A constant volume thermometer using helium gas records a pressure of `1.75 xx 10^4` Pa at normal freezing point of water, and a pressure of `2.39 xx 10^4` Pa at normal boiling point of water. Obtain from these observations the temperature of absolute zero on the Celsius scale.

Answer

273

Explanation

Let `T_1` and `T_2` be the absolute temperatures corresponding to normal freezing and normal boiling points (i.e., under 1 atmospheric pressure) of water. For a constant volume thermometer,
`(T_2)/(T_1)=(P_2)/(P_1)=(2.39 xx 10^4)/(1.75 xx 10^4)=(239)/(175)`
Now the temperature `t_c` on Celsius scale and temperature T on the Kelvin scale are related as
`T=t_c+T_0`
Where `T_0` is a constant. As the normal freezing point and normal boiling point of water are `0^@C` and `100^@C` respectively, so from the above equation, we get
`T_1=0+T_0` and `T_2=100 +T_0`
`therefore (T_0+100)/(T_0)=(239)/(175)`
or `(100)/(T_0)=(239)/(175)-1=(64)/(175)`
or `T_0=(100xx175)/(64)=273`
Hence the absolute zero (T = 0) on the Celsius scale will be
`t_c=T-T_0=0-273=-273`
The exact value of absolute zero is `-273.15`