Question

The viscosity of gases increases with temperature rather than the expected decrease we see with liquids.

If the constant in the relation for viscosity are a = 0.00000132 kg/(m s K^1/2) and b = 120 K then what is the viscosity of the gas (kg/(m s)) at T = 328 Ceclius?

Answer 1

According to the Principle of homogeneity of equations units on both sides of any equation should be the same.

The relation we want is in between viscosity and temperature, so both sides should have units of viscosity in the term.

We can develop an equation in terms of Temperature 'T' and given constants 'a' and 'b' such that its unit is consistent with that of viscosity.

Unit of 'a' suggest that if we multiply a temperature term raised to the power of 1/2 we get the unit of viscosity, but the unit of 'b' is K which suggest that it should be added with 'T'.

Hence we can develop a relation as:

viscosity = a(T + b)^1/2

So the viscosity at 328 Celcius is

=> 0.00000132 * ((273+328) + 120)^1/2 [273 is added to convert into Kelvin scale]

=> viscosity = 3.674x10^-5kg/m s