The four sets of lines in the hydrogen emission spectrum are known as Balmer, Brackett, Paschen, and Lyman. For each series, ass
ign the energy (infrared, ultraviolet, or visible), the nr value, and all possible n, values up to 7 Series Balmer Brackett Paschen Lyman Energy nf ni visible) infrared ultraviolet 1 2 3 6 7
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Answer:
A) Partial Pressure of dry air = 13.32 KPa
Partial Pressure of water vapour = 1.332 KPa
B) Humidity ratio; X = 0.0691
C) V_p = 0.8384 m³/Kg
Explanation:
A) We are given;
Temperature = 75°F
Relative Humidity = 45%
Now,to calculate the partial pressure, we will use the relationship;
Relative Humidity = (Partial Pressure/Vapour Pressure) × 100%
Making partial pressure the subject;
Partial Pressure = Relative Humidity × Vapour Pressure/100%
From the first table attached, at temperature of 75°F, the vapor pressure is 29.6 × 10^(-3) bar = 29.6 KPa
Thus;
Partial Pressure of dry air = (45 × 29.6)/100
Partial Pressure of dry air = 13.32 KPa
From online values, vapour pressure of water vapour at 75°F = 2.96 KPa
Thus;
Partial Pressure of water vapour = (45 × 2.96)/100 = 1.332 KPa
B) humidity ratio of moist air is given as;
X = 0.62198 pw / (pa - pw)
where;
pw = partial pressure of the water vapor in moist air
pa = atmospheric pressure of the moist air
Thus;
X = (0.62198 × 1.332)/(13.32 - 1.332)
X = 0.0691
C) Formula for moist air specific volume is;
V_p = (1 + (xRw/Ra) × RaT/p
Where;
V_p is specific volume
T is temperature = 75°F = 297.039 K
p is barometric pressure which in this case is standard sea level pressure = 101.325 KPa
pw is partial pressure of the water vapor in moist air = 1.332 KPa
Rw is individual gas constant for water = 0.4614 KJ/Kg.K
Ra is individual gas constant for air = 0.2869 KJ/Kg.K
V_p = (1 + (0.0691 * 0.4614/0.2869)) × 0.286.9 * 297.039/101.325
V_p = 0.8384 m³/Kg
