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Question: Calculate the mean free path in \({\text{C}}{{\text{O}}_2}\) at \({27^{\text{o}}}{\text{C}}\) and a ...

Calculate the mean free path in CO2{\text{C}}{{\text{O}}_2} at 27oC{27^{\text{o}}}{\text{C}} and a pressure of 106 mmHg{10^{ - 6}}{\text{ mmHg}}.
(molecular diameter =460pm = 460{\text{pm}})

Explanation

Solution

To solve this question, you must recall the formula for mean free path of gas particles. It is the average distance travelled by a gas particle between two successive collisions. We shall calculate the number of molecules present at the given pressure and then substitute the values in the given formula.

Formula used: The number of molecules per unit volume is given by,
N=nV×NAN = \dfrac{n}{V} \times {N_A}
The mean free path is given by:
λ=12πσ2N\lambda = \dfrac{1}{{\sqrt 2 \pi {\sigma ^2}N}}
Where, λ\lambda is the mean free path
σ\sigma is the molecular diameter
And NN is the number of molecules per unit molar volume of the gas.

Complete step by step solution:
We can write, 106 mmHg1091.01325 bar{10^{ - 6{\text{ }}}}{\text{mmHg}} \approx \dfrac{{{\text{1}}{{\text{0}}^{ - 9}}}}{{1.01325}}{\text{ bar}}
The number of molecules per unit volume is given by,
N=nV×NAN = \dfrac{n}{V} \times {N_A}
We know from the ideal gas equation, PV=nRTPV = nRT
Thus we get, nV=PRT\dfrac{n}{V} = \dfrac{P}{{RT}}
So we have the number of molecules per unit volume as,
N=PRT×NAN = \dfrac{P}{{RT}} \times {N_A}
Substituting the values, we get,
N=1091.01325×0.0821×300×6.023×1023N = \dfrac{{{{10}^{ - 9}}}}{{1.01325 \times 0.0821 \times 300}} \times 6.023 \times {10^{23}}
N=2.4×1010 molecules/cm3\Rightarrow N = 2.4 \times {10^{10}}{\text{ molecules/c}}{{\text{m}}^3}
The mean free path is given by,
λ=12πσ2N\lambda = \dfrac{1}{{\sqrt 2 \pi {\sigma ^2}N}}
Substituting the values, we get,
λ=11.414×3.14×(460×1010)2×2.4×1010\lambda = \dfrac{1}{{1.414 \times 3.14 \times {{\left( {460 \times {{10}^{ - 10}}} \right)}^2} \times 2.4 \times {{10}^{10}}}}
λ=4.435×103cm\therefore \lambda = 4.435 \times {10^3}{\text{cm}}.

Additional information:
The kinetic theory of gases provides an explanation for various experimental observations about a gas. Its postulates are:
a)Each gas is made up of large number of tiny particles
b)The volume of a molecule is negligible as compared to the volume of the gas.
c)There are no attractive forces present between the molecules of the gas.
d)The molecules are never stationary and keep moving in straight line motion until it collides.
e)All collisions are assumed to be completely elastic

Note:
a)Based on the kinetic theory of gases, the mean free path is directly proportional to temperature and inversely proportional to the pressure.
b)Larger is the size of a molecule, smaller is the mean free path.
c) Greater the number of molecules per unit volume, smaller is the mean free path
d)Larger the temperature, larger the mean free path
e)And, larger the pressure, lesser is the mean free path.