The viscosity η of a gas depends on the long-range attractive part...

Dimensionof\left[η\right] ≡ \left[\dfrac{Fr}{Av}\right]≡ \dfrac{ \left[MLT^{- 2}\right] [ L ]}{ \left[L^{2}\right]\left[LT^{- 1}\right]}<br /><br /> \left[η\right] ≡ \left[ML^{- 1}T^{- 1}\right]

Dimensions of $\mu = Fr^{n}$

\left[\mu\right] = \left[MLT^{- 2}\right]L^{n}$<br /><br />\left[\mu\right] =ML^{n + 1}T^{- 2}$

Let $‘η’$ depend on mass $m$ mean speed $v$ and constant $\mu$ as -

$η ∝ m^{a}v^{b}{\mu}^{c}$

\left[ML^{- 1}T^{- 1}\right] ∝ M^{a}{\left[LT^{- 1}\right]}^{b}{\left[ML^{n + 1}T^{- 2}\right]}^{c}$<br /><br />\left[ML^{- 1}T^{- 1}\right] ∝ M^{a + c}L^{b + c\left(n + 1\right)}T^{- b - 2c}$

Equating dimensions both sides

$a + c = 1 ⇒ c = 1 - a$

$b + c \left(n + 1\right) = - 1$

$⇒ b = - \left[1 + c\left(n + 1\right)\right]$

$- \left(b + 2c\right) = - 1$

$⇒ b = 1 - 2c$

$∴ 1 - 2c = - \left[1 + c \left(n + 1\right)\right]$

$2c - 1 = 1 + c \left(n + 1\right)<br /><br />$2c - c \left(n + 1\right)$= 2$

$c\left[2 - n - 1\right]$ = 2$<br /><br />$c\left[1 - n\right]$= 2$

$= c = -$\dfrac{2}{n - 1}$<br /><br />$a = 1 - c$<br /><br />$a = 1 + \dfrac{2}{n - 1} = \dfrac{n - 1 + 2}{n - 1}

$a =$\dfrac{n + 1}{n - 1}

$b = 1 - 2c$

$= 1 +$\dfrac{4}{n - 1} = \dfrac{n - 1 + 4}{n - 1}

$b =$\dfrac{n + 3}{n - 1}

$∴ η ∝$m^{\dfrac{n + 1}{n - 11}}v^{\dfrac{n + 3}{n - 1}}{\mu}^{- \dfrac{2}{n - 1}}