Using the results of the Arrhenius analysis (Ea=93.1kJ/molEa=93.1kJ/mol and A=4.36×1011M⋅s−1A=4.36×1011M⋅s−1), predict the rate
1 answer:
Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae ------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
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Answer:
isobaric expansion = 281.09 Btu
isothermal compression= 72 Btu
Explanation:
The first law of thermodynamics is:
where:
Q=heat transferred
W= work
U=internal energy
P=pressure, V= volume, T= temperature, n = moles, Cv= specific heat at constant volume.
In a isobaric process heat transferred is:
For an isothermal process (T2-T1 = 0) so
From the data we know that the energy transferred to the system in the isothermal compression by work was 72 Btu that is the heat transferred to the system.
For the first process
we have to properties at the beginning of the process : temperature (350°F) and specific volume (V/mass)
we use this information in the appropriate unit to find the pressure in thermodynamic tables.
T1= 176°C
v1= 1.49 m^3/kg
P=1.37 bar
in the second state we have
P=1.37 bar =137000Pa
with thee properties we check in the thermodynamic tables
T2= 255°C
we usually find Cp on tables for water but from the Mayer relation we have:
Cp for water vapor is: 33.12 J/mol*K
R=8.314 J/mol*K
Cv= 41.434 J/mol*K
replacing in the equation for Q
296569J =281.09 Btu
Answer:
a. 4.279 MPa
b. 3.198 MPa to 4.279 MPa
c. 0.939 MPa
d. Below 3.198 MPa
Explanation:
From the given parameters
M = 1.75 MPa
M at 1.6 MPa gives A/A* = 1.2502
M at 1.8 MPa gives A/A* = 1.4390
Therefore, by interpolation, we have M = 1.75 MPa gives A
However, we shall use M = 1.75 MPa and A
Similarly,
P/P₀ = 0.1878
a) Where the nozzle is choked at the throat there is subsonic flow in the following diverging part of the nozzle. From tables, we have
A/A* = 1.387. by interpolation M
Therefore P = P₀ × P
Which shows that the nozzle is choked for back pressures lower than 4.279 MPa
b) Where there is a normal shock at the exit of the nozzle, we have;
M₁ = 1.75 MPa, P₁ = 0.1878 × 5 = 0.939 MPa
Where the normal shock is at M₁ = 1.75 MPa, P₂/P₁ = 3.406
Where the normal shock occurs at the nozzle exit, we have
Where the shock occurs t the section prior to the nozzle exit from the throat, the back pressure was derived as = 4.279 MPa
Therefore the back pressure value ranges from 3.198 MPa to 4.279 MPa
c) At M = 1.75 MPa and P
d) Where the back pressure is less than 3.198 MPa according to isentropic flow relations supersonic flow will exist at the exit plane