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Thermodynamics filled in class notes_Part_59

# Thermodynamics filled in class notes_Part_59 - 125 4.5 VAN...

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Unformatted text preview: 125 4.5. VAN DER WAALS GAS Recalling that T1 = T2 and canceling the terms in a, one gets 1 q2 = = = v2 − b v1 − b RT1 ln J 200 kg K −140408 J . kg , (4.261) m3 kg 0.0585 (300 K ) ln 0.598 m3 kg − 0.001 − 0.001 m3 kg m3 kg , (4.262) (4.263) Since the process is isothermal, = J 200 kg K ln s2 − s1 = R ln 0.0585 0.598 m3 kg m3 kg − 0.001 − 0.001 v2 − b v1 − b 3 m kg m3 kg = −468.0 (4.264) , (4.265) J . kg K (4.266) Entropy drops because heat was transferred out of the system. Check the second law. Note that in this portion of the process in which the heat is transferred out of the system, that the surroundings must have Tsurr ≤ 300 K . For this portion of the process let us take Tsurr = 300 K . s2 − s1 J kg K J −468.0 kg K −468.0 1 q2 ≥ ? T −140408 ≥ (4.267) J kg 300 K ≥ −468.0 , (4.268) J kg K ok. (4.269) Next look at 2 → 3 2 q3 = u3 − u2 + 2 w3 , (4.270) T3 2 q3 = cv (T )dT + a T2 1 1 − v2 v3 v3 + P dv , (4.271) v2 T3 since isochoric 2 q3 = cv (T )dT, T2 1000 K 290 = 300 K = 294000 J . kg (4.272) J kg K + 0.2 J kg K 2 T dT, (4.273) (4.274) CC BY-NC-ND. 18 November 2011, J. M. Powers. 126 CHAPTER 4. MATHEMATICAL FOUNDATIONS OF THERMODYNAMICS Now look at the entropy change for the isochoric process: T3 s3 − s2 = = cv (T ) dT , T T2 T3 290 kgJ K T T2 J kg K J = 489 . kg K = 290 ln (4.275) J dT , + 0.2 kg K 2 1000 K 300 K + 0.2 J kg K 2 (4.276) ((1000 K ) − (300 K )), (4.277) (4.278) Entropy rises because heat transferred into system. In order to transfer heat into the system we must have a diﬀerent thermal reservoir. This one must have Tsurr ≥ 1000 K . Assume here that the heat transfer was from a reservoir held at 1000 K to assess the inﬂuence of the second law. s3 − s2 ≥ J kg K J 489 kg K ≥ 489 4.6 ≥ 2 q3 ? T 294000 (4.279) J kg , 1000 K J 294 , kg K (4.280) ok. (4.281) Redlich-Kwong gas The Redlich-Kwong equation of state is P= RT a − . v − b v (v + b)T 1/2 (4.282) It is modestly more accurate than the van der Waals equation in predicting material behavior. Example 4.11 For the case in which b = 0, ﬁnd an expression for u(T, v ) consistent with the Redlich-Kwong state equation. Here the equation of state is now P= a RT − 2 1/2 . v vT CC BY-NC-ND. 18 November 2011, J. M. Powers. (4.283) ...
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