Question

Chemical Unive ChE Engineering Department 205: Given by Dr. M. K. Hadi-kali Time allowed: 2h This final esam is closed book. Answer problems showint all the necessary steps) 2.One kg of air ul.-20 giaci -21 J mer Krois adia atically and reversibly fro bars and 120 C to 20 bars. calculate the final temperature and the work on joule) required process. Assume that air is an ideal gas. l 3. Calculate the molar volume and the compressibility factor of ethylene at 100 C and 20 bar 4. What is the standard heat of combustion of n-eptane gas at 25 c ir te combustion produces I s. Determine the heat transfer (in J molio when water is heated in steady-flow process from 2s 500 C at atmospheric pressure. 6. Show that the entropy change of an ideal gas, with a gas constant R and constant heat capacity undergoing a mechanical reversible work in a system from (T. Pa to P is given b expression: Gg 7. One kg of air (M. 29 g moll, C, -21 Jmoli K is compressed from 2sc, 10 bar) so bar) in a frictionless piston cylinder. Assuming that air behaves as an ideal gas, calculate the energy, enthalpy and entropy changes. 8. An inventor claims to have devised a cyclic engine which exchanges heat with reservoirs at 24 250 "C, and which produces 0.45 k of work for each kl of heat extracted from the hot reservoi claim believable? Why or why not? 9. A heat engine receives 800 k of heat from a high temperature source at 900 C during a converts 250 kJ of this heat to net work (W) and rejects the remaining quantity of energy t temperature sink at 30 C. o much beat is rejected to tbe low temperature reservoir? o Determine if this heat engine violates the second law of thermodynamics on the basi The Carnot principle, The Clausius inequality,

Answer 1

1.

For adiabatic compression, T2/T1= (P2/P1) ^(R/CP)

R= CP-CV , T1= 120 deg.c =120+273= 393 K, P2= 20 bar, P1= 1 (assumed not mentioned)

R= CP-CV= 29-21= 8 J/mole.K

T2/393= (20/1) ^(8/29) = 2.28

T2= 2.28* 393 =898 K =898-273= 625 deg.c

Work done = R*(T2-T1)(Y-1) , Y= CP/CV= 29/21= 1.38

Work done = 8.314*(625-393)/ (1.38-1) =5076 Joules/mole

But given mass of air =1 kg, moles of air = mass of air/molar mass = 1000/29 =34.48 moles

Work done = 5076*34.48 Joules =175021 Joules

2.

Critical properties of ethylene Tc= 282.4 K, Pc= 5.12 Mpa= 5.12*10 bar= 51.2 bar =51.2

Tr= T/TC= (100+273)/ 282.4=1.32 and Pr= P/PC= 20/51.2 = 0.391

From compressibility charts Z=0.9

From PV= ZRT

V= ZRT/P= 0.9* 0.0821L.atm/mole.K * 373/ (51.2*0.9869atm )=0.5475 m3/mole

3.

the combustion of n-heptane is given by

C₇H₁₆+11O₂-->7CO₂ + 8H₂O (g)

Standard heat of reaction = 7* heat of formation of CO2+ 8* heat of formation of water –{ heat of formation of C7H16+ heat of formation of O2)

4.

Since heat of formation of O2= 0

Standard hear of reaction = 7* (-393.52)+8*(-241.83) – (1*-224.83)=-4464.55 Kj/mole

When H2O product is in the form of gas

4.