9. Consider a container of volume 5.0 L, which is divided into two compartments
ID: 907934 • Letter: 9
Question
9. Consider a container of volume 5.0 L, which is divided into two compartments of equal size. In the left compartment there is nitrogen at 3.0 atm and 25 C. In the right compartment there is oxygen at 1.0 atm and 25 C. What are the changes in the entropy and the Gibbs free energy function of the system when the partition is removed and the gases are allowed to mix? Assume the gases to be ideal. 10. In a similar experiment (see the previous problem), the left compartment initially contains nitrogen at 3.0 atm, while the right compartment contains also nitrogen, but at 1.0 atm, all at 25 C. Find delta S and delta G in this case.Explanation / Answer
Moles of nitrogen gas in the first compartment n= PV/RT= 3*2.5/(25+298.15)*0.08206 =0.3065
Moles of oxygen gas in the second compartment = PV/RT= 1*2.5/(25+298.15)= 0.102181
Total moles after mixing = 0.3065+0.102181= 0.408727 moles
Volume after mixnig = 5 L temerature =25 Deg.C =25+273.15= 298.15K
P= nRT/V= 0.408727*0.08206*298.15/ 5=2 atm
Mole fraction of Nitrogen xa = 0.306545/0.408727= 0.75 Xb= 1-0.75= 0.25
delG=PV (xa* lnxa+ xblnxb)= 2*1.01325*105Pa*5 *10-3 m3 (0.75ln0.75+0.25ln0.25)=-569.78Joules
Entropy change due to mixing = -delG/T =569.78/298.15= 1.1911
10. Entopy change
Final pressure in this case
total moles of nitrogen is same as that of nitrogen and oxygen = 0.408727 moles
P= 2 atm( same as the previous case)
DelS/R = -ln(2/3) + (-ln(2/1)= -0.28768
delS= 0.28768*8,314 joules/K=-2.39179 J/K
delG= -TDelS= -298.15*(-2.39179)=713.112 Joules
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