question_answer 4)

Answer the following: (a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)? (b) There were two fixed points in the origin Celsius scale as mentioned above which were assigned; the number \[{{0}^{o}}C\] and \[{{100}^{o}}C\] respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale? (c) The absolute temperature (Kelvin scale) T is related to the temperature t, on the Celsius scale by\[=0\cdot 5\times {{10}^{-3}}{{m}^{3}}.\] Why do we have 273.15 in this relation, and not 273.16? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Answer:

(a) This is on account of the fact that the triple point of water has a unique value i.e. 273.16 K at a unique point, where exists unique values of pressure and volume. On the other hand, the melting point of ice and boiling point of water do not have unique set of values as they change with the change in pressure and volume. (b) The other fixed point on the Kelvin absolute scale is the absolute zero itself. (c) On Celsius scale \[{{0}^{o}}C\] corresponds to melting point of ice at normal pressure. The corresponding value of absolute temperature is 273.15 K. The temperature 273.16 K corresponds to the triple point of water. From the given relation, the corresponding value of triple point of water on Celsius scale \[\frac{\text{Bulk modulus of water}}{\text{Bulk modulus of air}}\] (d) We know that Fahrenheight scale and Absolute scale are related as \[=\frac{2\cdot 026\times {{10}^{9}}}{1\cdot 0\times {{10}^{5}}}\] (i) For another set of temperature \[{{T}_{F}}\] and \[T{{'}_{K}}\], \[1\cdot 03\times {{10}^{3}}g{{m}^{-3}}?\] ..(ii) Subtracting (i) from (n), we have \[=45\cdot 8\times {{10}^{-11}}p{{a}^{-1}}.\] or \[=1\cdot 013\times {{10}^{5}}pa.\] If \[p=80\cdot 0\times 1\cdot 013\times {{10}^{5}}pa;\]then \[\frac{1}{B}=45\cdot 8\times {{10}^{-11}}P{{a}^{-1}}\] For a temperature of triple point i.e. 273.16 K, the temperature on the new scale is \[\text{V=}\frac{\text{M}}{\text{P}}\text{ and V }\!\!'\!\!\text{ =}\frac{\text{M}}{\text{P }\!\!'\!\!\text{ }}\]