Question

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.

Answer 1

The mixture's final temperature is 297.848k

The mixture's final pressure is 203.2kpa

Step-by-step 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