Adiabatic Reversible Process: q=0 Ideal gas

Grains of sand Adiabatic walls Adiabatic Reversible Process: q=0 Ideal gas Small adiabatic change (dV,dT)) TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAAAA

isotherm : pV = constant • In an adiabatic process, p, V, and T all change monatomic gas g =(5/2)/(3/2) = 5/3 = 1.67 diatomic gas g =(7/2)/(5/2) = 7/5 = 1.4 • Adiabats (adiabatic p-V curves) are steeper than isotherms • Adiabatic and isothermal processes are reversible -- The p-V curve for an Adiabatic process adiabat : pVg = constant  = (+1)/ p V

 5/3 1.67 (monatomic), 7/5 =1.4 (diatomic), 8/6  1.33 (polyatomic) Adiabatic Process in an Ideal Gas • adiabatic(thermallyisolated system) 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. P An adiabata is “steeper” than an isothermal: in an adiabatic process, the work flowing out of the gas comes at the expense of its thermal energy  its temperature will decrease. 2 PV= NkBT2 1 PV= NkBT1 V2 V1 V The pressure at everypoint on the adiabata obeys the following equation:

Homework Problem Imagine that we rapidly compress a sample of air whose initial pressure is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature? Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity • poor approx. for a bike pump, works better for diesel engines • How does a diesel engine ignite? • What happens if we quickly vent the tire to ¼ the initial pressure?

Homework Problem Imagine that we rapidly compress a sample of air whose initial pressure is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature? Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity For adiabatic processes: also • poor approx. for a bike pump, works better for diesel engines • How does a diesel engine ignite? • What happens if we quickly vent the tire to ¼ the initial pressure?

2nd law of Thermodynamics • The first law came from the fact that we couldn’t create energy but doesn’t say anything about limits converting energy from one type to another. • From the first law the possibility of any transformation of heat to work or work to heat exists as long as the amount of heat equals the amount of work. • The first law also doesn’t say anything about whether a process will occur spontaneously or not.

Let’s revisit Joules Experiment Work can be directly transformed into an equal amount of heat through friction Same goes for electrical work.

Second Law of Themodynamics • Places limitations on the tranformation of heat into work. • If there were no limitations then by cooling the surroundings we could convert heat into work. For example the ocean and the earth have a basically unlimited supply of thermal energy.

Perpetual Motion Machine • The second law prohibits making such a machine. • The evidence for it is in part, that noone has ever been able to make one.

Second Law of Thermodynamics • Overall: NO thermodynamic cycle can have a thermal efficiency of 100% (i.e., cannot convert all heat into work) • Quick review: • 1st Law: Conservation/transformation of energy • 2nd Law: Limits the direction of processes & extent of heat-to-work conversions

The Second Law According to Lords Kelvin and Clausius A transformation whose only final result is to transform into work heat extracted from a source which is at the same temperature is impossible (Postulate of Lord Kelvin ) If heat flows by conduction from a body A to another body B, then a transformation whose only final result is to transfer heat from B to A is impossible. Postulate of Clausius  (These two happen to be equivalent).

2nd law • Let’s show that Kelvin’s and Clausius’ statements are equivalent. First, let’s assume that Kelvin’s postulate is not valid. Then we could extract heat from a cooler source Tc and convert into entirely to work then take that work and through friction heat up Th. The net result would be having heat flow from a cold source at Tc to a hot source Th Before we go on, let’s talk about the possibilities of transforming heat into work.

The Carnot cycle Carnot Cycle Named for Sadi Carnot (1796- 1832) Isothermal expansion Adiabatic expansion Isothermal compression Adiabatic compression

a • q1 b • T1 Q=0 Q=0 • d • T2 q2 c Carnot Cycle Pressure Volume

a • q1 b • T2 W Q=0 Q=0 • d • T1 q2 c Carnot Cycle Pressure Volume

The Carnot cycle for a gas might occur as visualized below. TL QL

Carnot Cycle For Any Cycle, all of the state variables remain unchanged since, for the whole process, there is no change of state. During an isothermal compression, the system gives up q1 to heat source T1 And during the expansion absorbs q2 from T2 then the total work is: We define the efficiency of the engine as:

Carnot Cycle We can now show that A violation of Clausius’ postulate violates Kelvin’s postulate. • Allow heat q2 to flow from t1 to t2 • Now go through Carnot cycle where the heat reservoir t2 gives up heat q2 to the system • The heat reservoir has given up and received q2 heat and remains in the same state. • Only heat has flowed from t1 and the overall work is done. This violates Kelvin’s postulate since only heat has flowed from t1 and work was done.

From b to c: adiabatic, q = 0, so that TVg-1 is constant. Thus, T1Vbg-1 = T2Vcg-1 or Similarly, d to a: adiabatic, q = 0, so that TVg-1 is constant. Thus, T2Vdg-1 = T1Vag-1 or Carnot Cycle From a to b: isothermal, so that DU = 0 and q = - w Thus, q1 = +nRT1ln(Vb/Va) Similarly, from c to d: isothermal, so that DU = 0 and Q = - W Thus, Q2 = +nRT2ln(Vd/Vc) = -nRT2ln(Vc/Vd) (-ve)

Which means that Carnot Cycle We see that: Now also: This is an important result. Temperature can be defined (on the absolute (Kelvin) scale) in terms of the heat flows in a Carnot Cycle.

Carnot Cycle We can also run the process in reverse and do work w and q2 and q1 will have opposite signs.

Efficiency of a heat engine The efficiency of a heat engine is defined simply as the fraction of the available energy that can be used for work. We already know how to relate the available energy qin to the enthalpy of reaction (or combustion for a fuel).

Efficiency of an engine cycle • When we analyse an engine cycle, we make use of the fact that, by returning to the same state, the internal energy change for a complete cycle is DE = 0. The First Law then tells us • qin = wout(DE = 0 = q – w) • If we can calculate the net heat in, then we know the work done. • A Carnot cycle is a theoretical model of the most efficient type of heat engine possible. It consists of adiabatic and isothermal steps. • A heating step, in which the working fluid (gas) absorbed heat at constant temperature, Thigh. • A working stroke in which the gas expands adiabatically – with no heat loss, cooling to Tlow. • A cooling step in which heat is released to surroundings at low temperature. • A compression step where the gas is compressed adiabatically until it reaches Thigh again.

Efficiency of the Carnot cycle Engine cycles may be represented on a P-V diagram. No heat flows in the expansion and compression steps, so we obtain for the maximum possible efficiency: It is not possible to convert all the heat into work, unless you can cool to and reject heat at absolute zero. The key is to maximise the ratio Thot/Tcold.

Sample Problems • Calculate the maximum possible efficiency of a heat engine operating between 343°C and 25°C. • hCarnot = 1 – (273+25)/(273+343) = 0.52 …only 52% efficiency! • What is the compression ratio P2/P1 when an ideal gas is compressed isothermally from V1= 250 cm3 to V2= 80cm3? • An ideal gas is described by the Equation of State, PV = nRT, where n is number of moles, R is the gas constant, and T is absolute temperature. • This is a closed, constant-temperature system; The number of moles is constant, which means the product nRT is constant. Hence PV is constant, or P1V1 = P2V2. • Therefore compression ratio P2/P1 = V1/V2 = 250/80 or 3.13.

Adiabatic Process in an Ideal Gas • adiabatic(thermallyisolated system) 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. P 2 to calculate W1-2 , we need to know P(V,T) for an adiabatic process PV= NkBT2 1 PV= NkBT1 V2 V1 V ( f – the # of “unfrozen” degrees of freedom )

Adiabatic Process in an Ideal Gas (cont.) P 2 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. PV= NkBT2 1 PV= NkBT1 V2 V1 V 1+2/3¼1.67 (monatomic), 1+2/5 =1.4 (diatomic), 1+2/6 \approx1.33 (polyatomic) (again, neglecting the vibrational degrees of freedom)

Problem Imagine that we rapidly compress a sample of air whose initial pressure is 105 Pa and temperature is 220C (= 295 K) to a volume that is a quarter of its original volume (e.g., pumping bike’s tire). What is its final temperature? Rapid compression – approx. adiabatic, no time for the energy exchange with the environment due to thermal conductivity For adiabatic processes: also - poor approx. for a bike pump, works better for diesel engines

Isothermal Reversible Change • In an isothermal reversible expansion, temperature does not change • T is not a function of V and can be removed from the integral • For ideal gas, w = -nRT ln (Vf/Vi) • If final volume > initial (expansion), w < 0 • System has done work on surroundings internal energy(U) decreased • If final volume < initial (compression), w > 0 • System has work done on it, U increased • Note that as T increases |w| increases

Adiabatic Reversible Process: q=0 Ideal gas