Answer:
Amount of Energy = 23,467.9278J
Explanation:
Given
Cv = 5/2R
Cp = 7/2R wjere R = Boltzmann constant = 8.314
The energy balance in the tank is given as
∆U = Q + W
According to the first law of thermodynamics
In the question, it can be observed that the volume of the reactor is unaltered
So, dV = W = 0.
The Internal energy to keep the tank's constant temperature is given as
∆U = Cv((45°C) - (25°C))
∆U = Cv((45 + 273) - (25 + 273))
∆U = Cv(20)
∆U = 5/2 * 8.314 * 20
∆U = 415.7 J/mol
Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa
The Initial mole is calculated as
(P * V)/(R * T)
Where P = P1 = 101.33kPa = 101330Pa
V = Volume of Tank = 0.1m³
R = 8.314J/molK
T = Initial Temperature = 25 + 273 = 298K
So, n = (101330 * 0.1)/(8.314*298)
n = 4.089891232222
n = 4.089
Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa
V = Volume of Tank = 0.1m³
R = 8.314J/molK
T = Initial Temperature = 25 + 273 = 298K
n = (1500000 * 0.1)/(8.314*298)
n = 60.54314465936812
n = 60.543
So, tue moles that entered the tank is ∆n
∆n = 60.543 - 4.089
∆n = 56.454
Amount of Energy is then calculated as:(∆n)(U)
Q = 415.7 * 56.454
Q = 23,467.9278J
, where n is the number of moles, m is the mass of iron in the reaction and M is the Atomic weight.
moles of Iron.
where 32 is the Atomical Weight of Oxygen (16 x 2).
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