Question

If $\eta$ is the coefficient of viscosity and T is temperature in kelvin, a linear plot is obtained for

• A. $\eta$ vs T
• B. $\eta$ vs 1/T
• C. log $\eta$ vs 1/T
• D. $\eta$ vs log T

Answer 1

Answer: (C)

log $\eta$ vs 1/T

Answer 2

The viscosity ($\eta$) of a liquid is highly dependent on temperature (T). For most liquids, viscosity decreases exponentially with increasing temperature. This relationship is often described by an Arrhenius-type equation:

$\eta = A e^{E_a / RT}$

where:

• $\eta$ is the coefficient of viscosity
• A is a pre-exponential factor (a constant)
• $E_a$ is the activation energy for viscous flow
• R is the ideal gas constant
• T is the absolute temperature in Kelvin

To obtain a linear plot from this exponential relationship, we take the logarithm of both sides of the equation. Taking the natural logarithm ($\ln$):

$\ln \eta = \ln (A e^{E_a / RT})$

$\ln \eta = \ln A + \ln (e^{E_a / RT})$

$\ln \eta = \ln A + \frac{E_a}{RT}$

This equation can be rearranged into the form of a straight line, $y = mx + c$:

$\ln \eta = \left(\frac{E_a}{R}\right) \left(\frac{1}{T}\right) + \ln A$

Here, if we plot:

• $y = \ln \eta$ (or $\log \eta$, as $\log_{10} \eta = \frac{\ln \eta}{\ln 10}$ also yields a linear plot)
• $x = \frac{1}{T}$

The plot will be a straight line with:

• Slope ($m$) $= \frac{E_a}{R}$
• Y-intercept ($c$) $= \ln A$

Therefore, a linear plot is obtained when log $\eta$ is plotted against 1/T.