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NCERT/CBSE PHYSICS CLASS 11 textbook http://www.TutorBreeze.com Contact for Online Tutoring in Physics, Math, Chemistry, English Solutions/Answers to NCERT/CBSE PHYSICS Class 11(Class XI)textbook Exercise and Additional exercise CHAPTER THIRTEEN Kinetic Theory EXERCISES (For simplicity in numerical calculations, take g = 10 m s-2) 13.10 Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N2 = 28.0 u).

Q.13.10 Maxwell correction for mean free path

(
) =

1 2πnd 2

d = diameter of molecule. n=

N number of molecules = V volume of gas

PV = nRT n=

N N.P = V RT

N = 6.023 × 1023 / mole P = 2atm = 2 × 1.013 × 105 N / m2 R = 8.3 Jk −1 mol −1 T = 273 + 17 = 290 n=

6.023 × 1023 × 2 × 2.013 × 105 = 5.1 × 1025 / m3 8.3 × 290

r = 1 Å = 1 × 10−10 m d = 2r = 2 × 10−10 m

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NCERT/CBSE PHYSICS CLASS 11 textbook http://www.TutorBreeze.com


=

1 2 × 3.142 × ( 2 × 10

r.m.s. velocity c = R = 8 .3

c=

Contact for Online Tutoring in Physics, Math, Chemistry, English −10

) × 5.07 × 10

25

= 1.0 × 10 −7 m

3RT M

J / mole / k ,

T = 290 k ,

3 × 8.3 × 290 = 5.1 × 102 −3 28 × 10

M = 28 × 10 −3 kg

m/s

collision frequency c 5.1 × 102 f = = = 5.1 × 109 /s −7 τ 28 × 10

Time between successive collisions, _________ =

Time taken for the collision t =

1 × 10 −7 m = 2 × 10 −10 s 5.1 × 102 m/s

d 2 × 10 −10 = = 4 × 0−13 s c 5.1 × 102

τ 2 × 10 −10 = = 500 t 4 × 10 −13

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