The presssure of the gas in a constant volume gas thermometer at steam point (373.15K) is `1.50 xx 10^4 Pa`. What will be the pressure of the triple point of water?

Explanation

The temperature in kelvin is defined as `T=P/P_tr xx 273.16 K`.
Thus,
373.15 = (1.50 xx 10^4 Pa)/(P_tr) xx 273.16`
or, ` P_tr = 1.50 xx 10^4 Pa xx (273.16)/(373.15)`
`=1.10 xx 10^4 Pa`.