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Question: The viscosity of a gas depends on the mass, the effective diameter and the mean speed of the molecul...

The viscosity of a gas depends on the mass, the effective diameter and the mean speed of the molecules. At room temperature, for He{\text{He}}, ηHe=2×10−5kgm−1s−1{\eta _{He}} = 2 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}} and for CH4{\text{C}}{{\text{H}}_{\text{4}}}, ηCH4=1⋅1×10−5kgm−1s−1{\eta _{C{H_4}}} = 1 \cdot 1 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}. The diameter of the He{\text{He}} atom is 2⋅1×10−10m2 \cdot 1 \times {10^{ - 10}}{\text{m}}. If the diameter of CH4{\text{C}}{{\text{H}}_{\text{4}}} is n×10−10mn \times {10^{ - 10}}{\text{m}}, find nn .
Given, the mean speed of the molecules of the gas v∝KBTmv \propto \sqrt {\dfrac{{{K_B}T}}{m}} where KB{K_B} is the Boltzmann’s constant, TT is the temperature and mm is the mass of the gas.
A) 2⋅12 \cdot 1
B) 4⋅04 \cdot 0
C) 3⋅23 \cdot 2
D) 3⋅03 \cdot 0

Explanation

Solution

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 nn .

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 He{\text{He}} is ηHe=2×10−5kgm−1s−1{\eta _{He}} = 2 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}} and that of CH4{\text{C}}{{\text{H}}_{\text{4}}} is ηCH4=1⋅1×10−5kgm−1s−1{\eta _{C{H_4}}} = 1 \cdot 1 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}} .
The diameter of the He{\text{He}} atom is given to be dHe=2⋅1×10−10m{d_{He}} = 2 \cdot 1 \times {10^{ - 10}}{\text{m}} and that of the CH4{\text{C}}{{\text{H}}_{\text{4}}} atom is given as dCH4=n×10−10m{d_{C{H_4}}} = n \times {10^{ - 10}}{\text{m}}. We have to determine nn .
The coefficient of viscosity η\eta is given to be depending on the mass mm, the effective diameter dd and the mean speed vv of the molecules.
So we represent the general relation by η=kmadbvc\eta = k{m^a}{d^b}{v^c} -------- (1) where k, a, b, ck{\text{, }}a{\text{, }}b{\text{, }}care some constants.
Step 2: Using dimensional analysis determine the values of a, b, ca{\text{, }}b{\text{, }}c.
The dimension of the coefficient of viscosity is η→[ML−1T−1]\eta \to \left[ {M{L^{ - 1}}{T^{ - 1}}} \right] .
The dimension of the mass of the molecule is m→[M]m \to \left[ M \right] .
The dimension of the effective diameter is d→[L]d \to \left[ L \right] .
The dimension of the mean speed of the molecules is v→[LT−1]v \to \left[ {L{T^{ - 1}}} \right] .
Rewriting equation (1) in terms of the dimensions of η,m,d,v\eta ,m,d,v we get, [ML−1T−1]=k[M]a[L]b[LT−1]c\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 ML−1T−1=kMaLb+cT−cM{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 MM we have, a=1a = 1
For TT we have −c=−1 - c = - 1
⇒c=1\Rightarrow c = 1
For LL we have b+c=−1b + c = - 1
⇒b=−1−c=−1−1=−2\Rightarrow b = - 1 - c = - 1 - 1 = - 2
Thus the values of the constants are a=1a = 1, b=−2b = - 2 and c=1c = 1 .
So the relation becomes η=kmvd2\eta = \dfrac{{kmv}}{{{d^2}}} --------- (3)
Step 3: Incorporate the given proportionality of the mean speed of the molecules to find the value of nn .
It is given that v∝KBTmv \propto \sqrt {\dfrac{{{K_B}T}}{m}} but as KB{K_B} is the Boltzmann’s constant and TT is the room temperature, we have v∝m−1/2v \propto {m^{ - 1/2}} ------(4)
Incorporating proportionality (4) into equation (3) we get, η=kmm−1/2d2=kmd2\eta = \dfrac{{km{m^{ - 1/2}}}}{{{d^2}}} = \dfrac{{k\sqrt m }}{{{d^2}}}
⇒ηd2m=constant\Rightarrow \eta {d^2}\sqrt m = {\text{constant}}
Then we have ηHedHe2mHe=ηCH4dCH42mCH4{\eta _{He}}{d_{He}}^2\sqrt {{m_{He}}} = {\eta _{C{H_4}}}{d_{C{H_4}}}^2\sqrt {{m_{C{H_4}}}}
⇒dCH4=dHeηHeηCH4(mHemCH4)1/4\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 ηHe=2×10−5kgm−1s−1{\eta _{He}} = 2 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}, ηCH4=1⋅1×10−5kgm−1s−1{\eta _{C{H_4}}} = 1 \cdot 1 \times {10^{ - 5}}{\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}, dCH4=n×10−10m{d_{C{H_4}}} = n \times {10^{ - 10}}{\text{m}}, dHe=2⋅1×10−10m{d_{He}} = 2 \cdot 1 \times {10^{ - 10}}{\text{m}}, mHe=4×10−3kg{m_{He}} = 4 \times {10^{ - 3}}{\text{kg}} and mCH4=16×10−3kg{m_{C{H_4}}} = 16 \times {10^{ - 3}}{\text{kg}} in equation (5) we get, n×10−10=2⋅1×10−102×10−51⋅1×10−5(16×10−34×10−3)1/4n \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}}
⇒n=2⋅1×2×10−51⋅1×10−5(164)1/4=4⋅0\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 nn as 4.

So the correct option is B.

Note: The atomic mass of Helium is 4⋅00u4 \cdot 00{\text{u}}, that of Carbon is 12⋅01u12 \cdot 01{\text{u}} and that of hydrogen is 1⋅01u1 \cdot 01{\text{u}} as obtained from a periodic table. So the molecular mass of Helium will be 4×10−3kg4 \times {10^{ - 3}}{\text{kg}} and that of CH4{\text{C}}{{\text{H}}_{\text{4}}} will be 12+(4×1)=16×10−3kg12 + \left( {4 \times 1} \right) = 16 \times {10^{ - 3}}{\text{kg}}. These values are substituted in equation (5) as mHe=4×10−3kg{m_{He}} = 4 \times {10^{ - 3}}{\text{kg}} and mCH4=16×10−3kg{m_{C{H_4}}} = 16 \times {10^{ - 3}}{\text{kg}} .