Saturated water vapor and the ideal gas law of state
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Dear sea enthusiasts, I have a question: Does low-pressure saturated water vapor conform to the ideal gas law? As an example: What is the volume of 1 kg of water vapor at normal pressure (101.33 KPa, 100°C, ρ=0.597 kg/m3) when the pressure is reduced to 3.3 KPa (the saturation temperature at this pressure is 24°C)? Solution 1: The molar volume Vm method 1. Molar volume of water vapor: Vm = 18 g/mol ÷ 0.597 g/L = 30.15 L/mol ； 2. At 101.33 KPa and 100°C, the volume of 1 kg of water vapor is: V1 = 1000 g ÷ 18 g/mol × 30.15 L/mol = 1675 L ； 3. Using the ideal gas law, the volume at a pressure of 3.3 KPa and a temperature of 24°C is calculated as follows: V2 = 101.33 × 1675 ÷ 100 × 24 ÷ 3.3 = 12343.8 L. Solution 2: Specific volume method. At 101.33 KPa and 100°C, the specific volume of steam is v = 1/ρ = 1 ÷ 0.597 g/L = 1.675 L/g ； Volume of 1000g of water vapor: V1=1.675L/g×1000g=1675L ； 2. Using the ideal gas law, the volume at a pressure of 3.3 KPa and a temperature of 24°C is calculated as follows: V2 = 101.33 × 1675 ÷ 100 × 24 ÷ 3.3 = 12343.8 L. Solution 3: Specific volume method under operating conditions (at 3.3 KPa and 24°C, the density of saturated steam is 0.0239 g/L). 1. At 3.3 KPa and 24°C, the specific volume of steam v = 1/ρ = 1 ÷ 0.0239 g/L = 41.8 L/g ； 2. Volume of 1000g of water vapor: V2 = 41.8 L/g × 1000 g = 41,800 L ； Question: Why is V2 in Solution 1 equal to V2 in Solution 2 but not equal to V2 in Solution 3? And the difference is huge? Is that algorithm correct? How to explain it? I would appreciate your guidance; thank you!
This post was last edited by Qingchengjun on 2017-10-9 at 13:57. The temperature in the gas equation of state is in K, and the pressure value is absolute pressure, not gauge pressure; it’s definitely incorrect to simply use 100 and 24 for multiplication or division. (101.33KPa, 100°C, ρ=0.597kg/m3) Is this steam that you came up with?
Dear Qingcheng Jun! Hello! Thank you for your tip! The meaning of temperature K was ignored! Thank you! Furthermore, 101.33 KPa, 100°C, ρ=0.579 kg/m3 – aren’t these the parameters for saturated steam at standard atmospheric pressure (absolute pressure of 101.33, gauge pressure of 0)? Do these steam parameters not make physical sense?
This post was last edited by Qingchengjun on 2017-10-9 at 14:51. It’s confusing – steam at 100°C and 101 kPa is generally not produced by ordinary boilers using this product. It’s all saturated steam, so the gas state equation can be used directly for calculations
Normal, it seems that tigers do take naps too! Thank you! :lol
Dear sea friends, as suggested by Qingcheng Jun, this issue is summarized as follows for sharing: What is the volume of 1 kg of water vapor at normal pressure (101.33 KPa, 100°C, ρ=0.597 kg/m3) when the pressure is reduced to 3.3 KPa (the saturation temperature at this pressure is 24°C)? Solution 1: The molar volume Vm method 1. Molar volume of water vapor: Vm = 18 g/mol ÷ 0.597 g/L = 30.15 L/mol ； 2. At 101.33 KPa and 100°C, the volume of 1 kg of water vapor is: V1 = 1000 g ÷ 18 g/mol × 30.15 L/mol = 1675 L ； 3. Using the ideal gas law to calculate the volume at a pressure of 3.3 KPa and a temperature of 24°C: V2 = 101.33 × 1675 ÷ (273 + 100) × (273 + 24) ÷ 3.3 = 40953 L. Solution 2: Specific volume method. At 101.33 KPa and 100°C, the specific volume of steam is v = 1/ρ = 1 ÷ 0.597 g/L = 1.675 L/g ； Volume of 1000g of water vapor: V1=1.675L/g×1000g=1675L ； 2. Using the ideal gas law, the volume at a pressure of 3.3 KPa and a temperature of 24°C is calculated as follows: V2 = 101.33 × 1675 ÷ (273 + 100) × (273 + 24) ÷ 3.3 = 40953 L. Solution 3: Specific volume method under operating conditions (at 3.3 KPa and 24°C, the density of saturated steam is 0.0239 g/L). 1. At 3.3 KPa and 24°C, the specific volume of steam v = 1/ρ = 1 ÷ 0.0239 g/L = 41.8 L/g ； 2. Volume of 1000g of water vapor: V2 = 41.8 L/g × 1000 g = 41,800 L. Solution 1: V2 = Solution 2: V2 ≈ Solution 3: V3! Insight: Chemical process calculations require careful attention ； Small mistakes can lead to big errors ； Physics is a rigorous and precise natural science; the same proposition can be proven using various methods, with the results matching almost perfectly! A very meaningful course; I love the field of chemical engineering!