Dew point and absolute humidity

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SUMMARY

The discussion focuses on calculating absolute moisture from room temperature and relative humidity, specifically at 200°C with a relative humidity of 60%. The vapor pressure at this temperature is established as 17.5 atm, leading to a calculated absolute moisture of approximately 10.3 g/m³. The calculations utilize the ideal gas law (PV=nRT) and require adjustments for pressure units, specifically converting atm to mm Hg. The final consensus confirms that the absolute moisture should be around 10 g/m³, correcting earlier miscalculations.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of vapor pressure and its relation to temperature
  • Familiarity with unit conversions, particularly between atm and mm Hg
  • Basic concepts of humidity and moisture content in gases
NEXT STEPS
  • Research the calculation of vapor pressure using Antoine's equation
  • Learn about the impact of temperature on relative humidity
  • Explore the use of psychrometric charts for humidity calculations
  • Study the implications of absolute humidity in meteorology and HVAC systems
USEFUL FOR

Students in thermodynamics, meteorologists, and professionals in HVAC design who require accurate calculations of humidity and moisture content in air.

Karol
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Homework Statement


Room temperature is 200C and relative humidity is 60%. what is the absolute moisture

Homework Equations


Vapor pressure at 200C=17.5[atm]
PV=nRT
Molecular weight of water=18[gr/mol]

The Attempt at a Solution


The vapor pressure at 200C:
$$\frac{x}{17.5}=0.6\rightarrow x=10.5[atm]$$
$$PV=nRT\rightarrow 10.5\cdot 1=n\cdot 0.08208\cdot 293\rightarrow n=0.437[mol]$$
$$0.437\cdot 18=7.86[\frac{gr}{m^3}]$$
It should be 10[gr/m3]
 
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Karol said:

Homework Statement


Room temperature is 200C and relative humidity is 60%. what is the absolute moisture

Homework Equations


Vapor pressure at 200C=17.5[atm]
That should be mm Hg.
PV=nRT
Molecular weight of water=18[gr/mol]

The Attempt at a Solution


The vapor pressure at 200C:
$$\frac{x}{17.5}=0.6\rightarrow x=10.5[atm]$$
Again, that should be mm Hg
$$PV=nRT\rightarrow 10.5\cdot 1=n\cdot 0.08208\cdot 293\rightarrow n=0.437[mol]$$
$$0.437\cdot 18=7.86[\frac{gr}{m^3}]$$

The gas constant value you used here for liters. You also need to use a pressure of 10.5/760 atm.

Chet
 
$$\frac{x}{0.023}=0.6\rightarrow x=0.0138[atm]$$
$$PV=nRT\rightarrow 0.0138\cdot 1[liter]=n\cdot 0.08208\cdot 293\rightarrow n=0.000574[\frac{mol}{liter}]=0.574[\frac{mol}{m^3}]$$
$$0.574\cdot 18=10.3[\frac{gr}{m^3}]$$
 
Karol said:
$$\frac{x}{0.023}=0.6\rightarrow x=0.0138[atm]$$
$$PV=nRT\rightarrow 0.0138\cdot 1[liter]=n\cdot 0.08208\cdot 293\rightarrow n=0.000574[\frac{mol}{liter}]=0.574[\frac{mol}{m^3}]$$
$$0.574\cdot 18=10.3[\frac{gr}{m^3}]$$
Excellent!

Chet
 

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