What is the vapor pressure of water at 105°C?

1 Answer
Feb 8, 2017

You can do this with the Antoine equation:

$\log P = A - \frac{B}{C + T}$

where:

• $P$ is the vapor pressure of water.
• $T$ is the water temperature in celsius.
• $A , B ,$ and $C$ are the Antoine constants for water.

When the water temperature is in the range of $1 - {100}^{\circ} \text{C}$:

$A = 8.07131 , B = 1730.63 , C = 233.426$

When the water temperature is in the range of $99 - {374}^{\circ} \text{C}$:

$A = 8.14019 , B = 1810.94 , C = 244.485$

If you calculate using this way you will get answer in $\text{torr}$:

$P = {10}^{A - \frac{B}{C + T}}$

$= {10}^{8.14019 - \frac{1810.94}{105 + 244.485}}$

$= {10}^{8.14019 - 5.18174}$

$= {10}^{2.95845}$

$=$ $\text{908.76 torr}$

This is the water pressure at ${105}^{\circ} \text{C}$.