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HW3_solution

# HW3_solution - Homework 3 solution 2.24(a T=300K...

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Homework 3 solution 2.24 (a) T=300K, n=PV/RT=0.1e6 Pa*0.1m 3 /8.314/300 =4.01 mols . (b) T 2 =T 1 (P 2 /P 1 ) R/Cp =300(1/10) 2/7 =155.4K n 2 =P 2 V/RT 2 =0.1E6 Pa*0.1m 3 /8.314/155.4 =7.74 mols 2.27 Energy balance: ; out out in in dn H dn H dnU = Since n is constant (n=P i V/RT i =0.1E6*0.002/8.314/300=0.08 mols), dt dt dn H H ndU in ) / )( ( = (Attn: n and n in are different here.) dt dt dn T T C dT nC in P V ) ( ) ( = ; Integrate from initial temperature T i to T, dn/dt is constant mass flow rate t dt dn nC C T T T T V P in i in ) ( ln = = t dt dn nC C T T T T V P i in in ) ( exp ) ( T(t)=350-50*exp(-0.4062t) K (t in min) P(t)=nRT(t)/V=332.56*T(t)=0.1164-0.01663*exp(-1.7452t) MPa Steady-state temperature is 350 K. It will be 345 K within 5 K of steady-state value: = ) ( / ln dt dn nC C T T T T t V P f in i in f =1.319 min=79.1s 3.1 T 1 =15 o C=288 K; V 1 =8.314*288/0.1=23944.34 cm 3 /mol T 2 =30 o C=303 K; V 2 =V 1 *T 2 /T 1 =255191.42 cm 3 /mol n=P 1 V 1 /RT 1 =0.1013*60e6/8.314/288.15=2537 mol Energy balance: dU=dQ+dW=dQ-PdV S-balance: dS=C V dT/T+RdV/V = 1 2 1 2 ln ln P P R T T C n S P id gas =2538.389*(7/2)*8.314*ln(303/288)=3750.2767 J/K

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res P res gas res rev res res T T T nC T Q T Q S ) ( 1 2 = = = =-3430.2438 J/K univ S =320.033 J/K 3.4 (a)n U=nC V (T 2 -T 1 )=W EC n and the initial temperature are not given, but P and V are given.
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