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Two Absolute scales X and Y

Question :- Two absolute scales X and Y assigned numerical values 200 and 450 to the triple point of water. What is the relation between $T_{X}$ and $T_{Y}$ ?

(A) 9T_x = 4T_Y

(B) 4T_x = 9T_Y

(D) None of these

Solution :-

We are going to use the formula,

$\displaystyle \frac{{{T}_{C}}-0}{100-0}=\frac{{{T}_{F}}-32}{212-32}=\frac{T-273}{373-273}=\frac{{{T}_{R}}-0}{80-0}=\frac{{{T}_{{{R}_{N}}}}-492}{672-492}=\underbrace{\frac{T-{{T}_{L}}}{{{T}_{U}}-{{T}_{L}}}}_{For\text{ }Unknown\text{ }Scale}$

Here both X and Y are unknown scales, so the above formula becomes

$\displaystyle \underbrace{\frac{{{T}_{X}}-{{T}_{L}}}{{{T}_{U}}-{{T}_{L}}}}_{Unknown\text{ }Scale\text{ X}}=\underbrace{\frac{{{T}_{Y}}-{{T}_{L}}}{{{T}_{U}}-{{T}_{L}}}}_{Unknown\text{ }Scale\text{ Y}}$ …..(1)

Here T_U = Upper Fixed Point (U.F.P.) [boiling point of water] &

T_L = Lower Fixed Point (L.F.P.) [melting point of ice]

We know the triple point of water on Kelvin scale, $T_{k} = 273.1 6K ≅ 273K = L.F.P.$

A.T.Q., triple point of water on absolute scale X, $T_{X} = 200 X$

⇒ $273K = 200 X$

$\displaystyle 1K=\frac{200}{273}X$

$\displaystyle \Rightarrow 373K=\left( \frac{200}{273}\times 373 \right)X$ = $U.F.P.$

So $\displaystyle {{\left( {{T}_{L}} \right)}_{X}}=200X$ and $\displaystyle {{\left( {{T}_{U}} \right)}_{X}}=\left( \frac{200}{273}\times 373 \right)X$

Similarly for scale Y,

$\displaystyle {{\left( {{T}_{L}} \right)}_{Y}}=450Y$ and $\displaystyle {{\left( {{T}_{U}} \right)}_{Y}}=\left( \frac{450}{273}\times 373 \right)Y$

Using equation (1)

$\displaystyle \frac{{{T}_{X}}-200}{\left( \frac{373}{273}\times 200 \right)-\left( 200 \right)}=\frac{{{T}_{Y}}-450}{\left( \frac{373}{273}\times 450 \right)-\left( 450 \right)}$