When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is
a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4 10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).Answer: 3.05 x 10-8 kg/m2-s
1 answer:
Answer:
See attachments for step by step explanation towards getting answer.
Explanation:
Given that;
College Engineering 10+5 pts
When -iron is subjected to an atmosphere of hydrogen gas, the concentration of hydrogen in the iron, CH (in weight percent), is a function of hydrogen pressure, (in MPa), and absolute pH2temperature (T) according to(5.14)Furthermore, the values of D0 and Qd for this diffusion system are 1.4 10-7 m2/s and 13,400 J/mol, respectively. Consider a thin iron membrane 1 mm thick that is at 250C. Compute the diffusion flux through this membrane if the hydrogen pressure on one side of the membrane is 0.15 MPa (1.48 atm), and on the other side 7.5 MPa (74 atm).
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Answer:
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Explanation:
Given that
Heat loss from room (Q)= 60 KJ/min
Work supplied to the room(W) = 1.2 KW = 1.2 KJ/s
We know that 1 W = 1 J/s
Sign convention for heat and work:
1. If heat is added to the system then it is taken as positive and if heat is rejected from the system then it is taken as negative.
2. If work is done by the system then it is taken as positive and if work is done on the system then it is taken as negative.
So
Q = -60 KJ/min
In 10 min Q = -600 KJ
W = -1.2 KJ/s
We know that
1 min = 60 s
10 min = 600 s
So W = -1.2 x 600 KJ
W = -720 KJ
WE know that ,first law of thermodynamics
Q = W + ΔU
-600 = - 720 + ΔU
ΔU = 120 KJ
The net amount of energy change of the air in the room during a 10-min period is 120 KJ.
Answer:
Q = 62 ( since we are instructed not to include the units in the answer)
Explanation:
Given that:
Q = ???
Now the gas expands at constant pressure until its volume doubles
i.e if
Using Charles Law; since pressure is constant
mass of He =number of moles of He × molecular weight of He
mass of He = 3 kg × 4
mass of He = 12 kg
mass of Ar =number of moles of Ar × molecular weight of Ar
mass of He = 7 kg × 40
mass of He = 280 kg
Now; the amount of Heat Q transferred =
From gas table
∴ Q =
Q =
Q = 62 MJ
Q = 62 ( since we are instructed not to include the units in the answer)