Thermodynamics Davidson College APSI

Ideal Gas Equations P 1 V 1 / T 1 = P 2 V 2 / T 2 PV = n R T (using moles) P V = N k B T (using molecules)  P: pressure (Pa)  V: volume (m 3 )  N: number of molecules  k B : Boltzman’s constant 1.38 x J/K  T: temperature (K)

System Boundary For our purposes, the system will almost always be an ideal gas. (gas)

First Law of Thermodynamics U (E int ) W Q UU

T1T1 T2T2 T3T3 Gas “isotherms” Isothermal Process (constant temperature) P V PV = nRT  T = 0 (constant T) Initial State of Gas Final State of Gas Isothermal Process

Isobaric Process (constant pressure) P V T1T1 T2T2 T3T3 PV = nRT  P = 0 (constant P) Isobaric Expansion Isobaric Contraction

Isometric Process (constant volume) P V T1T1 T2T2 T3T3  V = 0 (constant V) PV = nRT

Adiabatic process (insulated) P V T isotherm adiabat Q = 0 (no heat enters or leaves) Temperature, pressure, and volume all change in an adiabatic process. PV = nRT

p Work done BY gas VV p W gas = p  V W env = -p  V Positive work Negative work

p Work done ON gas VV p W gas = p  V W ext = -p  V Negative since  V is negative Positive since  V is negative

W CD = p 1  V Work (isobaric) P V P2P2 V1V1 A V2V2 B P1P1 C D W AB > W CD Where we are considering work done BY the gas W AB = p 2  V

W ACD Work is path dependent P V P2P2 V1V1 A V2V2 B P1P1 C D W ABD > W ACD Where we are considering work done BY the gas W ABD

Work done by cycle P V P2P2 V1V1 A V2V2 B P1P1 C D Work done by the gas is equal to the area circumscribed by the cycle. W ABCD Work done by gas is positive for clockwise cycles, negative for counterclockwise cycles. Work done by environment is negative of work done by gas.

Problem Calculate the heat necessary to change the temperature of one mole of an ideal gas from 300K to 500K A) at constant volume. B) at constant pressure (assume 1 atmosphere).

Second Law of Thermodynamics No process is possible whose sole result is the complete conversion of heat from a hot reservoir into mechanical work. (Kelvin-Planck statement.) No process is possible whose sole result is the transfer of heat from a cooler to a hotter body. (Clausius statement.)

Heat Transfer Heat Source (High Temperature) Heat Sink (Low Temperature) QHQH QCQC Q H = Q C

Heat Engines Engine Heat Source (High Temperature) Heat Sink (Low Temperature) QHQH QCQC W Q H = Q C + W

Work and Heat Engines Q H = W + Q C  Q H : Heat that is put into the system and comes from the hot reservoir in the environment.  W: Work that is done by the system on the environment.  Q C : Waste heat that is dumped into the cold reservoir in the environment. Engine Heat Source (High Temperature) Heat Sink (Low Temperature) QHQH QCQC W

Efficiency of Heat Engine Efficiency = W/Q H = (Q H - Q C )/Q H  W: Work done by engine on environment  Q H : Heat absorbed from hot reservoir  Q C : Waste heat dumped to cold reservoir Efficiency is often given as percent efficiency.

Adiabatic vs Isothermal Expansion Volume Pressure Initial State Isothermal Expansion Adiabatic Expansion In an adiabatic expansion, no heat energy can enter the gas to replace energy being lost as it does work on the environment. The temperature drops, and so does the pressure.

Carnot Cycle P V Isothermal expansion Adiabatic expansion Isothermal compression Adiabatic compression Q H = Q C + W Efficiency = W/Q H

Efficiency of Carnot Cycle For a Carnot engine, the efficiency can be calculated from the temperatures of the hot and cold reservoirs. Carnot Efficiency = (T H - T C )/T H  T H : Temperature of hot reservoir (K)  T C : Temperature of cold reservoir (K)