Question

Calculate the viscosity of air at (a) 273 K, (b) 298 K, (c) 1000 K. Take
σ ≈ 0.40 nm2
. (The experimental values are 173 µP at 273 K, 182 µP at 20°C,
and 394 µP at 600°C

Answer 1

M = 29*10^-3 kg/mol

σ = 0.4*10^-18 m² (collision cross-section)

$\eta=(M*R*T)$ ^0.5/($\pi$^0.5*N_A* σ)

(a) $\eta=(29*10$ ^-3*8.314*273)^0.5/(3.14^0.5*6.02*10²³* 0.4*10^-18) = 1.90*10^-5 kg/(m*s) = 190 µP

(b) $\eta=(29*10$ ^-3*8.314*298)^0.5/(3.14^0.5*6.02*10²³* 0.4*10^-18) = 1.99*10^-5 kg/(m*s) = 199 µP

(c) $\eta=(29*10$ ^-3*8.314*1000)^0.5/(3.14^0.5*6.02*10²³* 0.4*10^-18) = 3.64*10^-5 kg/(m*s) = 364 µP