Dew point and absolute humidity

AI Thread Summary
The discussion centers on calculating absolute moisture at a room temperature of 200°C with a relative humidity of 60%. The initial calculations used an incorrect vapor pressure unit, leading to a result of 7.86 g/m³, which was later corrected to 10.3 g/m³. Participants highlighted the importance of using the correct gas constant and pressure conversions in the calculations. The final consensus confirms that the absolute moisture is approximately 10.3 g/m³. Accurate unit conversions and the application of the ideal gas law are crucial for such calculations.
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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