Question

The viscosity η of a gas depends on the long-range attractive part of the intermolecular force, which varies with molecular separation ‘r’ according to F = μr^–n where n is a number and μ is a constant. If η  is a function of mass ‘m’ of the molecules, their mean speed v, and the constant μ, then which of following is correct –

• A. η∝ mⁿ⁺¹ vⁿ⁺³μ^n–2
• B. η∝$m^{\frac{n + 1}{n–1}}$ $v ^ { \frac { \mathrm { n } + 3 } { \mathrm { n } - 1 } }$ $\mu^{\frac{–2}{n–1}}$
• C. η∝mⁿ v^–nμ^–2
• D. η∝ mv μ^–n

Answer 1

Answer: (B)

η∝$m^{\frac{n + 1}{n–1}}$ $v ^ { \frac { \mathrm { n } + 3 } { \mathrm { n } - 1 } }$ $\mu^{\frac{–2}{n–1}}$

Answer 2

Dimension of [η] ≡ $\left\lbrack \frac{Fr}{Av} \right\rbrack$ ≡ $\frac{\lbrack MLT^{- 2}\rbrack\lbrack L\rbrack}{\lbrack L^{2}\rbrack\lbrack LT^{- 1}\rbrack}$

[η] ≡ [ML^–1T^–1]

Dimensions of µ = Frⁿ

[µ] = [MLT^–2]Lⁿ

[µ] = MLⁿ⁺¹T^–2

Let ‘η’ depend on mass ‘m’ mean speed ‘v’ and constant ‘µ’ as –

η ∝ m^av^bµ^c

[ML^–1T^–1] ∝ M^a[LT^–1]^b [MLⁿ⁺¹T^–2]^c

[ML^–1T^–1] ∝ M^a+c L^{b+c(n + 1)} T^–b–2c

equating dimensions both sides

a + c = 1⇒     c = 1 – a

b + c (n + 1) = –1⇒     b = – [1 + c (n + 1)]

– (b + 2c) = – 1⇒     b = 1 – 2c

$\therefore$1 – 2c = – [ 1 + c (n + 1)]

2c –1 = 1 + c (n + 1)

2c – c (n + 1) = 2

c [2 – n – 1] = 2

c [1 – n] = 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 = $\frac{n + 3}{n - 1}$

∴ η ∝ $m^{\frac{n + 1}{n - 1}}v^{\frac{n + 3}{n - 1}}µ^{- \frac{2}{n - 1}}$