Question

An ideal gas mixture contains 0.25 kg of CO2, 0.45 kg of O2, and 0.5 kg of CO at 25C, 2 bar. The mixture is heated at a constant volume process to a final temperature of 180C. For the mixture, determine:

a. The gravimetric analysis (list of mass fractions) and molar analysis (list of molar fractions)

b. The apparent molecular weight of the mixture, in kg/kmol

c. The volume, in m3

d. The final pressure, in bar

e. The amount of energy transfer by heat

f. The change in entropy, in kJ/K

Answer 1

Mass fraction of a substance = mass of a substance/total mass of the mixture. The total mass of the mixture is equal to 1.2 kg. So mass fraction of CO2=0.25/1.2 = 0.208 or 20.8%. Similarly mass fraction of O2 = 0.45/1.2 =37.5% and mass fraction of of CO=0.5/1.2 = 41.6% .

Similarly to find the mole fractions we need to find the number of moles = mass/molar mass of substance

For CO2 number of moles = 0.25/44 = 0.0056 mole and for O2 = 0.45/32 = 0.014 and for CO = 0.5/28 = 0.018. So the total number of moles = 0.0056+0.014+0.018= 0.0376.

Mole fraction of substance = number of moles/total number of moles.

So mole fraction of CO2= 0.0056/0.0376 = 14.89%; mole fraction of O2= 0.014/0.0376 = 37.23% and mole fraction of CO= 0.018/0.0376 = 47.87%.

Apparent molar mass is weighed average of all gases present in the mixture = 0.208*44+0.375*32+0.416*28 = 32.8 kg/kmol

In accroding to Ideal gas PV=nRT which gives 2 bar * V= 0.0376*8.314*298K . so we will get volume as 46.58L

In according to ideal gas equation at constant volume P1/T1=P2/T2 which gives 2 bar/298K= P2/453K which gives P2 as 3.04 bar.

but we know that q=CdT and for the mixture of gases q= mass of the O2 * heat capacity * temperature change+ mass of CO2 * heat capacity * temperature change + mass of CO * heat capacity * temperature change. By this way we can find the total heat energy transfered.

Now entropy change can be found using delta S= q/T