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’ a constant If η in a function of mass ‘m’ of the molecules, their mean speed v, and the constant µ, use method of dimensions to show that 
. If n ∝ T s , where T is the temperature and ‘S’ a number and for helium gas S = 0.68, determine n for helium. If S = 0.98 for carbon dioxide, determine n for CO 2 . Are your answers sensible in terms of the molecular structure of the two gases.
Text Solution
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Sol. Dimension of [ η ] ≡ 
≡ 
[ η ] ≡ [ML –1 T –1 ]
Dimensions of µ = Fr n [µ] = [MLT
–2 ]L n [µ] = ML
n+1 T –2 Let ‘ η ’ depend on mass ‘m’ mean speed ‘ ν ’ and constant ‘µ’ as –
η ∝ m
a v b µ c [ML
–1 T –1 ] ∝ M a [LT –1 ] b [ML n+1 T –2 ] c [ML
–1 T –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
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 = 
a = 1 – c
a = 
= 
a = 
b = 1 – 2c = 
= 
b = 
∴ η ∝ 
ν ∝ T
1/2 with T the temperature
given η ∝ T s hence S = 
If S = 0.68 for helium, n = 12.1 and if S = 0.98 for CO 2 n = 5.2.
Helium has closed shell atomic structure and CO 2 is a linear molecule. It is possible that the attractive force between helium atoms will fall off with distance more rapidly than that between carbon dioxide molecules.
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