Dew point and absolute humidity

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Homework Help Overview

The discussion revolves around calculating absolute moisture given a room temperature of 200°C and a relative humidity of 60%. Participants are exploring the relationship between vapor pressure, molar calculations, and the conversion of units in the context of humidity and moisture content.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants attempt to calculate absolute moisture using vapor pressure and the ideal gas law, while also addressing unit conversions and the correct application of constants. Some question the use of atmospheric pressure versus mm Hg in their calculations.

Discussion Status

There are multiple calculations presented, with some participants providing corrections and alternative approaches. Guidance has been offered regarding the appropriate units and constants to use, but no consensus has been reached on the final values.

Contextual Notes

There is a noted discrepancy in the expected results, with participants highlighting potential errors in unit conversions and assumptions about pressure values. The original poster's calculations are challenged, leading to further exploration of the problem.

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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