= 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 $-m\tau_c v_y\,du_x/dy$. The resulting transverse <momentum flux> is
$$
\Pi_{xy}\simeq-nm\tau_c\langle v_y^2\rangle\frac{du_x}{dy}.
$$
Thus the <dynamic viscosity> is $\eta\simeq\rho\tau_c\langle v_y^2\rangle$. With isotropy, a representative collision time $\tau_c\sim\lambda/\langle v\rangle$, and a chosen thermal speed average, this has the form
$$
\boxed{\eta=K\rho\langle v\rangle\lambda.}
$$
The 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 $p=\rho\langle v^2\rangle/3$ then gives
$$
\frac p\eta=\frac{\langle v^2\rangle}{3K\langle v\rangle^2}\frac{\langle v\rangle}{\lambda}.
$$
For the <Maxwell speed distribution>, $\langle v^2\rangle/\langle v\rangle^2=3\pi/8$, so
$$
\boxed{\frac p\eta=\frac\pi{8K}\,\nu_{\rm coll},\qquad \nu_{\rm coll}\simeq\frac{\langle v\rangle}{\lambda}.}
$$
This 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>, $\langle\mathbf v\rangle=\mathbf u$, from the mean thermal speed $\langle|\mathbf v-\mathbf u|\rangle$. 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,
$$
u_\varphi(y)\simeq\omega R\left(1-\frac yh\right),\qquad 0\le y\le h.
$$
Across either end gap, at cylindrical radius $s$, the corresponding profile is $u_\varphi(s,z)\simeq\omega s(1-z/h)$, 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> $2\pi\eta\omega R^3H/h$. \b[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 is
$$
d\tau_{\rm end}=s\left(\frac{\eta\omega s}{h}\right)2\pi s\,ds,
\qquad \tau_{\rm end}=\frac{\pi\eta\omega R^4}{2h}.
$$
Adding the side and both ends gives the <end-face torque in narrow-gap cylindrical viscometry>:
$$
\boxed{\tau=C\eta\omega,\qquad C=\frac\pi h\left(2R^3H+R^4\right).}
$$
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 $d_{\rm Ar}$, $\lambda=1/(\sqrt2n\pi d_{\rm Ar}^2)$ and $\rho=nm_{\rm Ar}$. Therefore
$$
d_{\rm Ar}=\left[\frac{K m_{\rm Ar}\langle v\rangle}{\sqrt2\pi\eta}\right]^{1/2},\qquad
\langle v\rangle=\sqrt{\frac{8k_BT}{\pi m_{\rm Ar}}}.
$$
Using $m_{\rm Ar}=40m_u$, $T=273.15\,\mathrm K$ and the simple estimate $K=1/3$ gives
$$
\boxed{\langle v\rangle\simeq380\,\mathrm{m\,s^{-1}},\qquad d_{\rm Ar}\simeq3.0\times10^{-10}\,\mathrm m.}
$$
This 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.
Back to article page
