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A theoretical colloidal system has its viscosity dependent on temperature like how the rate constant can be dependent with temperature. The following data were obtained for the viscosity of this colloid as a function of temperature.

T (Celsius): -45, -25, -10, 0, 20, 30

Viscosity (Pa-s): 7010, 263, 35, 12, 1.5, 0.7

Calculate the viscosity of this colloidal system (Pa-s) at 298 K.

choices are;

a) 0.90

b) 0.94

c) 0.96

d) 0.98

e) 1.00

Answer 1

Given that;

Viscosity of the colloidal solution is dependent on temperature like how the rate constant can be dependent with temperature.

The dependence of rate constant on temperature is given by the Arrhenius equation:

k = A exp (-Ea / RT)

Let us assume that viscosity has the same functional form. So viscosity as a function of temperature can be written as:

μ = α exp (- β / T)

where;

α and β are some constant

T is the absolute temperature in K.

We have;

μ = α exp (- β / T)

Taking natural logarithm of both the sides:

ln (μ) = ln (α) - β / T

which is the equation of a straight line.

Slope = - β

Intercept = ln (α)

From the given data; we will plot a graph between ln (μ) and 1 / T. Then we will fit a straight line through it.

The plot is as follows:

T (°C)	T(K)	μ (Pa.s)	1 / T (K-1)	ln (μ)
-45	228.15	7010	0.004383081306	8.85509298
-25	248.15	263	0.004029820673	5.572154032
-10	263.15	35	0.003800114003	3.555348061
0	273.15	12	0.003660992129	2.48490665
20	293.15	1.5	0.003411222923	0.4054651081
30	303.15	0.7	0.003298697015	-0.3566749439

The equation of the fitted line is:

ln (μ) = - 28.6 + 8501 / T

We have to find the viscosity of the polymer at T = 298 K.

Substituting this in the equation:

ln (μ) = - 28.6 + 8501 / 298

ln (μ) = -0.0731544

μ = 0.94

Hence option b) is correct.