A sample of helium gas at 27.0 °C and 3.60 atm pressure is cooled in the same container to a temperature of -73.0 °C. What is th
1 answer:
Answer: The new pressure, if volume and amount of gas do not change is 2.40 atm
Explanation:
To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:
Putting values in above equation, we get:
Thus the new pressure, if volume and amount of gas do not change is 2.40 atm
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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:
Part a: <em>Units of k is </em><em> where reaction is first order in A and second order in B</em>
Part b: <em>Units of k is </em><em> where reaction is first order in A and second order overall.</em>
Part c: <em>Units of k is </em><em> where reaction is independent of the concentration of A and second order overall.</em>
Part d: <em>Units of k is </em><em> where reaction reaction is second order in both A and B.</em>
Explanation:
As the reaction is given as
where as the rate is given as
where x is the order wrt A and y is the order wrt B.
Part a:
x=1 and y=2 now the reaction rate equation is given as
Now the units are given as
The units of k is
Part b:
x=1 and o=2
x+y=o
1+y=2
y=2-1
y=1
Now the reaction rate equation is given as
Now the units are given as
The units of k is
Part c:
x=0 and o=2
x+y=o
0+y=2
y=2
y=2
Now the reaction rate equation is given as
Now the units are given as
The units of k is
Part d:
x=2 and y=2
Now the reaction rate equation is given as
Now the units are given as
The units of k is
Answer:
Highest speed: He
Lowest speed: CO2
Explanation:
The rms speed (average speed) of the molecules/atoms in an ideal gas is given by:
where
R is the gas constant
T is the absolute temperature of the gas
M is the molar mass of the gas, which is the mass of the gas per unit mole
From the equation, we see that at equal temperatures, the speed of the molecules in the gas is inversely proportional to the molar mass: the higher the molar mass, the lower the speed, and vice-versa.
In this problem, we have 5 gases:
(CO2) (O2) (He) (N2) (CH4)
Their molar mass is:
CO2: 44 g/mol
O2: 16 g/mol
He: 4 g/mol
N2: 14 g/mol
CH4: 16 g/mol
The gas with lowest molar mass is Helium (He): therefore, this is the gas with greatest average speed.
The gas with highest molar mass is CO2: therefore, this is the gas with lowest average speed.