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Question: Calculate the mean free path of nitrogen molecule at \({{27}^{0}}C\) when pressure is 1.0 atm. Given...

Calculate the mean free path of nitrogen molecule at 270C{{27}^{0}}C when pressure is 1.0 atm. Given, diameter of nitrogen molecule is 1.5A01.5{{A}^{0}}, kB=1.38×10−23JK−1{{k}_{B}}=1.38\times {{10}^{-23}}J{{K}^{-1}}. If the average speed of nitrogen molecules is 675ms−1675ms{}^{-1}. The time taken by the molecule between two successive collisions is?
a.) 0.6ns
b.) 0.4ns
c.) 0.8ns
d.) 0.3ns

Explanation

Solution

In this question we have been given quantities in different units, so to solve this question our first step is to convert all the given quantities in such a way that it is suitable for us to use it in our formula directly. Most probably we change the units according to the units given in our constant.

Complete step by step solution:
We have been given with a temperature of 270C{{27}^{0}}C, but we need to convert it into kelvin
So, T=27+273T=27+273
T = 300 K
Now, the pressure is given in atm., but we need to convert it into Nm−2N{{m}^{-2}}
So, Pressure (P) = 1.01×105Nm−21.01\times {{10}^{5}}N{{m}^{-2}}
Again, the diameter is given in Armstrong, but we need to convert it into meters
So, Diameter (D) = 1.5×10−10m1.5\times {{10}^{-10}}m
Now, according to the question we need to find the Time taken by the molecule between two successive collisions
But to find the time we need to find the distance, and in this case, we will be using λ\lambda to denote the distance.
λ=KBT2πd2p\lambda =\dfrac{{{K}_{B}}T}{\sqrt{2}\pi {{d}^{2}}p} …….. (1)
We have been given that kB=1.38×10−23JK−1{{k}_{B}}=1.38\times {{10}^{-23}}J{{K}^{-1}}
Now putting the values in above equation (1), we get,
λ=(1.38×10−23JK−1)(300)2(3.14)(1.5×10−10m)2(1.01×105Nm−2)\lambda =\dfrac{(1.38\times {{10}^{-23}}J{{K}^{-1}})(300)}{\sqrt{2}(3.14){{(1.5\times {{10}^{-10}}m)}^{2}}(1.01\times {{10}^{5}}N{{m}^{-2}})}
λ=4.1×10−7m\lambda =4.1\times {{10}^{-7}}m
Hence, we got the distance as λ=4.1×10−7m\lambda =4.1\times {{10}^{-7}}m
Now time can be defined as t=distan⁡cespeedt=\dfrac{dis\tan ce}{speed}
t=λvt=\dfrac{\lambda }{v}
t=4.1×10−7m675ms−1t=\dfrac{4.1\times {{10}^{-7}}m}{675m{{s}^{-1}}}
t=0.6×10−9st=0.6\times {{10}^{-9}}s
Hence, the correct option will be option (a).

Note: The answer in the above question is the time interval between two successive collisions and at subatomic level the collisions occur very fast and hence this is the reason that we got the time interval in terms of nanoseconds. This time in nano seconds is also finite in case of sub atomic level.