1. What is the Ideal Gas Law for Water Vapor?

The ideal gas law for water vapor estimates the density of water vapor given pressure and temperature. The specific gas constant for water vapor is 461.5 J/(kg·K).

2. How Does the Calculator Work?

The calculator uses the ideal gas equation for water vapor:

Where:

• \( \rho \) — Density (kg/m³)
• \( P \) — Pressure (Pascals)
• \( T \) — Temperature (Kelvin)
• 461.5 — Specific gas constant for water vapor (J/(kg·K))

Explanation: The equation shows that density increases with pressure and decreases with temperature.

3. Importance of Water Vapor Density Calculation

Details: Calculating water vapor density is important for atmospheric studies, HVAC design, industrial processes, and understanding humidity effects.

4. Using the Calculator

Tips: Enter pressure in Pascals and temperature in Kelvin. All values must be positive (pressure > 0, temperature > 0).

5. Frequently Asked Questions (FAQ)

Q1: Why 461.5 for water vapor?
A: This is the specific gas constant (R) for water vapor, calculated as the universal gas constant (8.314 J/(mol·K)) divided by the molar mass of water (0.018 kg/mol).

Q2: How accurate is this for real water vapor?
A: This is an ideal gas approximation. For more accurate results at high pressures or near condensation, more complex equations of state should be used.

Q3: What are typical water vapor densities?
A: At room temperature (298K) and saturation pressure (~3173 Pa), density is about 0.023 kg/m³. Values vary greatly with temperature and humidity.

Q4: How to convert Celsius to Kelvin?
A: Add 273.15 to Celsius temperature to get Kelvin (K = °C + 273.15).

Q5: What pressure units can be used?
A: The calculator uses Pascals. For other units: 1 atm = 101325 Pa, 1 bar = 100000 Pa, 1 mmHg ≈ 133.322 Pa.