Question

The viscosity of a gas depends on the mass, the effective diameter and the mean speed of the molecules. At room temperature, for ${\text{He}}$, ${\eta _{He}} = 2 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}$ and for ${\text{C}}{{\text{H}}_{\text{4}}}$, ${\eta _{C{H_4}}} = 1 \cdot 1 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}$. The diameter of the ${\text{He}}$ atom is $2 \cdot 1 \times {10^{ - 10}}{\text{m}}$. If the diameter of ${\text{C}}{{\text{H}}_{\text{4}}}$ is $n \times {10^{ - 10}}{\text{m}}$, find $n$ .
Given, the mean speed of the molecules of the gas $v \propto \sqrt {\dfrac{{{K_B}T}}{m}} $ where ${K_B}$ is the Boltzmann’s constant, $T$ is the temperature and $m$ is the mass of the gas.
A) $2 \cdot 1$
B) $4 \cdot 0$
C) $3 \cdot 2$
D) $3 \cdot 0$

Answer 1

Hint:Here we have to first obtain a relation for the coefficient of viscosity in terms of the mass, diameter and mean speed of the molecules using the method of dimensions. The mean speed of the molecules is given to be dependent on the temperature and mass of the molecule. So incorporating this dependency into our obtained relation for the coefficient of viscosity will provide us with the value of $n$ .

Complete step by step answer.
Step 1: List the parameters given in the question and state the general relation for the coefficient of viscosity.
The coefficient of viscosity of ${\text{He}}$ is ${\eta _{He}} = 2 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}$ and that of ${\text{C}}{{\text{H}}_{\text{4}}}$ is ${\eta _{C{H_4}}} = 1 \cdot 1 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}$ .
The diameter of the ${\text{He}}$ atom is given to be ${d_{He}} = 2 \cdot 1 \times {10^{ - 10}}{\text{m}}$ and that of the ${\text{C}}{{\text{H}}_{\text{4}}}$ atom is given as ${d_{C{H_4}}} = n \times {10^{ - 10}}{\text{m}}$. We have to determine $n$ .
The coefficient of viscosity $\eta $ is given to be depending on the mass $m$, the effective diameter $d$ and the mean speed $v$ of the molecules.
So we represent the general relation by $\eta = k{m^a}{d^b}{v^c}$ -------- (1) where $k{\text{, }}a{\text{, }}b{\text{, }}c$are some constants.
Step 2: Using dimensional analysis determine the values of $a{\text{, }}b{\text{, }}c$.
The dimension of the coefficient of viscosity is $\eta \to \left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$ .
The dimension of the mass of the molecule is $m \to \left[ M \right]$ .
The dimension of the effective diameter is $d \to \left[ L \right]$ .
The dimension of the mean speed of the molecules is $v \to \left[ {L{T^{ - 1}}} \right]$ .
Rewriting equation (1) in terms of the dimensions of $\eta ,m,d,v$ we get, $\left[ {M{L^{ - 1}}{T^{ - 1}}} \right] = k{\left[ M \right]^a}{\left[ L \right]^b}{\left[ {L{T^{ - 1}}} \right]^c}$
On simplifying, the above equation becomes $M{L^{ - 1}}{T^{ - 1}} = k{M^a}{L^{b + c}}{T^{ - c}}$ --------- (2)
Now we equate the corresponding powers on both sides of the equation.
For $M$ we have, $a = 1$
For $T$ we have $ - c = - 1$
$ \Rightarrow c = 1$
For $L$ we have $b + c = - 1$
 $ \Rightarrow b = - 1 - c = - 1 - 1 = - 2$
Thus the values of the constants are $a = 1$, $b = - 2$ and $c = 1$ .
So the relation becomes $\eta = \dfrac{{kmv}}{{{d^2}}}$ --------- (3)
Step 3: Incorporate the given proportionality of the mean speed of the molecules to find the value of $n$ .
It is given that $v \propto \sqrt {\dfrac{{{K_B}T}}{m}} $ but as ${K_B}$ is the Boltzmann’s constant and $T$ is the room temperature, we have $v \propto {m^{ - 1/2}}$ ------(4)
Incorporating proportionality (4) into equation (3) we get, $\eta = \dfrac{{km{m^{ - 1/2}}}}{{{d^2}}} = \dfrac{{k\sqrt m }}{{{d^2}}}$
$ \Rightarrow \eta {d^2}\sqrt m = {\text{constant}}$
Then we have ${\eta _{He}}{d_{He}}^2\sqrt {{m_{He}}} = {\eta _{C{H_4}}}{d_{C{H_4}}}^2\sqrt {{m_{C{H_4}}}} $
$ \Rightarrow {d_{C{H_4}}} = {d_{He}}\sqrt {\dfrac{{{\eta _{He}}}}{{{\eta _{C{H_4}}}}}} {\left( {\dfrac{{{m_{He}}}}{{{m_{C{H_4}}}}}} \right)^{1/4}}$ ----------- (5)
Substituting for ${\eta _{He}} = 2 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}$, ${\eta _{C{H_4}}} = 1 \cdot 1 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}$, ${d_{C{H_4}}} = n \times {10^{ - 10}}{\text{m}}$, ${d_{He}} = 2 \cdot 1 \times {10^{ - 10}}{\text{m}}$, ${m_{He}} = 4 \times {10^{ - 3}}{\text{kg}}$ and ${m_{C{H_4}}} = 16 \times {10^{ - 3}}{\text{kg}}$ in equation (5) we get, $n \times {10^{ - 10}} = 2 \cdot 1 \times {10^{ - 10}}\sqrt {\dfrac{{2 \times {{10}^{ - 5}}}}{{1 \cdot 1 \times {{10}^{ - 5}}}}} {\left( {\dfrac{{16 \times {{10}^{ - 3}}}}{{4 \times {{10}^{ - 3}}}}} \right)^{1/4}}$
$ \Rightarrow n = 2 \cdot 1 \times \sqrt {\dfrac{{2 \times {{10}^{ - 5}}}}{{1 \cdot 1 \times {{10}^{ - 5}}}}} {\left( {\dfrac{{16}}{4}} \right)^{1/4}} = 4 \cdot 0$
Thus we obtain the value of $n$ as 4.

So the correct option is B.

Note:The atomic mass of Helium is $4 \cdot 00{\text{u}}$, that of Carbon is $12 \cdot 01{\text{u}}$ and that of hydrogen is $1 \cdot 01{\text{u}}$ as obtained from a periodic table. So the molecular mass of Helium will be $4 \times {10^{ - 3}}{\text{kg}}$ and that of ${\text{C}}{{\text{H}}_{\text{4}}}$ will be $12 + \left( {4 \times 1} \right) = 16 \times {10^{ - 3}}{\text{kg}}$. These values are substituted in equation (5) as ${m_{He}} = 4 \times {10^{ - 3}}{\text{kg}}$ and ${m_{C{H_4}}} = 16 \times {10^{ - 3}}{\text{kg}}$ .