Answer:
Explanation:
The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.
Properties of water at initial and final state are:
State 1 - (Liquid-Vapor Mixture)
State 2 - (Liquid-Vapor Mixture)
The mass stored in the vessel is:
The heat transfer require to the process is:
Answer:
(a) The heat transferred is 2552.64 kJ
(b) The entropy change of the mixture is 1066.0279 J/K
Explanation:
Here we have
Molar mass of H₂ = 2.01588 g/mol
Molar mass of N₂ = 28.0134 g/mol
Number of moles of H₂ = 500/2.01588 = 248 moles
Number of moles of N₂ = 1200/28.0134 = 42.8 moles
P·V = n·R·T
V₁ = n·R·T/P = 290.8×8.3145×300/100000 = 7.25 m³
Since the volume is doubled then
V₂ = 2 × 7.25 = 14.51 m³
At constant pressure, the temperature is doubled, therefore
T₂ = 600 K
If we assume constant specific heat at the average temperature, we have
Heat supplied = m₁×cp₁×dT₁ + m₂×cp₂×dT₂
cp₁ = Specific heat of hydrogen at constant pressure = 14.50 kJ/(kg K
cp₂ = Specific heat of nitrogen at constant pressure = 1.049 kJ/(kg K
Heat supplied = 0.5×14.50×300 K+ 1.2×1.049×300 = 2552.64 kJ
b)
Where:
and
are the mole fractions of Hydrogen and nitrogen respectively.
Therefore, = 248 /(248 + 42.8) = 0.83
= 42.8/(248 + 42.8) = 0.1472
∴ = 1066.0279 J/K