Relative Humidity Changes in a Closed Room

[slaw155]

Homework Statement

A room with volume 2000 cubic metres has air at T=25degrees Celsius with relative humidity 80%. Density of water vapor in saturated air at 25deg Celsius is 22.8g/m^3. Temperature and pressure of room remains constant. What mass of water vapor must be removed from this air to reduce relative humidity to 50%?

Homework Equations

relative humidity/100 = water vapor pressure/saturated water pressure

The Attempt at a Solution

using above formula and density definition we get (18.3x2000)-(11.4x2000) = 6.9kg.
However the textbook answer is double this? Where have I gone wrong?

 

[jbriggs444]

using density = mass/volume we get (18.3x2000)-(11.4x2000) = 6.9kg.

Try that computation again.