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The viscosity of mercury decreases with the rise in temperature (Table 21.1a). Check whether relation (34.10) is valid for mercury. Calculate the activation energy. |
Answer» Solution :To check the exponential dependence of viscosity on temperature, find the dependence of the logarithm of viscosity on the reciprocal of the temperature. To do this, compile a NEW table (Table 21.1b) using the data of Table 21.1a. Using the data of the new table, plot  a graph on millimetre graph paper (FIG. 21.1). Almost all the points are SEEN to fall on the straight line `y=a+bx, or log (10^(8) eta)=a+10^(0)+(b)/(T)=a+(B)/(T)`  Taking antilogarrithms we obtain `10^(3)eta=Axx10^(B//T)` the distinction from formula (34.10) being only in numerical coefficients. To find the activation energy e, use two points lying on the straight line: `x_(1)=2.75, y_(1)=0.100 and x_(2)=3.60, y_(2)=0.213` The TEMPERATURES corresponding to these points are `T_(1)=10^(3)//x_(1)=364 K and T_(2)=10^(3)//x_(2)=278K ` and the corresponding visocsities are `eta_(1)=10^(-3)xx10 y_(1)=1.26xx10^(-3)` Pa, * s and `eta_(2)=1.63xx10^(-3)Pa*s`. The ratio or the viscosities is `eta_(2):eta_(1)=e^(epsi_(0)//kT_(e)), epsi^(epsi_(0)//hT_(1))=exp {(epsi_(0)(T_(1)-T_(2)))/(kT_(1)T_(2))}` Hence we obtain after taking the logarithms `(epsi_(0)(T_(1)-T_(2)))/(kT_(1)T_(2))=ln(eta_(2))/(eta_(1))` or `epsi_(0)=(kT_(1)T_(2)ln(eta_(2)//eta_(1)))/(T_(1)-T_(2))-(2.3kT_(1)T_(2)log(eta_(2)//eta_(1)))/(T_(1)-T_(2))` |
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