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 STATE The STATE of a thermodynamic system is determined by four factors: Absolute Pressure P in Pascals Temperature T in Kelvins Volume V in cubic meters Number of moles, n, of working gas

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 U =Q + W But W = - P V QIN QOUT Work Out Work In +U -U HEAT IN = Wout + INCREASE IN INTERNAL ENERGY HEAT OUT = Wout + DECREASE IN INTERNAL ENERGY

ISOBARIC EXAMPLE (Constant Pressure): 400 J B A P V1 V2 VA VB TA T B = 400 J heat minus 120 J of work, increases 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 = 0 400 J of energy is absorbed by gas as 400 J of work is done on gas. T = U = 0 Isothermal Work

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

MOLAR HEAT CAPACITY OPTIONAL TREATMENT The molar heat capacity C is defined as the heat per unit mole per Celsius degree. Check with your instructor to see if this more thorough treatment of thermodynamic processes is required.

SPECIFIC HEAT CAPACITY Remember the definition of specific heat capacity as the heat per unit mass required to change the temperature? For example, copper: c = 390 J/kgK

MOLAR SPECIFIC HEAT CAPACITY The “mole” is a better reference for gases than is the “kilogram.” Thus the molar specific heat capacity is defined by: C = Q n T For example, a constant volume of oxygen requires 21.1 J to raise the temperature of one mole by one kelvin degree.

SPECIFIC HEAT CAPACITY CONSTANT VOLUME How much heat is required to raise the temperature of 2 moles of O2 from 0oC to 100oC? Q = nCv T Q = (2 mol)(21.1 J/mol K)(373 K - 273 K) Q = +4220 J

SPECIFIC HEAT CAPACITY CONSTANT VOLUME (Cont.) Since the volume has not changed, no work is done. The entire 4220 J goes to increase the internal energy, U. Q = U = nCv T = 4220 J Thus, U is determined by the change of temperature and the specific heat at constant volume. U = nCv T

SPECIFIC HEAT CAPACITY CONSTANT PRESSURE We have just seen that 4220 J of heat were needed at constant volume. Suppose we want to also do 1000 J of work at constant pressure? Same Q = U + W Q = 4220 J + J Cp > Cv Q = 5220 J

HEAT CAPACITY (Cont.)  U = nCvT For constant pressure Heat to raise temperature of an ideal gas, U, is the same for any process. U = nCvT For constant pressure Q = U + W nCpT = nCvT + P V Cp > Cv Cp Cv 

REMEMBER, FOR ANY PROCESS INVOLVING AN IDEAL GAS: PAVA PBVB TA T B = PV = nRT Q = U + W U = nCv T

Example Problem: A 2-L sample of Oxygen gas has an initial temp- erature and pressure of 200 K and 1 atm. The gas undergoes four processes: AB: Heated at constant V to 400 K. BC: Heated at constant P to 800 K. CD: Cooled at constant V back to 1 atm. DA: Cooled at constant P back to 200 K.

PV-DIAGRAM FOR PROBLEM PB 2 L 1 atm 200 K 400 K 800 K How many moles of O2 are present? Consider point A: PV = nRT

PROCESS AB: ISOCHORIC B PB What is the pressure at point B? A PA P B = 2 L 1 atm 200 K 400 K 800 K What is the pressure at point B? PA P B TA T B = 1 atm P B 200 K 400 K = P B = 2 atm or 203 kPa

PROCESS AB: Q = U + W W = 0 Q = U = nCv T Analyze first law for ISOCHORIC process AB. B A PB 2 L 1 atm 200 K 400 K 800 K W = 0 Q = U = nCv T U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K) Q = +514 J U = +514 J W = 0

PROCESS BC: ISOBARIC B C PB D What is the volume at point C (& D)? 1 atm 200 K 400 K 800 K D 4 L What is the volume at point C (& D)? VB V C TB T C = 2 L V C 400 K 800 K = V C = V D = 4 L

FINDING U FOR PROCESS BC. 2 L 1 atm 200 K 400 K 800 K 4 L 2 atm Process BC is ISOBARIC. P = 0 U = nCv T U = (0.122 mol)(21.1 J/mol K)(800 K - 400 K) U = +1028 J

FINDING W FOR PROCESS BC. 2 L 1 atm 200 K 400 K 800 K 4 L 2 atm Work depends on change in V. P = 0 Work = P V W = (2 atm)(4 L - 2 L) = 4 atm L = 405 J W = +405 J

FINDING Q FOR PROCESS BC. 2 L 1 atm 200 K 400 K 800 K 4 L 2 atm Analyze first law for BC. Q = U + W Q = +1028 J + 405 J Q = +1433 J Q = 1433 J U = 1028 J W = +405 J

PROCESS CD: ISOCHORIC B A PB What is temperature at point D? PC P D 2 L 1 atm 200 K 400 K 800 K C D What is temperature at point D? PC P D TC T D = 2 atm 1 atm 800 K TD = T D = 400 K

PROCESS CD: Q = U + W W = 0 Q = U = nCv T PB 2 L 1 atm 200 K 400 K 800 K Analyze first law for ISOCHORIC process CD. W = 0 Q = U = nCv T U = (0.122 mol)(21.1 J/mol K)(400 K - 800 K) Q = -1028 J U = -1028 J W = 0

FINDING U FOR PROCESS DA. Process DA is ISOBARIC. A D 2 L 1 atm 200 K 400 K 800 K 4 L 2 atm P = 0 U = nCv T U = (0.122 mol)(21.1 J/mol K)(400 K - 200 K) U = -514 J

FINDING W FOR PROCESS DA. 2 L 1 atm 200 K 400 K 800 K 4 L 2 atm Work depends on change in V. P = 0 Work = P V W = (1 atm)(2 L - 4 L) = -2 atm L = -203 J W = -203 J

FINDING Q FOR PROCESS DA. 2 L 1 atm 200 K 400 K 800 K 4 L 2 atm Analyze first law for DA. Q = U + W Q = -514 J - 203 J Q = -717 J Q = -717 J U = -514 J W = -203 J

PROBLEM SUMMARY DQ = DU + DW For all processes:

NET WORK FOR COMPLETE CYCLE IS ENCLOSED AREA B C 2 L 1 atm 4 L 2 atm +404 J B C 2 L 1 atm 4 L 2 atm Neg -202 J 2 L 4 L B C 1 atm 2 atm Area = (1 atm)(2 L) Net Work = 2 atm L = 202 J

ADIABATIC EXAMPLE: Example 2: A diatomic gas at 300 K and 1 atm is compressed adiabatically, decreasing its volume by 1/12. (VA = 12VB). What is the new pressure and temperature? ( = 1.4) A B PB VB VA PA PAVA PBVB TA T B = PAVA = PBVB  Q = 0

ADIABATIC (Cont.): FIND PB Q = 0 PAVA = PBVB  A B PB VB 12VB 1 atm 300 K Solve for PB: PB = 32.4 atm or 3284 kPa

ADIABATIC (Cont.): FIND TB Q = 0 A B 32.4 atm VB 12VB 1 atm 300 K Solve for TB TB=? (1 atm)(12VB) (32.4 atm)(1 VB) (300 K) T B = TB = 810 K

ADIABATIC (Cont.): If VA= 96 cm3 and VA= 8 cm3, FIND W Q = 0 A B 32.4 atm 1 atm 300 K 810 K Since Q = 0, W = - U 8 cm3 96 cm3 W = - U = - nCV T & CV= 21.1 j/mol K PV RT n = Find n from point A PV = nRT

ADIABATIC (Cont.): If VA= 96 cm3 and VA= 8 cm3, FIND W PV RT n = = (101,300 Pa)(8 x10-6 m3) (8.314 J/mol K)(300 K) n = 0.000325 mol & CV= 21.1 j/mol K A B 32.4 atm 1 atm 300 K 810 K 8 cm3 96 cm3 T = 810 - 300 = 510 K W = - U = - nCV T W = - 3.50 J

HEAT ENGINES Qhot Wout Qcold Absorbs heat Qhot Performs work Wout A heat engine is any device which through a cyclic process: Cold Res. TC Engine Hot Res. TH Qhot Wout Qcold Absorbs heat Qhot Performs work Wout Rejects heat Qcold