Condensation point and freezing point of argon in KELVIN.
1 answer:
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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
Answer:
a) No. of moles of hydrogen needed = 5.4 mol
b) Grams of ammonia produced = 27.2 g
Explanation:
a)
No. of moles of nitrogen = 1.80 mol
1 mole of nitrogen reacts with 3 moles of hydrogen
1.80 moles of nitrogen will react with
= 1.80 × 3 = 5.4 moles of hydrogen
b)
No. of moles of hydrogen = 2.4 mol
It is given that nitrogen is present in sufficient amount.
3 moles of hydrogen produce 2 moles of
2.4 moles of hydrogen will produce
=
Molar mass of ammonia = 17 g/mol
Mass in gram = No. of moles × Molar mass
Mass of ammonia in g = 1.6 × 17
= 27.2 g