Chapter 3

1. Chapter 3 The first law of thermodynamics

2. 3-1 Thermodynamic systems • A thermodynamic system is any collection of objects that is convenient to regard as a unit, and that any have the potential to exchange energy with its surroundings. النظام فى الديناميكا الحراريةيتكون من مجموعة من الأشياء والتى تبدو كوحدة واحدة ويمكن لكل منها تبادل الطاقة مع الوسط المحيط.

3. Continue • A thermodynamic process is a procedure by which there are changes in the state of a thermodynamic system. الاجراء فى الديناميكا الحراريةهو طريقة يمكن بها تغيير حالة النظام مثل الاجراء الادياباتكى.

4. Continue • We describe the energy relations in any thermodynamics process in terms of the quantity of heat Q added to the system and the work W done by the system. كمية الحرارة المضافة الى النظام Q والشغل المبذول من النظام W هما اللذان يصفان تغير الطاقة فى أى اجراء ثرمودينامكى .

5. Sings for heat and work in thermodynamics

6. Insulation Lead shot W Q Thermal reservoir Control knob 3-2 Work done during volume changes A simple example of a thermodynamics system is a quantity of gas enclosed in a cylinder with a movable piston as shown. يعتبر الغازالمحبوس داخل اسطوانة بها مكبس متحرك هى أبسط نظام للديناميكا الحرارية

7. Continue dW = F . ds = ( P A ) (ds) = P ( A ds ) = P dV, W = ∫ dW = ∫ P dv

8. i f W > 0 0 (a) pressure Volume P-V diagram A plot of the pressure of the gas versus its volume is called a P-V diagram. In Fig. a, the curve indicates that the pressure decreases as the volume increases. The work W done by the gas is represented by the shaded area under the curve between points i and f. لتوضيح كيفية تغير الحجم مع الضغط نستخدم P-V diagram. فى الشكل (a) يتناقص الضغط مع زيادة الحجم. الشغل المبذول من الغاز يمثل بالمساحة تحت المنحنى.

9. Pressure i f W > 0 W > 0 Pressure Volume (c) (b) i W > 0 f W > 0 0 Volume Continue In Fig. b,c the change take place in two steps and the work is represented by the area under the curve. فى الشكلين b, cالتغير فى الحجم والضغط حدث على مرحلتين والشغل المبذول يمثل بالمساحة تحت المنحنى.

10. Pressure W > 0 Volume (d) i f Continue Fig.d shows a thermodynamics cycle in which the system is taken from some initial state i to some other state f and then back to i. عندما يبدأ النظام من حالة i وينتهى الى نفس الحالة i يسمى هذا الاجراء دورة cycle A system can be taken from a given initial state to a given final state by an infinite number of processes. We say that heat and work are path-dependent quantities. مما سبق يمكن أن نستنتج أن النظام يمكن أن ينتقل من حالة ابتدائية الى حالة نهائية بعدد لانهائى من الاجراءت.وأن كميتا الحرارة والشغل يعتمدان على المسار.

11. Isobaric process In an isobaric process the expansion or compression occurs at constant pressure.

12. Isothermal process The system expands at constant temperature. P V = n R T , P = n RT /V

13. Example 3-1 Three moles of helium are initially at 20ºC and a pressure of 1 atm. What is the work done by the gas if the volume is doubled a) at constant pressure, or b) isothermally?

14. 3-3 The first law of thermodynamics The quantity ( Q – W ) represent a change in the property which is called the internal energy ∆U, which depends only on the initial and final states and does not depend at all on how the system gets from one to the other. ∆U = Q –W This equation represents the first law of thermodynamics.

15. 3-4 Some special cases of the first law of thermodynamics 1- Adiabatic processes : is one that occurs so rapidly or occurs in a system that is so well insulated that no transfer of energy as heat occurs between the system and its environment, ( Q = 0 ). By applying the first law ∆U = –W ( adiabatic process )

16. Continue 2- Constant – volume processes ( isochoric process ) : If the volume of a system is held constant, that system can do no work, ( W = 0 ). By applying the first law ∆U = Q

17. Continue 3- Cyclical processes :There are processes in which the system is restored to its initial state, ( ∆U = 0 ). By applying the first law Q = W

18. Continue Free expansions : These are adiabatic processes in which no transfer of heat occurs between the system and its environment and no work is done on or by the system, Q = W = 0. By applying the first law ∆U = 0

19. Insulation steam Lead shot water W Q Thermal reservoir Control knob Example 3-2 Let 1.00 kg of liquid water at 100ºC be converted to steam at 100 ºC by boiling at standard atmospheric pressure ( which is 1.01x105 Pa ) in the arrangement shown in the Fig. The volume of that water changes from an initial value of 1x10-3m3 as a liquid to 1.67m3 as steam.

20. Continue ( a) How much work is done by the system during this process? Solution The pressure is constant at 1.01x105 Pa = ( 1.01x105 Pa ) ( 1.67m3 – 1x10-3m3 ) = 1.69x105 J = 169 kJ (b) How much energy is transferred as heat during the process? Solution Because the change is from liquid to gaseous phase Q = m Lv = ( 2256 kJ/kg ) ( 1.00 kg ) = 2256 kJ ≈ 2260 kJ (c) What is the change in the system’s internal energy during the process. Solution ∆U = Q - W = 2256 kJ – 169 kJ = 2090 kJ = 2.09 MJ

21. Example 3-3 You propose to eat a 900 Cal hot fudge sundae and then run up several fights of stairs to work off the energy you have taken in. How high do you have to climb? Assume that your mass is 60.0 kg. Solution W = Q 1 kcal = 4190 J, Q = 900 kcal = 3.77x106 J Then, potential energy = heat m g h = W h = Q / ( m g ) = 3.77x106J / ( 60.0kg)(9.8m/s2) = 6.4x103 m

22. P a Pa Pb b V 0 Va Vb Example 3-4 A PV-diagram for a cyclic process is shown in the Fig. The total work is W = -500J. Find the change in internal energy and the heat added during this process. Solution For any cyclic process ∆U = 0 So, Q = W = -500 J That is 500 Joules of heat must come out of the system.

23. Px104Pa 8.0 3.0 Vx10-3m3 2.0 5.0 b d c a Example 3-5 A series of thermodynamic processes is shown in the pV- diagram. In process a b 150J is added to the system, and in process b d, 600J of heat is added. Find a) The internal energy change in process a b. b) The internal energy change in process a b d and c) the total heat added in process a c d. c