Water properties – saturated vapor pressure at the normal boiling point At 100°C (the normal boiling point of water), the saturated vapor pressure of water is approximately equal to what pressure in bar?

Introduction / Context:
The normal boiling point of a liquid is defined as the temperature at which its vapor pressure equals the standard atmospheric pressure. For water, this reference point is foundational in steam tables, heat-transfer calculations, and unit operations design.

Given Data / Assumptions:

• Normal boiling point of water is 100°C at 1 atmosphere.
• Standard atmospheric pressure = 1.013 bar ≈ 101.325 kPa.

Concept / Approach:
At the normal boiling point, phase equilibrium requires p_sat,water = p_atm. Thus the correct numerical value for saturated vapor pressure at 100°C is the atmospheric pressure in consistent units.

Step-by-Step Solution:

Verification / Alternative check:
Steam tables list p_sat at 100°C as 101.325 kPa = 1.01325 bar, confirming the choice.

Why Other Options Are Wrong:

• 0.1013 bar: That is about 0.1 atm (10.13 kPa), much too low.
• 10.13 or 101.3 bar: Exceedingly high; 101.3 bar corresponds to about 100 atm.

Common Pitfalls:
Mixing up kPa and bar; note 1 bar = 100 kPa, not 10 kPa.

Final Answer:
1.013