When the bulb of standard gas thermometer is placed in melting ice, the pressure recorded is 83 cm of Hg and when immersed in boiling water, it is 113 cm of Hg. When the bulb is placed in another boiling liquid, the pressure recorded is 143 cm of Hg. Find the boiling point of the liquid.

When the bulb of standard gas thermometer is placed in melting ice, th

1.	When the bulb of standard gas thermometer is placed in melting ice, the pressure recorded is 83 cm of Hg and when immersed in boiling water, it is 113 cm of Hg. When the bulb is placed in another boiling liquid, the pressure recorded is 143 cm of Hg. Find the boiling point of the liquid.
Answer» Solution :Pressure of the gas at `0^(@)C=83` cm of Hg `(P_(0))` , Pressure of the gas at `100^(@)C=113` cm of Hg `(P_(100))` Pressure of the gas at the BOILING point of the LIQUID `(t^(0)C)=143` cm of Hg `(P_(t))` At `0^(0)C` and `100^(0)C`, the pressure of the gas are considered Pressure coefficient, `beta=(P_(100)-P_(0))/(P_(0) times 100)`, At `0^(0)C" and "t_(0)C` the pressure of the gas are considered, `beta=(P_(t)-P_(0))/(P_(0)t)" "THEREFORE (P_(100)-P_(0))/(P_(0) times 100)=(P_(1)-P_(0))/(P_(0)t) rArr t=(P_(t)-P_(0))/(P_(100)-P_(0)) times 100=(143-83)/(113-83) times 100=63/30 times 100=200^(0)C`