1.

The viscosity of molten sodium is 4.5 xx 10^(-4) Nm^(-2)sat 473 K and 2.12 xx 10^(-4) Nm^(-2) s at 873 K. Calculate the activation energy for the viscous flow. Also, calculate the viscosity at 673 K.

Answer»

Solution :We have,
`log eta_(1)/eta_(2) = E/(2.303 R) ((T_(2)-T_(1))/(T_(1)T_(2)))`
`log(4.5 xx 10^(-4))/(2.12 xx 10^(-4)) = E/(2.303 xx 8.314) ((873-473)/(473 xx 873))`
E=6461 J/mole
=6.461 kJ/mole
Further, let the viscosity at 673 K be `eta_(2)`
Again, we have, `T_(1) = 473 K, eta_(1) = 4.5 xx 10^(-4)`
`T_(2) = 673 K, eta_(2)`= ?
`log (4.5 xx 10^(-4))/eta_(2) = 6461/(2.303 xx 8.314) ((673-473)/(473 xx 673))`
`eta_(2) = (2.76 xx 10^(-4) Nm^(-2)s)`


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