The mass fractions of a mixture of gases are 15 percent nitrogen, 5 percent helium, 60 percent methane, and 20 percent ethane. This mixture is enclosed in a 10 m3 rigid, wellinsulated vessel at 200 kPa and 20oC. A paddle wheel in the vessel is turned until 100 kJ of work have been done on the mixture. Calculate the mixture’s final pressure and temperature.

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

1

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answer:

yes

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