Answer:
The mixture's final temperature is 297.848k
The mixture's final pressure is 203.2kpa
Explanation:
Mass fraction of Nitrogen= 0.15 MfN2
Mass fraction of Helium= 0.05 MfHe
Mass fraction of Methane= 0.6 Mf CH4
Mass fraction of Ethane = 0.2 C2H6
Volume of the tank= 10m^3
Initial mixture pressure= Pm= 200kpa
Initial mixture temperature = 20°C = 20 +273.15= 293.15
Workdone,w = 100KJ
From the property table, molar mass and specific heat of constant are given below:
MN2= 28.013 kg/kmol
MHe= 4.003 "
MCH4= 16.043 "
MC2H6= 30.07 "
CPN2= 1.039 kJ/kgmol
CPHe= 5.1926
CPCH4 = 2.2537
CPC2H6 = 1.7662
For an ideal gas, the molar mass of the mixture is computed as follows:
Mm= mm/Nm
= mm/Σm1/M1
The molar masses of the mixture would bw
Mm= 1/ (mfN2/N2 + mfHe/He + mfCH4/CH4 + mfC2H6/C2H6)
Mm = 1/ (0.15/28.13 + 0.05/4.003 + 0.6/16.043 + 0.2/30.07)
Mm= 16.156kg/kmol
The specific heat at constant pressure of a mixture is computed as:
Cpm= Σk, i=1 mfiCp.i
=MfN2CP.N2+MfHeCP.He+MfCH4.CP.CH4+MfC2H6.CPC2H6
=0.15×1.039 + 0.05 × 5.1926 + 0.6 × 2.2537 + 0.2×1.7662
=2.121kJ/kg-K
Apparent constant gas of mixture can be calculated as:
Rm= RM/Mm
= 8.314/16.156
= 0.5146kJ/kg-K
The specific heat of a mixture at constant volume:
Cv.m = Cp.m - Rm
=2.121 - 0.5146
= 1.6064kJ/kg-K
The mass of a mixture present in the vessel is computed using ideal Gass equation
Mm= P1Vm/RmT1
= 200 × 10 / 0.5146 × 293.15
=13.25kg
From the first law of thermodynamics, we have:
Q - W = ΔE
The vessel is well insulated, so the heat of transfer Q=0
Neglecting potential and kinetic energy, change in energy becomes internal.
Hence,
ΔE= U2 - U1
= mmCv.m = T2 - T1
Substitute the values known into the first law of thermodynamics
0 - W = Cvm (T2 - T1)
Therefore, work supplied to the system is given by:
W = mmCvm (T2 - T1)
100 = 13.25 × 1.6064 × ( T2 - 293.15)
T2= 293. 15 + 4.698
T2= 297.848k
Therefore, the final mixture temperature is 297.848k
The final pressure is expressed as
P2= P1. T2/T1
P2= 200 × 297.89/293.15
P2= 203.2kpa
The mixture's final pressure is 203.2kpa
