Thermodynamics

THERMODYNAMICS Thermodynamics is the study of energy relationships that involve heat, mechanical work, and other aspects of energy and heat transfer. Central Heating

A THERMODYNAMIC SYSTEM A system is a closed environment in which heat transfer can take place. (For example, the gas, walls, and cylinder of an automobile engine.) Work done on gas or work done by gas

INTERNAL ENERGY OF SYSTEM The internal energy U of a system is the total of all kinds of energy possessed by the particles that make up the system. Usually the internal energy consists of the sum of the potential and kinetic energies of the working gas molecules.

TWO WAYS TO INCREASE THE INTERNAL ENERGY, U. WORK DONE ON A GAS (Positive) HEAT PUT INTO A SYSTEM (Positive)

TWO WAYS TO DECREASE THE INTERNAL ENERGY, U. Wout hot HEAT LEAVES A SYSTEM Q is negative Qout hot -U Decrease WORK DONE BY EXPANDING GAS: W is positive

THERMODYNAMIC PROCESS Increase in Internal Energy, U. Wout Initial State: P1 V1 T1 n1 Final State: P2 V2 T2 n2 Heat input Qin Work by gas

The Reverse Process Decrease in Internal Energy, U. Win Qout Work on gas Loss of heat Qout Win Initial State: P1 V1 T1 n1 Final State: P2 V2 T2 n2

THE FIRST LAW OF THERMODYAMICS: The net heat put into a system is equal to the change in internal energy of the system plus the work done BY the system. Q = U + W final - initial) Conversely, the work done ON a system is equal to the change in internal energy plus the heat lost in the process.

SIGN CONVENTIONS FOR FIRST LAW +Wout +Qin Heat Q input is positive U Work BY a gas is positive -Win U Work ON a gas is negative -Qout Heat OUT is negative Q = U + W final - initial)

APPLICATION OF FIRST LAW OF THERMODYNAMICS Example 1: In the figure, the gas absorbs 400 J of heat and at the same time does 120 J of work on the piston. What is the change in internal energy of the system? Wout =120 J Qin 400 J Q = U + W Apply First Law:

Example 1 (Cont.): Apply First Law Qin 400 J Wout =120 J DQ is positive: +400 J (Heat IN) DW is positive: +120 J (Work OUT) Q = U + W U = Q - W U = Q - W = (+400 J) - (+120 J) = +280 J  U = +280 J

Example 1 (Cont.): Apply First Law Energy is conserved: Qin 400 J Wout =120 J The 400 J of input thermal energy is used to perform 120 J of external work, increasing the internal energy of the system by 280 J The increase in internal energy is:  U = +280 J

FOUR THERMODYNAMIC PROCESSES: Isochoric Process: V = 0, W = 0 Isobaric Process: P = 0 Isothermal Process: T = 0, U = 0 Adiabatic Process: Q = 0 Q = U + W

ISOCHORIC PROCESS: CONSTANT VOLUME, V = 0, W = 0 Q = U + W so that Q = U +U -U QIN QOUT No Work Done HEAT IN = INCREASE IN INTERNAL ENERGY HEAT OUT = DECREASE IN INTERNAL ENERGY

ISOCHORIC EXAMPLE: 400 J B A P2 V1= V2 P1 PA P B TA T B = No Change in volume: B A P2 V1= V2 P1 PA P B TA T B = 400 J heat input increases internal energy by 400 J and zero work is done. Heat input increases P with const. V

ISOBARIC PROCESS: CONSTANT PRESSURE, P = 0 Q = U + W But W = P V QIN QOUT Work Out Work In +U -U HEAT IN = Wout + INCREASE IN INTERNAL ENERGY HEAT OUT = Win + DECREASE IN INTERNAL ENERGY

ISOBARIC EXAMPLE (Constant Pressure): 400 J B A P V1 V2 VA VB TA T B = 400 J heat does 120 J of work, increasing the internal energy by 280 J. Heat input increases V with const. P

Work = Area under PV curve ISOBARIC WORK B A P V1 V2 VA VB TA T B = 400 J PA = PB Work = Area under PV curve

ISOTHERMAL PROCESS: CONST. TEMPERATURE, T = 0, U = 0 Q = U + W AND Q = W QIN QOUT Work Out Work In U = 0 U = 0 NET HEAT INPUT = WORK OUTPUT WORK INPUT = NET HEAT OUT

ISOTHERMAL EXAMPLE (Constant T): B A PA V2 V1 PB U = T = 0 Slow compression at constant temperature: ----- No change in U. PAVA = PBVB

ISOTHERMAL EXPANSION (Constant T): B A PA VA VB PB PAVA = PBVB TA = TB U = T = 0 Isothermal Work 400 J of energy is absorbed by gas as 400 J of work is done by gas. T = U = 0

ADIABATIC PROCESS: NO HEAT EXCHANGE, Q = 0 Q = U + W ; W = -U or U = -W Work Out Work In U +U Q = 0 W = -U U = -W Work done at EXPENSE of internal energy INPUT Work INCREASES internal energy

ADIABATIC EXAMPLE: B A PA V1 V2 PB Expanding gas does work with zero heat loss. Work = -DU Insulated Walls: Q = 0

ADIABATIC EXPANSION: Q = 0 A PA PAVA PBVB B = PB TA T B VA VB 400 J of WORK is done, DECREASING the internal energy by 400 J: Net heat exchange is ZERO. Q = 0 Gamma is the ratio of the specific heat capacities at constant pressure and constant volume gamma=Cp/Cv

ADIABATIC EXPANSION/COMPRESSION According to the 1st Law of Thermodynamics: ΔU = Q – Wby but Q = 0, since walls are insulated ΔU = – Wby For a monatomic ideal gas: The red curve shows an adiabatic expansion of an ideal gas. The blue curves are isotherms at Ti and Tf. Adiabatic curves can be approximated as linear.

To Be Continued…