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.

A constant volume thermometer using the gas reads a pressure of 1.75xx

1.	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.
Answer» <P> Solution :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`