Thermodynamics filled in class notes_Part_39

Thermodynamics filled in class notes_Part_39 - 85 3.3....

Info iconThis 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: 85 3.3. IDEAL MIXTURES OF IDEAL GASES Mass conservation gives m2 = m1 = mA + mB . (3.123) − 1W2 , (3.124) One also has the first law U2 − U1 U2 − U1 U2 m2 u 2 (mA + mB )u2 0 0 T2 T2 = 1Q2 = 0, = U1 , (3.125) (3.126) = m A u A1 + m B u B 1 , = m A u A1 + m B u B 1 , = mA (uA1 − u2 ) + mB (uB 1 − u2 ), = mA cvA (T1 − T2 ) + mB cvB (T1 − T2 ), mA cvA T1 + mB cvB T1 , = mA cvA + mB cvB = T1 . (3.127) (3.128) (3.129) (3.130) (3.131) (3.132) The final pressure by Dalton’s law then is P2 = = = = = = PA2 + PB 2 , mB RB T2 mA RA T2 + , V2 V2 mA RA T1 mB RB T1 + , V2 V2 (mA RA + mB RB ) T1 , V2 substitute for V2 from Eq. (3.122) (mA RA + mB RB ) T1 , T (mA RA + mB RB ) P1 1 P1 . (3.133) (3.134) (3.135) (3.136) (3.137) (3.138) (3.139) So the initial and final temperatures and pressures are identical. Now the entropy change of gas A is sA2 − sA1 = = cP A ln = PA2 T A2 − RA ln , T A1 PA1 T2 yA2 P2 − RA ln , T1 yA1 P1 yA2 P1 T1 , −RA ln T1 yA1 P1 cP A ln cP A ln (3.140) (3.141) (3.142) =0 = = yA2 P1 (1)P1 −RA ln yA2 . −RA ln , (3.143) (3.144) Likewise sB 2 − sB 1 = −RB ln yB 2 . (3.145) CC BY-NC-ND. 18 November 2011, J. M. Powers. 86 CHAPTER 3. GAS MIXTURES So the change in entropy of the mixture is ∆S = mA (sA2 − sA1 ) + mB (sB 2 − sB 1 ), = −mA RA ln yA2 − mB RB ln yB 2 , = − (nA MA ) (3.146) R MA (3.147) ln yA2 − (nB MB ) =m A = ≥ −R nA ln nA nA + nB (3.149) nB nA + nB +nB ln ≤0 ≤0 0. We can also scale Eq. (3.149) by Rn to get =∆ s ∆s R (3.148) =RB −R(nA ln yA2 + nB ln yB 2 ), 1 ∆S Rn ln yB 2 , =m B =RA = R MB nA nB = − n ln yA2 + n ln yB 2 , =yA2 , (3.150) (3.151) (3.152) =y B 2 = − (yA2 ln yA2 + yB 2 ln yB 2 ) , y y = − (ln yAA2 + ln yBB2 ) , 2 2 yA2 yB 2 = − ln (yA2 yB 2 ) . (3.153) (3.154) (3.155) For an N -component mixture, mixed in the same fashion such that P and T are constant, this extends to N ∆S = −R = −R nk ln yk , k =1 N nk ln k =1 (3.156) nk N i=1 ni ≥ 0, (3.157) ≤0 N = −R k =1 N = −Rn = −R nk n ln yk , n (3.158) nk ln yk , n (3.159) k =1 N m M yk ln yk , k =1 CC BY-NC-ND. 18 November 2011, J. M. Powers. (3.160) ...
View Full Document

This note was uploaded on 11/26/2011 for the course EGN 3381 taught by Professor Park-sou during the Fall '11 term at FSU.

Ask a homework question - tutors are online