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)

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