t = a ln p + b ...(Given)
On Celsius scale :
Ice point = 0°C, and
Steam point = 100°C
\(\therefore\) From given conditions, we have
0 = a ln 1.5 + b ...(i)
and 100 = a ln 7.5 + b ...(ii)
i.e., 0 = a × 0.4054 + b ...(iii)
and 100 = a × 2.015 + b ...(iv)
Subtracting (iii) from (iv), we get
100 = 1.61a
or a = 62.112
Substituting this value in eqn. (iii), we get
b = – 0.4054 × 62.112= – 25.18
∴ When p = 3.5 the value of temperature is given b
t = 62.112 ln (3.5) – 25.18= 52.63°C.