1. IB Physics Topic 10 – Thermodynamic Processes Mr. Jean

2. The plan: Video clip of the day –https://www.youtube.com/watch?v=nBVL3lp9f Gohttps://www.youtube.com/watch?v=nBVL3lp9f Go –https://www.youtube.com/watch?v=s3tjOEtoA rghttps://www.youtube.com/watch?v=s3tjOEtoA rg Returning Exams Thermodynamic processes

3. Thermodynamic Processes Isobar, expansion at constant pressure, work is done Isochoric pressure change, W = 0 Isothermal compression W = Q, U is constant Adiabatic expansion; no heat, Q = 0 P V 12 3 4 The area enclosed by the cycle is the total work done, W The work done, W, in a cycle is + if you travel clockwise

4. Heat Engines and Refrigerators Engines use a working fluid, often a gas, to create motion and drive equipment; the gas moves from 1 state (P, V, n, & T define a state) to another in a cycle Stirling designed this engine in the early 18 th century – simple and effective The Stirling Cycle: 2 isotherms 2 isochores The Stirling Engine

5. Isobaric expansion of a piston in a cylinder The work done is the area under the process W = P  V The work done W = Fd = PAd = P  V 4 stroke engine

6. Isochoric expansion of a piston in a cylinder Thus  U = Q – W = Q The work done W = 0 since there is no change in volume

7. Adiabatic expansion of an ideal gas Thus  U = Q – W = 0, that is adiabatic expansion against no resistance does not change the internal energy of a system The work done W = 0 here because chamber B is empty and P = 0

8. How much work is done by the system when the system is taken from: (a) A to B (900 J) (b) B to C (0 J) (c) C to A (-1500 J) EXAMPLE Each rectangle on the graph represents 100 Pa-m³ = 100 J (a) From A  B the area is 900 J, isobaric expansion (b) From B  C, 0, isovolumetric change of pressure (c) From C  A the area is -1500 J

9. 10 grams of steam at 100 C at constant pressure rises to 110 C: P = 4 x 10 5 Pa  T = 10 C  V = 30.0 x 10 -6 m 3 c = 2.01 J/g What is the change in internal energy? EXAMPLE  U = Q – W  U = mc  T – P  V  U = 189 J So heating the steam produces a higher internal energy and expansion

10. Aluminum cube of side L is heated in a chamber at atmospheric pressure. What is the change in the cube's internal energy if L = 10 cm and  T = 5 °C? EXAMPLE  U = Q – W Q = mc  T m =  V 0 V 0 = L 3 W = P  V  V =  V 0  T  U = mc  T – P  V  U =  V 0 c  T – P  V 0  T  U = V 0  T (  c – P  ) c Al = 0.90 J/g°C  Al = 72(10 -6 ) °C -1  U = L³  T (  c – P  ) P atm = 101.5 kPa  Al = 2.7 g/cm ³  U = 0.10³(5)((2700)(900) – 101.5(10³)(72(10 -6 ))  U = 12,150 JNB: P  is neglible

11. Isobar Isochore Isotherm P V 1, (P 1,V 1 ) T 1 2, (P 2,V 2 ) T 2 3, (P 3,V 3 ) T 3 4, (P 4,V 4 ) T 4 1. P 2 = P 1 = 1000 kPa Isotherm 2. T 4 = T 1 = 400 K 3. T 3 = T 2 = 600 K 4. P 3 = P 2 V 2 /V 3 = 625 kPa 5. P 4 = P 1 V 1 /V 4 = 250 kPa W = Area enclosed = P 1  V 12 +  (P 2 +P 3 )  V 23 +  (P 1 +P 4 )  V 41 = (15 + 12.188 – 18.75)(10³) = 8.44 kJ Find the work done for a cycle if P 1 = 1000 kPa, V 1 = 0.01 m³, V 2 = 0.025 m³, V 3 = V 4 = 0.04 m³, T 1 = 400 K, T 2 = 600K, n = 2 mol EXAMPLE W = Area enclosed +  (P 2 +P 3 )  V 23 = P 1  V 12 –  (P 1 +P 4 )  V 41

12. Isobar Isochore Isotherm P V 1, (P 1,V 1 ) T 1 2, (P 2,V 2 ) T 2 3, (P 3,V 3 ) T 3 4, (P 4,V 4 ) T 4 Find the internal energy for each state if P 1 = 1000 kPa, V 1 = 0.01 m³, V 2 = 0.025 m³, V 3 = V 4 = 0.04 m³, T 1 = 400 K, T 2 = 600K, n = 2 mol 1. P 2 = P 1 = 1000 kPa Isotherm 2. T 4 = T 1 = 400 K 3. T 3 = T 2 = 600 K 6. U 1 =  nRT 1 = 9972 J 7. U 4 = U 1 = 9972 J 9. U 3 = U 2 = 14958 J 8. U 2 =  nRT 2 = 14958 J 4. P 3 = P 2 V 2 /V 3 = 625 kPa 5. P 4 = P 1 V 1 /V 4 = 250 kPa EXAMPLE

13. Isobar Isochore Isotherm P V 1, (P 1,V 1 ) T 1 2, (P 2,V 2 ) T 2 3, (P 3,V 3 ) T 3 4, (P 4,V 4 ) T 4 Find the thermal energy change Q for each state if P 1 = 1000 kPa, V 1 = 0.01 m³, V 2 = 0.025 m³, V 3 = V 4 = 0.04 m³, T 1 = 400 K, T 2 = 600K, n = 2 mol 1. P 2 = P 1 = 1000 kPa Isotherm 2. T 4 = T 1 = 400 K 3. T 3 = T 2 = 600 K 6. U 1 =  nRT 1 = 9972 J 7. U 4 = U 1 = 9972 J 9. U 3 = U 2 = 14958 J 8. U 2 =  nRT 2 = 14958 J 10. Q 12 =  U 12 + W 12 = 34986 J 12. Q 34 =  U 34 = -4986 J 13. Q 41 = W 41 (U 41 = 0) W41 =  (P 4 +P 1 )  V 41 = - 18.75 kJ 4. P 3 = P 2 V 2 /V 3 = 625 kPa 5. P 4 = P 1 V 1 /V 4 = 250 kPa EXAMPLE 11. Q 23 = W 23 (  U 23 = 0) W 23 =  (P 2 +P 3 )  V 23 = 12.188 kJ Q 12 Q 34 Q 41

14. Heat Engines and Refrigerators The Wankel Rotary engine is a powerful and simple alternative to the piston engine used by Nissan and invented by the German, Wankel in the 1920s The Wankel Cycle: 2 adiabats 2 isochores The Wankel Engine