Question

The viscosity $\eta$ of a gas depends on the long-range attractive part of the intermolecular force, which varies with molecular separation $r$ according to $F \, = \, \mu r^{- n}$ where $n$ is a number and $\mu $ is a constant. If $\eta$ is a function of the mass $m$ of the molecules, their mean speed $v$ , and the constant $\mu $ then which of following is correct -

• A. $\eta \propto m^{n + 1}v^{n + 3}\mu ^{n - 2}$
• B. $\eta \propto m^{\frac{n + 1}{n - 1}} \, v^{\frac{n + 3}{n - 1}} \, \mu ^{\frac{- 2}{n - 1}}$
• C. $\eta \propto m^{n}v^{- n}\mu ^{- 2}$
• D. $\eta \propto mv \, \mu ^{- n}$

Answer 1

Answer: (B)

$Dimension \, of \, \left[\eta\right]\equiv \left[\frac{F r}{A v}\right]\equiv \frac{\left[M L T^{- 2}\right] \left[\right. L \left]\right.}{\left[L^{2}\right] \left[L T^{- 1}\right]}$
$ \, \left[\eta\right]\equiv \left[M L^{- 1} T^{- 1}\right]$
Dimensions of $\mu =Fr^{n}$
$\left[\mu \right]=\left[M L T^{- 2}\right]L^{n}$
$\left[\mu \right]=ML^{n + 1}T^{- 2}$
Let $‘\eta’$ depend on mass $m$ mean speed $v$ and constant $\mu $ as -
$\eta \propto m^{a}v^{b}\mu ^{c}$
$\left[M L^{- 1} T^{- 1}\right] \propto M^{a}\left[L T^{- 1}\right]^{b}\left[M L^{n + 1} T^{- 2}\right]^{c}$
$\left[M L^{- 1} T^{- 1}\right] \propto M^{a + c}L^{b + c \left(n + 1\right)}T^{- b - 2 c}$
Equating dimensions both sides
$a+c=1\Rightarrow c=1-a$
$b+c\left(n + 1\right)=-1$
$\Rightarrow b=-\left[1 + c \left(n + 1\right)\right]$
$-\left(b + 2 c\right)=-1$
$\Rightarrow b=1-2c$
$\therefore 1-2c=-\left[1 + c \left(n + 1\right)\right]$
$2c-1=1+c\left(n + 1\right)$
$2c-c\left(n + 1\right)=2$
$c\left[2 - n - 1\right]=2$
$c\left[1 - n\right]=2$
$=c=-\frac{2}{n - 1}$
$a=1-c$
$a=1+\frac{2}{n - 1}=\frac{n - 1 + 2}{n - 1}$
$a=\frac{n + 1}{n - 1}$
$b=1-2c$
$=1+\frac{4}{n - 1}=\frac{n - 1 + 4}{n - 1}$
$b=\frac{n + 3}{n - 1}$
$\therefore \eta \propto m^{\frac{n + 1}{n - 1}}v^{\frac{n + 3}{n - 1}}\mu ^{- \frac{2}{n - 1}}$