Question

Pressurized tanks are commonly used in gas propulsion for space applications. Two rigid, insulated tanks are interconnected by a valve. Initially 0.79 kmol of nitrogen at 200 kPa and -18oC fills one tank. The other tank contains 0.21 kmol of oxygen at 100 kPa and 27oC. The valve is opened and the gases are allowed to mix until a final equilibrium state is attained. During this process, there are no heat or work interactions between the tank contents and the surroundings. Determine the following:

a. The volume of each gas

b. The final temperature of the mixture

c. The final pressure of the mixture

d. The amount of entropy produced in the mixing process

Answer 1

a) Volume of nitrogen in the 1st gas = P/nRT where P= 200Kpa= 200/101.3 atm =1.974 atm n= 0.79 kmol, T= -18deg.c= -18+273.15=255.15K, R= 0.08206 L.atm/mole.K   V= 1.974/(0.79*255.15*0.08206)=8.379L

Volume of Oxygen in the second gas = (100/101.3)/0.21*(27+273.15)*0.08206=0.1909 L

b) when mixed, Final volume= 8.379+0.1909= 8.5699L ( since the tanks are rigid)

since the system is insulated, there is no heat transfer and no work is done by or on the system. Hence from first law of thermodynamics Change in internal energy= 0

Internal energy before mixing= internal energy after mixing

moles of N2* CVN2* T1+Moles of oxygen* CVO2* T2= (n1+n2)*CV*T

since both N2 and O2 are diatomic CVN2=CVO2= CV= specific heat at constant volume= 2.5R =2.5*8.314=20.785 Joules/mole T1= 255.15K , T2= 27+273.15= 300.15K and T= final temperature

0.79*255.15+0.21*300.15=(0.79+0.21)*T

T= 264.6 K

The equilibrium temperature = 264.6 K

Final pressure = (P1V1+P2V2)/ V= (1.974*8.379+0.987*0.1909)/8.5699=1.95 atm =197 Kpa

d) entropy change of mixing = moles of nitrogen*CVN2* ln (T/T1)+ moles of oxygen*CVO2* ln (T/T2)

=0.79*2.5*8.314ln(264.6/255,15)+ 0.21*2.5*8.314*ln(264.6/300.15)=0.0469 J/K.