Question

A constant-volume gas thermometer registers an absolute pressure corresponding to 325 mm of mercury when in contact with water at the triple point. What pressure does it read when in contact with water at the normal boiling point?

Answer 1

Answer:

$P = 444$ mm of Hg

Explanation:

As we know that volume is constant so by ideal gas equation we have

$PV = nRT$

now for constant volume condition we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2}$

here we know that

$P_1 = 325$ mm of Hg

$T_1 = 273 K$

$T_2 = 373 K$

now from above equation

$\frac{325}{P} = \frac{273}{373}$

$P = 444$ mm of Hg