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
A heat amount of 23.461 kilojoules is lost from the tank.
According to the Equation of State for Ideal Gases, the mass of air ([tex]m[/tex]), in kilograms, within the tank can be found by using the following form:
[tex]m = \frac{P\cdot V\cdot M}{R_{u}\cdot T}[/tex] (1)
Where:
Heat losses ([tex]Q[/tex]) occur in a isochoric process, that is, a process at constant volume, it is caused by the heat released by the air flowing to the tank. The formula is represented by the following application of the definition of sensible heat:
[tex]Q = \frac{5}{2}\cdot (m_2-m_1) \cdot R_{u}\cdot (T_{1}-T_{2})[/tex] (2)
Where:
[tex]m_{1}, m_{2}[/tex] - Initial and final masses of the air within the tank, in kilograms.
[tex]T_{1}, T_{2}[/tex] - Initial and final temperatures of the air inflow, in Kelvin.
By (1) and key information from statement we have the following expression for the air introduced in the tank:
[tex]m_{2}-m_{1} = \frac{V\cdot M}{R_{u}\cdot T}\cdot (P_{2}-P_{1})[/tex] (3)
Where [tex]P_{1}[/tex] and [tex]P_{2}[/tex] are the initial and final pressures of the air inside the tank, in kilopascals.
If we know that [tex]P_{1} = 101.33\,kPa[/tex], [tex]P_{2} = 1500\,kPa[/tex], [tex]R_{u} = 8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}[/tex], [tex]T = 298.15\,K[/tex], [tex]T_{1} = 318.15\,K[/tex], [tex]T_{2} = 298.15\,K[/tex], [tex]V = 0.1\,m^{3}[/tex] and [tex]M = 28.97\,\frac{kg}{kmol}[/tex], then the heat losses are:
[tex]m_{2} - m_{1} = \left[\frac{(0.1\,m^{3})\cdot \left(28.97\,\frac{kg}{kmol} \right)}{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right)\cdot (298.15\,K)} \right]\cdot (1500\,kPa-101.325\,kPa)[/tex]
[tex]m_{2}-m_{1} = 1.635\,kg[/tex]
[tex]Q = \frac{5}{2}\cdot \left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K} \right) \cdot \left(\frac{1}{28.97} \,\frac{kmol}{kg} \right) \cdot (1.635\,kg)\cdot (318.15\,K-298.15\,K)[/tex]
[tex]Q = 23.461\,kJ[/tex]
A heat amount of 23.461 kilojoules is lost from the tank.
To learn more on ideal gases, we kindly invite to check this verified question: https://brainly.com/question/8711877
