To solve this problem, we can use the concept of dew point temperature and the formula for relative humidity.
First, we need to find the dew point temperature at the original conditions of the room (30°C and 62.5% relative humidity). We can use a dew point calculator or a psychrometric chart for this. Assuming atmospheric pressure of 101.3 kPa, we find that the dew point temperature is approximately 22.2°C.
Next, we can use the dew point temperature and the new temperature of 22°C to find the new absolute humidity (the amount of water vapor per unit volume of air) using the formula:
A2 = A1 x (P2/P1) x (T2/T1)
where A1 is the absolute humidity at the original conditions, P1 is the atmospheric pressure at the original conditions, T1 is the temperature at the original conditions, and A2, P2, and T2 are the corresponding values at the new conditions.
Assuming that the atmospheric pressure remains constant at 101.3 kPa, we have:
A2 = A1 x (T2/T1)
= (0.622 x e_s(T2))/(P - e_s(T2)) x (T2 + 273.15 K)/(T1 + 273.15 K)
where e_s(T) is the saturation vapor pressure at temperature T, which can be approximated using the Antoine equation:
log10(e_s/Torr) = A - B/(T/°C + C)
with coefficients A = 8.07131, B = 1730.63, and C = 233.426 for water vapor.
Using these values and the given temperatures, we find that:
e_s(30°C) = 4242.6 Torr
e_s(22°C) = 1995.6 Torr
A1 = 0.0155 kg/m3 (from the dew point temperature of 22.2°C)
Therefore:
A2 = 0.0155 x (1995.6/4242.6) x (295.15/303.15)
= 0.0088 kg/m3
Finally, we can use the formula for relative humidity:
RH = A/As x 100%
where A is the absolute humidity at the new conditions and As is the saturation absolute humidity at the same temperature, which can be found using the same Antoine equation as before:
e_s(22°C) = 1995.6 Torr
As = 0.622 x e_s(22°C)/(101.3 - e_s(22°C))
= 0.0114 kg/m3
Therefore:
RH = A/As x 100%
= 0.0088/0.0114 x 100%
= 77.2%
So the relative humidity in the room after the water vapor was removed is approximately 77.2%.