To get the humidity change: 0.8-0.5=0.3
Humidity change times density of water vapour gives the density of water vapour.
0.3*22.8= 6.84
This times volume gives the mass of water vapour to be removed.
2000*6.84=13.7
A concert hall of volume 2000 m3
contains air at a temperature of 25 ◦C and relative humidity 80 %. Given
that the density of water vapour in saturated air at 25 ◦C is 22.8 g/m3 and that the temperature and
pressure of the concert hall remain constant what mass of water vapour must be removed from this air to
reduce the relative humidity to 50 %?
1 answer
1 answer