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# Estimate the Mean Free Path and Collision Frequency of a Nitrogen Molecule in a Cylinder Containing Nitrogen at 2.0 Atm and Temperature 17 °C. 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 - CBSE (Science) Class 11 - Physics

#### Question

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. 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).

#### Solution 1

Mean free path = 1.11 × 10–7 m

Collision frequency = 4.58 × 109 s–1

Successive collision time ≈ 500 × (Collision time)

Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 105 Pa

Temperature inside the cylinder, T = 17°C =290 K

Radius of a nitrogen molecule, r = 1.0 Å = 1 × 1010 m

Diameter, d = 2 × 1 × 1010 = 2 × 1010 m

Molecular mass of nitrogen, M = 28.0 g = 28 × 10–3 kg

The root mean square speed of nitrogen is given by the relation:

v_"rms" = sqrt((3RT)/M)

Where

R is the universal gas constant = 8.314 J mole–1 K–1

:. v_"rms" = sqrt((3xx8.314xx290)/(28xx10^(-3))) = 508.26 "m/s"

The mean free path (l) is given by the relation:

l = kT/(sqrt2xx d^2xx P)

Where,

k is the Boltzmann constant = 1.38 × 10–23 kg m2 s–2K–1

:. l = (1.38xx10^(-23)xx290)/(sqrt2 xx 3.14 xx (2xx10^(-10))^2 xx 2.026 xx 10^5)

= 1.11 xx 10^(-7) m

Collision frequency  = v_"rms"/l

=508.26/(1.11xx10^(-7)) = 4.58 xx 10^(9) s^(-1)

Collision time is given as:

T = d/v_"rms"

= (2xx10)^(-10)/508.26 = 3.93 xx 10^(-13) s

Time taken between successive collisions:

T' = 1/v_"rms"

= (1.11 xx 10^(-7))/(508.26 "m/s") = 2.18 xx 10^(-10) s

:. "T'"/T = (2.18xx 10^(-10))/(3.93xx10^(-13)) = 500

Hence, the time taken between successive collisions is 500 times the time taken for a collision.

#### Solution 2

Here P = 2.0 " atm" = 2 xx 1.013 xx 10^5 Pa = 2.026 xx 10^5 Pa

T = 17 ^@C = 17 + 273 = 290

Radius R = 1.0 Å = 1 xx 10^(-10) m = Molecular mass = 28 u

:. m = 28 xx 1.66 xx 10^(-27) = 4.65 xx 10^(-26) kg

Also R = 8.31 J mol^(-1) K^(-1), k = 1.38 xx 10^(-23) JK^(-1)

Now for one mole of a gas, PV = RT => V = RT/P = (8.31 xx 290)/(2.026 xx 10^5)

=> V = 1.189 xx 10^(-2) m^3

:. Number of molecules per unit volume ,  n = N/V

:. n = (6.023 xx 10^(23))/(1.189xx10^(-2)) = 5.06 xx 10^(25) m^(-3)

Now, mean free path, lambda  = 1/(sqrt(2)pind^2) = 1/(sqrt(2)pin(2r)^2)

= 1/(1.414 xx 3.14xx 5.06xx 10^25xx (2xx1xx10^(-10))^2

= 1.1 xx 10^(-7)

Also v_"rms" = sqrt((3RT)/M) = sqrt((3xx8.31xx290)/(28xx10^(-3))) = 5.08 xx 10^2 ms^(-1)

:. Collision frequency, v = v_"rms"/lambda = (5.08 xx 10^2)/(1.1 xx 10^(-7)) = 4.62 xx 10^9 s^(-1)

Time between successive collisions = 1/v = 1/(4.62 xx 10^9) = 2.17 xx 10^(-10)s

Also the collision time =  d/v_"rms" = (2xx1xx10^(-10))/(5.08 xx 10^2)s = 3.92 xx 10^(-13) s

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Solution Estimate the Mean Free Path and Collision Frequency of a Nitrogen Molecule in a Cylinder Containing Nitrogen at 2.0 Atm and Temperature 17 °C. 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 Concept: Mean Free Path.
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