A constant volume thermometer using the gas reads a pressure of `1.75xx 10^(4)` Pa at normal freezing of water and reads` 2.39 xx 10^(4)` pa at normal boiling point of water. Obtained from the observation the temperature of absolute zero on celcius scale.

Explanation

Let temperature on kelvin scale of freezing point of water =`=T_(1) K=2.93xx10^(4) Pa`
Let temperature on kelvin scale of boiling point of water `=T_(2)=K=1.75xx10^(4) Pa`
As volume is constant so by universal gas law
`PV=nRT`= as volume and mass of gas in thermometer constant, so
`P prop T or (P)/(T)`= constant
`(P_(1))/(T_(1))=(P_(2))/(T_(2))= or (T_(2))/(T_(1))=(P_(2))/(T_(1))=(2.39xx10^(4))/(1.75xx10^(4))=(239)/(175)`
s `T=t_(c)+ tT_(0) " "...(i)`
`T_(1)=0 +T_(0) " T_(2)= 100+T_(0)`
`:. (100+T_(0))/(T_(0))=(239)/(175)`
`(100)/(T_(0))+(T_(0))/(T_(0))=(239)/(175)`
`(100)/(T_(0))=(239-175)/(175)`
`:.(100)/(T)=(64)/(175)`
`64T_(0)=100xx175`
`T_(0)=(100xx175)/(64)=273`
From (1)
`t_(c)=T-T_(0)=0-273=-273K`