Work in Thermodynamic Processes Energy can be transferred to a system by heat and/or work The system will be a volume of gas always in equilibrium Consider a cylinder with a movable piston As piston is pressed a distance Δy, work is done on the system reducing the volume  W = -F Δy = - P A Δy

Work in Thermodynamic Processes – Cont. Work done compressing a system is defined to be positive Since ΔV is negative (smaller final volume) & A Δy = V  W = - P ΔV Gas compressed  Won gas = pos. Gas expands  Won gas = neg.

Work in Thermodynamic Processes – Cont. Can only be used if gas is under constant pressure An isobaric process (iso = the same) P1 = P2 Represented on a pressure vs. volume graph – a PV diagram Area under any curve = work done on the gas If volume decreases – work is positive (work is done on the system)

THERMODYNAMICS First Law of Thermodynamics Energy is conserved Heat added to a system goes into internal energy, work or both ΔU = Q + W Heat added to system  internal energy  Q is positive Work done to the system  internal energy  W is positive (again)

First Law – Cont. A system will have a certain amount of internal energy (U) It will not have certain amounts of heat or work These change the system U depends only on state of system, not what brought it there ΔU is independent of process path (like Ug)

Isothermal Process Temperature remains constant Since P = N kB T / V = constant / V An isotherm (line on graph) is a hyperbola

Isothermal Process – Cont. Moving from 1 to 2, temperature is constant, so P & V change Work is done = area under curve Internal energy is constant because temperature is constant  Q = -W Heat is converted into mechanical work

Isometric (isovolumetric) Process The volume is not allowed to change V1 = V2 Since no change in volume, no work is done  ΔU = Q Heat added must go into internal energy  it Heat extracted is at the expense of internal energy  it

Isometric Process – Cont. The PV diagram representation No change in volume Area under curve = 0 That is, no work done The process moves from one isotherm to another

Isobaric Process As heat is added to system, pressure is required to be constant The ratio of V / T = constant Some of the heat does work and the rest causes a change in temperature Thus moving to another isotherm Recall: changes in temperature = changes in internal energy  ΔU = Q + W

Adiabatic Process No heat is transferred into or out of system Q = 0  ΔU = W All work done to a system goes into internal energy increasing temperature All work done by the system comes from internal energy & system gets cooler

Adiabatic Process – Cont. Either system is insulated to not allow heat exchange or the process happens so fast there is no time to exchange heat

Adiabatic Process – Cont. Since temperature changes, we move isotherms

The Second Law of Thermodynamics & Heat Engines Heat will not flow spontaneously from a colder body to a warmer OR: Heat energy cannot be transferred completely into mechanical work OR: It is impossible to construct an operational perpetual motion machine

Heat Engines Any device that converts heat energy into work Takes heat from a high temperature source (reservoir), converts some into work, then transfers the rest to surroundings (cold reservoir) as waste heat

Heat Engines – Cont. Consider a cylinder and piston Surround by water bath & allow to expand along an isothermal The heat flowing in (Q) along AC equals the work done by the gas as it expands (W) since ΔU = 0 To return to A along same isothermal, work is done on the gas and heat flows out Work expanding = work compressing

Heat Engines – Cont. A cycle naturally can have positive work done In going from A to B work is done by gas, temperature  (ΔU ) and heat enters system B to C No work done, T , ΔU , & heat leaves system C to A ΔU = 0, heat leaving = work done The work out = the net heat in (ΔU = 0)

Heat Engines – Cont. Thermal Efficiency Used to rate heat engines efficiency = work out / heat in  e = Wout / Qin Qin = heat into heat engine Qout = heat leaving heat engine For one cycle, energy is conserved  Qin = W + Qout Since system returns to its original state ΔU = 0

The Carnot Engine Any cyclic heat engine will always lose some heat energy What is the maximum efficiency? Solved by Sadi Carnot (France) (Died at 36) Must be reversible adiabatic process

The Carnot Engine – Cont. Carnot Cycle A four stage reversible process 2 isotherms & 2 adiabats Consider a hypothetical device – a cylinder & piston Can alternately be brought into contact with high or low temperature reservior High temp – heat source Low temp – heat sink – heat is exhausted

The Carnot Engine – Cont. Step 1: an isothermal expansion, from A to B Cylinder receives heat from source Step 2: an adiabatic expansion, from B to C

The Carnot Engine – Cont. Step 3: an isothermal compression, C to D Ejection of heat to sink at low temp Step 4: an adiabatic compression, D to A Represents the most efficient (ideal) device Sets the upper limit

Entropy A measure of disorder A messy room > neat room Pile of bricks > building made from them A puddle of water > ice came from All real processes increase disorder  increase entropy  of entropy of one system can be reduced at the expense of another

Entropy – Cont. Entropy of the universe always increases The universe only moves in one direction – towards  entropy This creates a “direction of time flow” Nature does not move systems towards more order

Entropy – Cont. As entropy , energy is less able to do work The “quality” of energy has been reduced Energy has “degraded” Nature proceeds towards what is most likely to happen