Past exam of the thermal physics course of the University of Oxford / 2016 / 7 / Solution

Let the gas have a small bulk velocity $u_x(y)$, an approximately isotropic local thermal velocity distribution, and a mean free path $\lambda$ short compared with the flow-gradient scale. A molecule transports the mean streamwise momentum acquired near its previous collision. To first order, its nonequilibrium momentum relative to the destination layer is approximately $-m{\tau }_c{v}_{y}d{u}_x/dy$. The resulting transverse momentum flux isThus the dynamic viscosity is $\eta \simeq \rho {\tau }_c⟨v_y^2⟩$. With isotropy, a representative collision time ${\tau }_c\sim \lambda /⟨v⟩$, and a chosen thermal speed average, this has the formThe dimensionless $K$ depends on how speeds and collision times are averaged; a simple one-speed estimate gives $K=1/3$. Assumptions include dilute binary collisions, a locally equilibrated distribution, weak shear, and negligible boundary slip. The kinetic theory identity $p=\rho ⟨v^2⟩/3$ then givesFor the Maxwell speed distribution, $⟨v^2⟩/⟨v{⟩}^2=3\pi /8$, soThis explains both the units of inverse time and the relation to molecular collision frequency, up to the averaging convention.

In the cylinder apparatus, distinguish the vector mean molecular velocity, $⟨\mathbf{v}⟩=\mathbf{u}$, from the mean thermal speed $⟨|\mathbf{v}-\mathbf{u}|⟩$. For a slow nearly isothermal Couette flow, the thermal speed is approximately position-independent, whereas the bulk mean is azimuthal and changes from the local rotating-wall speed to zero at the fixed wall. In the side gap, with $y$ measured radially from the inner wall,Across either end gap, at cylindrical radius $s$, the corresponding profile is $u_{\phi }(s,z)\simeq \omega s(1-z/h)$, with $z$ measured away from the moving end face. The no-slip boundary condition and small $h/R,h/H$ underlie these linear profiles.

The side-wall shear stress has magnitude $\eta \omega R/h$, producing resisting torque $2\pi \eta \omega R^{3}H/h$. The original diagram also has a gap at both circular end faces. At either face, the stress is $\eta \omega s/h$ and its ring contribution isAdding the side and both ends gives the end-face torque in narrow-gap cylindrical viscometry:The motor supplies the opposite of this viscous resisting torque. Corner flow and the small axle area are neglected. Retaining only the cylindrical wall would omit the $\pi R^4/h$ term shown by the actual closed-ended geometry.

For hard-sphere Ar atoms of collision diameter $d_{\mathrm{Ar}}$, $\lambda =1/(\sqrt{2}n\pi d_{\mathrm{Ar}}^2)$ and $\rho =n{m}_{\mathrm{Ar}}$. ThereforeUsing $m_{\mathrm{Ar}}=40m_u$, $T=273.15\mathrm{K}$ and the simple estimate $K=1/3$ givesThis is a kinetic estimate with a model-dependent numerical prefactor, rather than a precise atomic size measurement. Since the density cancels and the hard-sphere collision diameter is taken constant, the prediction is $\eta \propto \sqrt{T}$. The collision cross-section of a real atom can vary with speed, so its exact temperature dependence need not follow this power law.