A rigid and insulated tank is divided into two compartments by a partition. Comp
ID: 2250432 • Letter: A
Question
A rigid and insulated tank is divided into two compartments by a partition. Compartment A contains O2 at 20 C and 150 kPa, and Compartment B contains N2 at 30 °C and 250 kPa. The volume of Compartment B is 1 m3. The volume of Compartment A is twice the volume of Compartment B. The partition is removed, and the two gases mix until an equilibrium state is reached. Determine (a) the mass and number of moles of N2 in Compartment B. (b) the mass fraction of the two gases (c) the mole faction of the two gases; (d) the average molar mass of the gas mixture; (e) the final mixture temperature; (f) the final mixture pressure. If the volume of Compartment B is 1 L (instead of 1 m) while all the other information is the same (e.g, the volume of Compartment A is still twice the volume of Compartment B), which of the answers in (a) to () will change and which will not?Explanation / Answer
a) Mass and number of moles of nitrogen in COmpartment B.
Mass = m =PV/RT
Pressure = P = 250 Kpa
Temp= 30 C = 303 K
R = 0.296 KJ/kg-k.
V = volume = 1 m3
Mass = m = 250*1/(0.296*303) = 2.78 kg.
No of moles = ?
Mass = no.of moles * molecular weight
No.of moles = mass / molecular weight = 2.78*1000 / 28 = 99 moles.
b) Mass fraction of two gases = ?
We know the mass of nitrogen gas =m2 = 2.78 kg.
Mass of oxygen = ?
Mass of oxygen = m1 = PV/RT
Pressure of oxygen compartment = 150 Kpa
Volume = 2 m3
T = 20 C = 293 K.
R = 0.260
mass = m1 = 150*2 / 0.260*293 = 3.93 Kg.
Total mass = m1+m2 = 2.78+3.93 = 6.71 kg.
Mass fraction of oxygen gas = m1/m1+m2 = 3.93 / (3.93+2.78) = 0.585
mass fraction of nitrogen gas = m2 / (m1+m2) = 2.78 / (2.78+3.93) = 0.413.
d) The mole fraction of two gases.
No of moles of nitrogen gas = 99 moles.
No of moles of oxygen gas = 3.93*1000 / (32) = 123 moles.
Mole fraction of oxygen = 123 / (123+99) = 0.5540
Mole fraction of nitrogen = 99/(123+99) = 0.445
d) Average molar mass of the gas mixture.
Average molar mass = sum of molar fraction * molecular weight
Average molar mass = 0.554*32+0.445*28 = 30.18.
e)
Final mixture temp?
Let the final temp = T
Final temp = T = (m1C1T1+m2C2T2) / (m1C1+m2C2)
C1 = specific heat of oxygen = 0.918 KJ / kg-k
C2 = specific heat of nirogen = 1.040 KJ / kg-K.
Final temp=T = (3.93*0.918*293 + 2.78*1.040*303) / (3.93*0.918+2.78*1.040) =297.5 K.
f)
FInal mixture pressure = ?
P = (mRTfinal) / (Vfinal)
mass = m1 + m2 = 3.93+2.78 = 6.71 kg.
Final gas characteristic constant = R = ?
Final temp = Tfinal = 297.5 K.
Final volume = V = V1+V2 = 1+2 = 3 m3
Final gas constnat=R = ((m1R1T1+m2R2T2) / (m1T1+m2T2)
R = (3.93*0.260*303 + 2.78*0.298*293) / (3.93*303+2.78*293) = 0.275
Final pressure = P = 6.71*0.275*297.5 / (3) = 183 Kpa.
if the volume of the compartment B is 1 L instead of 1 m3 , and all the other information is the same, The answers from A,B,C,E,F is going to change. However, D is not going to change. If volume os reduces, mass reduces. So it alters all calculations.
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