A space air is at a temperature of 75 oF, and the relative humidity (RH) is 45%. Using calculations, find: (a) the partial press
ures of the dry air and water vapor, (b) the humidity ratio of the moist air, and (c) the specific volume of the moist air. Assume standard sea-level pressure.
1 answer:
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
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Given that, an experiment to measure the enthalpy change for the reaction of aqueous copper(II) sulfate, CuSO4(aq) and zinc, Zn(s) was carried out in a coffee cup calorimeter; the heat of the reaction in the whole system is calculated to be 2218.34 kJ
Heat of reaction (i.e enthalpy of reaction) is the quantity of heat that is required to be added or removed when a chemical reaction is taken place in order to maintain all of the compounds present at the same temperature.
The formula used to calculate the heat of the reaction can be expressed as follows:
Q = mcΔT
where:
- Q = quantity of heat transfer
- m = mass
- c = specific heat of water = 4.18 kJ/g °C (constant)
- ΔT = change in temparature
From the information given:
- The initial temperature (T₁) = 25° C
- The final temperature (T₂) = 91.5° C
∴
The change in temperature i.e. ΔT = T₂ - T₁
ΔT = 91.5° C - 25° C
ΔT = 66.5° C
The number of moles of CuSO₄ = 1.00 mol/dm³ × 50.0 cm³
= 0.05 moles
- Since the molar mass of CuSO₄ = 159.609 g/mol
Then;
Using the relation:
By crossing multiplying;
mass of CuSO₄ = number of moles of CuSO₄ × molar mass of CuSO₄
mass of CuSO₄ = 0.05 moles × 159.609 g/moles
mass of CuSO₄ = 7.9805 grams
∴
Using the formula from above:
Q = mcΔT
Q = 7.9805 g × 4.18 kJ/g °C × 66.5° C
Q = 2218.34 kJ
Therefore, we can conclude that the heat of the reaction is 2218.34 kJ
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Answer:
Explanation:
Scientific notation is the way of writing numbers which are either large or small. The number is written in the scientific notation when the number is between 1 and 10 and then multiplied by the power of 10. Engineering notation is the same version of the scientific notation but the number can be between 1 and 1000 and in this exponent of the ten is divisible by three.
For example, is to be written as
in engineering notation.
The given number:
0.00000009345 can be written as
Answer upto 4 significant digits =