1. Thermodynamics : Temperature, Heat Transfer, and First Law of Thermodynamics, Yohanes Edi Gunanto Dept. of Math. Educ. UPH

2. Temperature

3. Apakah Suhu = Panas ? Apa yang dimaksud dengan Suhu ? Apa yang dimaksud dengan Panas ?

4. Apa yang terjadi ketika benda padat, cair dan gas menerima panas ?

5. Ketika menerima panas, molekul- molekul bergerak makin lama makin cepat ! Inilah yang disebut dengan pemuaian Jadi apakah pemuaian itu ?

6. Mengapa termoemeter dapat digunakan untuk mengetahui panas suatu benda ?

7. Bandingkan skala-skala termometer di bawah ini

8. Bagaimana cara panas berpindah? Panas berpindah dari benda bersuhu panas ke benda bersuhu dingin Panas berpindah dengan cara konveksi, konduksi dan radiasi

9. Konduksi Perpindahan panas tanpa memindahkan penghantarnya

10. Konveksi Perpindahan panas dengan memindahkan perantaranya

11. Radiasi Perpindahan panas tanpa memerlukan perantara

12. THERMODYNAMICS BASICS

13. Zeroth Law A B C If A and B and B and C are in thermal equil, then A and C are in thermal equil. [ie. At same T]

14. First Law of Thermodynamics Conservation of Energy for Thermal Systems

15. Joule Equivalent of Heat James Joule showed that mechanical energy could be converted to heat and arrived at the conclusion that heat was another form of energy. He showed that 1 calorie of heat was equivalent to 4.184 J of work. 1 cal = 4.184 J

16. Energy Mechanical Energy: KE, PE, E Work is done by energy transfer. Heat is another form of energy. Need to expand the conservation of energy principle to accommodate thermal systems.

17. 1 st Law of Thermodynamics Consider an example system of a piston and cylinder with an enclosed dilute gas characterized by P,V,T & n.

18. 1 st Law of Thermodynamics What happens to the gas if the piston is moved inwards ?

19. 1 st Law of Thermodynamics If the container is insulated the temperature will rise, the atoms move faster and the pressure rises. Is there more internal energy in the gas?

20. 1 st Law of Thermodynamics External agent did work in pushing the piston inward. W = Fd =(PA)  x W =P  V xx

21. 1 st Law of Thermodynamics Work done on the gas equals the change in the gases internal energy, W =  U xx

22. 1 st Law of TD Let’s change the situation: Keep the piston fixed at its original location. Place the cylinder on a hot plate. What happens to gas?

23. Heat flows into the gas. Atoms move faster, internal energy increases. Q = heat in Joules  U = change in internal energy in Joules. Q =  U

24. 1 st Law of TD What if we added heat and pushed the piston in at the same time? F

25. 1 st Law of TD Work is done on the gas, heat is added to the gas and the internal energy of the gas increases! Q = W +  U F

26. 1 st Law of TD Some conventions: For the gases perspective: heat added is positive, heat removed is negative. Work done on the gas is positive, work done by the gas is negative. Temperature increase means internal energy change is positive.

27. First Law of Thermodynamics “Energy cannot be created or destroyed. It can only be changed from one form into another.” Rudolf Clausius 1850

28. First Law of Thermodynamics Conservation of Energy Says Nothing About Direction of Energy Transfer

29. 1 st Law of TD Example: 25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas?

30. P o = 202,600 Pa, V o = 0.025 m 3, T o = 300 K, P f = 202,600 Pa, V f =0.020 m 3, T f = n = PV/RT. W = -P  V  U = 3/2 nR  T Q = W +  U W =-P  V = -202,600 Pa (0.020 – 0.025)m 3 =1013 J energy added to the gas.  U =3/2 nR  T=1.5(2.03)(8.31)(-60)=-1518 J Q = W +  U = 1013 – 1518 = -505 J heat out

31. Quasistatic Processes in an Ideal Gas isochoric ( V = const ) isobaric ( P = const ) V P V 1,2 PV= Nk B T 1 PV= Nk B T 2 1 2 V P V1V1 PV= Nk B T 1 PV= Nk B T 2 1 2 V2V2 (see the last slide)

32. Isothermal Process in an Ideal Gas W i-f > 0 if V i >V f (compression) W i-f < 0 if V i <V f (expansion) isothermal ( T = const ) : V P PV= Nk B T V1V1 V2V2 W

33. Adiabatic Process in an Ideal Gas adiabatic (thermally isolated system) ( f – the # of “unfrozen” degrees of freedom ) The amount of work needed to change the state of a thermally isolated system depends only on the initial and final states and not on the intermediate states. to calculate W 1-2, we need to know P (V,T) for an adiabatic process V P V1V1 PV= Nk B T 1 PV= Nk B T 2 1 2 V2V2 Adiabatic exponent

34. Adiabatic Process in an Ideal Gas (cont.) V P V1V1 PV= Nk B T 1 PV= Nk B T 2 1 2  1+2/3  1.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6  1.33 (polyatomic) (again, neglecting the vibrational degrees of freedom) An adiabata is “steeper” than an isotherma: in an adiabatic process, the work flowing out of the gas comes at the expense of its thermal energy  its temperature will decrease. V2V2 Prove

35. Summary of quasi-static processes of ideal gas Quasi-Static process UU QW Ideal gas law isobaric (  P=0) isochoric (  V=0) 0 isothermal (  T=0) 0 adiabatic (Q=0) 0

36. TUHAN MEMBERKATI