11 chapter Kinetic Theory

  • Molecular nature of matter  
  • Behaviour of gases  
  • Kinetic theory of an ideal gas  
  • Law of equipartition of energy  
  • Specific heat capacity  
  •   Mean free path  
3. What is Kinetic Theory

The gas  consists of rapidly moving atoms or molecules. Kinetic theory explains the behaviour of gas based on these rapidly moving molecules.

4. A fluid is filled inside a container. Is there exist a pressure at the center of the container?

Pressure exists everywhere in a fluid, not only the walls of the container. Any layer of gas inside the volume of a container is in equilibrium because the pressure is the same on both sides of the layer.

5. How can Students download class 11 physics chapter 13 ncert solutions ?

To download chapter-wise solutions for class 11th Physics click on the given link.

https://school.careers360.com/ncert/ncert-solutions-class-11-physics 

6. what is behaviour of gases class 11 physics

The behaviour of gases in Class 11 Physics involves studying how gases respond to changes in factors like pressure, temperature, and volume, as described by gas laws such as Boyle's, Charles's, Gay-Lussac's, and Avogadro's laws. This understanding helps predict and explain the behaviour of gases in various situations.

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0.34\; J

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2,000 \; J - 5,000\; J

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200 \, \, J - 500 \, \, J

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2\times 10^{5}J-3\times 10^{5}J

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20,000 \, \, J - 50,000 \, \, J

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less than 3

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more than 3 but less than 6

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more than 6 but less than 9

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more than 9

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