An insulated tank having a total volume of 0.6 m3 is divided into two compartments. Initially one compartment contains 0.4 m3 of
hydrogen at 127 oC, 2 bar and the other contains nitrogen at 27 oC, 4 bar. The gases are allowed to mix until an equilibrium state is attained. Assume the ideal gas model with constant specific heats, determine
a. The final temperature, in oC.
b. The final pressure, in bar.
c. The amount of entropy produced, in KJ/K.
a. The final temperature, in oC.
b. The final pressure, in bar.
c. The amount of entropy produced, in KJ/K.
1 answer:
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Answer:
The work transfer per unit mass is approximately 149.89 kJ
The heat transfer for an adiabatic process = 0
Explanation:
The given information are;
P₁ = 1 atm
T₁ = 70°F = 294.2611 F
P₂ = 5 atm
γ = 1.5
Therefore, we have for adiabatic system under compression
Therefore, we have;
The p·dV work is given as follows;
Therefore, we have;
Taking air as a diatomic gas, we have;
The molar mass of air = 28.97 g/mol
Therefore, we have
The work done per unit mass of gas is therefore;
The work transfer per unit mass ≈ 149.89 kJ
The heat transfer for an adiabatic process = 0.
Answer:
Nₐ₀-Nₐ = 1.33
Nₐ₀ = 2.5
Conversion X = 1.33/2.5 = <u>0.533</u>
Explanation:
A + 2B + 4C ⇒ 2X + 3Y
Given a stream containing 50% A, 25% B and 25% C, to get the limiting reactant, lets take a simple basis
Say stream is 10 moles, this give
A = 5moles
B = 2.5mole
C = 2.5moles
from the balanced equation above,
1mole of A ⇒ 4moles of C
∴ 5moles of A ⇒ (5x4)/1 ⇒ 20moles of C
also;
2mole of B ⇒ 4moles of C
∴ 2.5moles of B ⇒ (2.5x4)/2 ⇒ 5moles of C
so clearly from above reactant C is the limiting reactant.
<em>Note: To get conversion of a process, we must use the limiting reactant. this is because ones it is used up, the reaction comes to an end</em>
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Formula to obtain conversion is:
Conversion = (Amount of A used up)/(Amount of A fed into the system)
where, Nₐ₀-Nₐ = is the amount in moles of A used up
Nₐ₀ = amount in moles of A fed into the system
The next question is what mole of reactant C will give 0.1mole fraction of Y
Recall our basis = 10moles
<em>from conservation of mass law</em>, 10mole of product must come out which 0.1 moles fraction is Y
therefore amount Y in the product is = 0.1x10 = 1mole
if 3moles of Y ⇒ 4mole of C
∴ 1mole of Y ⇒ (1x4)/3 ⇒ 1.33moles of C
calculating the conversion of limiting reactant C that will give 0.1mole fraction of Y
Nₐ₀-Nₐ = 1.33
Nₐ₀ = 2.5
Conversion X = 1.33/2.5 = <u>0.533</u>