IB Physics Topic 10 – Thermodynamic Processes Mr. Jean

Combining the gas laws gives Therefore, for n moles of an ideal gas; PV = (a constant)×T PV = nRT P = Pressure V = Volume n = number of moles R = Universal gas constant T = Temperature (KELVIN)

The work done by this force is w = Fs = PAs,since F=PA but As is the change in the volume occupied by the gas, ΔV. therefore; W = P  V Deduce an expression for the work involved in a volume change of a gas at constant pressure.

We can add energy to a gas by heating Q (temperature gradient) Or by working (mechanical energy) = W Q = ΔU + W Q = Heat energy added to the gas ΔU = Internal energy increase of the gas W = Work done by the gas. State the first law of thermodynamics. Students should be familiar with the terms system and surroundings. They should also appreciate that if a system and its surroundings are at different temperatures and the system undergoes a process, the energy transferred by non- mechanical means to or from the system is referred to as thermal energy (heat).

1.Change of p (and T) at constant volume; an isovolumetric change. 2. Change of V (and T) at constant pressure; an isobaric change. 3. Change in p and V at constant temperature; an isothermal change. 4. Change in p and V in an insulated container (no heating of the gas); an adiabatic change.

The product of pressure and volume represents a quantity of work. This is represented by the area below a p-V curve. Therefore, the area enclosed by the four curves represents the net work done by the engine during one cycle.

Second Law of Thermodynamics: It is impossible to extract an amount of heat Q H from a hot reservoir and use it all to do work W. Some amount of heat Q C must be exhausted to a cold reservoir. Entropy is a measure of the disorder (of the energy) of a system Every time we change energy from one form to another, we increase the entropy of the Universe even though local entropy may decrease.

Thermodynamic Processes A system can change its state A state is a unique set of values for P, V, n, & T (so PV = nRT is also called a “State Equation”) When you know the state of a system you know U since U =  NkT =  nRT =  PV, for a monatomic gas A “process” is a means of going from 1 state to another There are 4 basic processes with n constant Isobaric, a change at constant pressure Isochoric or isovolumetric, a change at constant volume, W = 0 Isothermal, a change at constant temperature (  U = 0, Q = W) Adiabatic, a process is one in which no heat is gained or lost by the system. “iso” means “same”

Thermodynamic Processes Isobar Isochore Isotherm Adiabat P V (P 1,V 1 ) T 1 (P 2,V 2 ) T 2 (P 3,V 3 ) T 3 (P 4,V 4 ) T 4 The trip from 1  2  3  4  1 is call a “thermodynamic cycle” Each part of the cycle is a process All state changes can be broken down into the 4 basic processes T 3 = T 4 Q = 0

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 The area enclosed by the cycle is the total work done, W The work done, W, in a cycle is + if you travel clockwise

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

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

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

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

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 J

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 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

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 = 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

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 = ( – 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 = 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

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 = 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 = J 8. U 2 =  nRT 2 = J 4. P 3 = P 2 V 2 /V 3 = 625 kPa 5. P 4 = P 1 V 1 /V 4 = 250 kPa EXAMPLE

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 = 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 = J 8. U 2 =  nRT 2 = J 10. Q 12 =  U 12 + W 12 = J 12. Q 34 =  U 34 = J 13. Q 41 = W 41 (U 41 = 0) W41 =  (P 4 +P 1 )  V 41 = 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 = kJ Q 12 Q 34 Q 41

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