{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Thermodynamics filled in class notes_Part_69

Thermodynamics filled in class notes_Part_69 - 145 4.11...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 145 4.11. EQUILIBRIUM IN A TWO-COMPONENT SYSTEM The pressure at equilibrium is P2 = = = (nN2 + nN )RT , V (0.888154 kmole + 0.223693 kmole) 8.314 498.84 111.185 kP a. (4.465) kJ kmole K (6000 K ) , (4.466) (4.467) The pressure has increased because there are more molecules with the volume and temperature being equal. The molar concentrations ρi at equilibrium, are ρN = kmole 0.223693 kmole mole = 4.484 × 10−4 = 4.484 × 10−7 , 3 3 498.84 m m cm3 (4.468) ρN 2 = kmole 0.888154 kmole mole = 1.78044 × 10−3 = 1.78044 × 10−6 . 3 3 498.84 m m cm3 (4.469) Now consider the heat transfer. One knows for the isochoric process that 1Q2 = U2 − U1 . The initial energy is given by U1 = nN2 uN2 , (4.470) = nN2 (hN2 − RT ), (4.471) = (1 kmole) 2.05848 × 105 = 1.555964 × 105 kJ. kJ kJ − 8.314 kmole kmole K (6000 K ) , (4.472) (4.473) The energy at the final state is U2 = = = = nN 2 u N 2 + nN u N , nN2 (hN2 − RT ) + nN (hN − RT ), (4.474) (4.475) (0.888154 kmole) 2.05848 × 105 (4.476) kJ kJ (6000 K ) − 8.314 kmole kmole K kJ kJ +(0.223693 kmole) 5.9727 × 105 (6000 K ) , − 8.314 kmole kmole K 2.60966 × 105 kJ. (4.477) (4.478) So 1Q2 = = = U2 − U1 , 2.60966 × 105 kJ − 1.555964 × 105 kJ, 1.05002 × 105 kJ. (4.479) (4.480) (4.481) Heat needed to be added to keep the system at the constant temperature. This is because the nitrogen dissociation process is endothermic. One can check for second law satisfaction in two ways. Fundamentally, one can demand that Eq. (4.377), dS ≥ δQ/T , be satisfied for the process, giving 2 S2 − S1 ≥ 1 δQ . T (4.482) CC BY-NC-ND. 18 November 2011, J. M. Powers. 146 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS For this isothermal process, this reduces to S2 − S1 1Q2 ≥ T (nN2 sN2 + nN sN )|2 − (nN2 sN2 + nN sN )|1 yN2 P Po yN2 P − nN2 so 2 − R ln T,N Po PN2 nN2 so 2 − R ln T,N Po PN2 − nN2 so 2 − R ln T,N Po nN2 so 2 − R ln T,N nN2 so 2 − R ln T,N − nN2 so 2 − R ln T,N nN2 RT Po V nN2 RT Po V + nN + nN + nN + nN + nN yN P Po yN P so − R ln T,N Po PN so − R ln T,N Po PN so − R ln T,N Po so − R ln T,N + nN so − R ln T,N nN RT Po V so − R ln T,N nN RT Po V =0 1Q2 ≥ T , (4.483) , (4.484) , (4.485) , (4.486) . (4.487) 2 1 1Q2 ≥ T 2 1 2 1Q2 ≥ ≥ T 1Q2 T 1 Now at the initial state nN = 0 kmole, and RT /Po /V has a constant value of kJ 8.314 kmole K (6000 K ) RT 1 = =1 , Po V (100 kP a)(498.84 m3 ) kmole (4.488) so Eq. (4.487) reduces to nN2 so 2 − R ln T,N nN 2 1 kmole nN 1 kmole nN 2 − R ln 1 kmole + nN so − R ln T,N − n N 2 so 2 T,N 2 1 ≥ 1Q2 T , (4.489) ((0.888143) (292.984 − 8.314 ln (0.88143)) + (0.223714) (216.926 − 8.314 ln (0.223714)))|2 − ((1) (292.984 − 8.314 ln (1)))|1 19.4181 kJ kJ ≥ 17.5004 . K K ≥ 105002 , 6000 (4.490) Indeed, the second law is satisfied. Moreover the irreversible entropy production of the chemical reaction is 19.4181 − 17.5004 = +1.91772 kJ/K. For the isochoric, isothermal process, it is also appropriate to use Eq. (4.394), dA|T,V ≤ 0, to check for second law satisfaction. This turns out to give an identical result. Since by Eq. (4.307), A = U − T S , A2 − A1 = (U2 − T2 S2 ) − (U1 − T1 S1 ). Since the process is isothermal, A2 − A1 = U2 − U1 − T (S2 − S1 ). For A2 − A1 ≤ 0, one must demand U2 − U1 − T (S2 − S1 ) ≤ 0, or U2 − U1 ≤ T (S2 − S1 ), or S2 −S1 ≥ (U2 −U1 )/T . Since 1Q2 = U2 −U1 for this isochoric process, one then recovers S2 −S1 ≥ 1Q2 /T. CC BY-NC-ND. 18 November 2011, J. M. Powers. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern