Thermodynamic Isolated systems can exchange neither nor matter with the .

reservoir reservoir

Heat

Work

Open systems can exchange Closed systems exchange energy both matter and energy with but not matter with the environment. the environment. Quasi-static processes

Quasi-static processes : near equilibrium

Initial state, final state, intermediate state: p, V & T well defined

Sufficiently slow processes = any intermediate state can considered as at . Thermal equilibrium means that It makes sense to define a .

Examples of quasi-static processes: - isothermal: T = constant - isochoric: V = constant - isobaric: P = constant - adiabatic: Q = 0 Work in

Expansion: work on positive, work on gas negative Compression: work on piston negative, work on gas positive Work during a change

dW = F.dx = p.Adx = pdV

V2 ⇒ W = ∫ pdV V1 Work in pV diagrams

Work done equals under curve in pV diagram

Careful with the signs… 1st Law of Thermodynamics

Heat is negative Work is negative when it leaves when it is done on the the system

Heat is positive Work is positive when it enters when it is done by the system the system ∆U = Q −W 1st Law of Thermodynamics

Heat is negative Work is negative when it leaves when it is done on the system the system

Heat is positive Work is positive when it enters when it is done by the system the system dU = dQ − pdV Conservation of energy State Functions

(a) (b) (c)

a. isochoric a. isobaric isothermal b. isobaric b. isochoric

V = − = − W = f pdV W p2 (V2 V1) W p1(V2 V1) ∫ Vi

• The work done by a system depends on the initial and final states and on the path  it is not a .

• Amount of heat transferred also depends on the initial, final, and intermediate states  it is not a state function either. State Functions ∆ = − = − U Q W U2 U1

The U is a state function: the energy gain (loss) only depends on the initial and final states, and not on the path. Even though Q and W depend on the path, ∆U does not! CPS question This pV –diagram shows two ways to take a system from state a (at lower left) to state c (at upper right): • via state b (at upper left), or • via state d (at lower right) For which path is W > 0? A. path abc only B. path adc only C. both path abc and path adc D. neither path abc nor path adc E. The answer depends on what the system is made of. CPS question A system can be taken from state a to state b along any of the three paths shown in the pV – diagram. If state b has greater internal energy than state a, along which path is the absolute value | Q| of the the greatest?

A. path 1 B. path 2 C. path 3 D. | Q| is the same for all three paths. E. not enough information given to decide Thermodynamic Processes:

•Adiabatic: no heat transfer (by insulation or by very fast process)

→→→ Q=0 U2 –U1 = -W

•Isochoric: constant volume process (no work done)

→→→ W=0 U2 –U1 = Q

•Isobaric: constant process

→→→ p=const. W = p (V 2 –V1)

•Isothermal: constant temperature process (heat may flow but very slowly so that thermal equilibrium is not disturbed)

∆∆∆U=0, Q =-W only for . Generally ∆∆∆U, Q, W not zero any energy entering as heat must leave as work Thermodynamic processes First Law for Several Types of Processes

Isolated systems: Cyclic processes Adiabatic processes

P i, f

Q = W = 0 V initial state = final state → ∆U = 0 Q = 0 ∆U = 0 →U = U → ∆U = −W i f → Q = W Expansion: U decreases The internal energy Compression: U increases of an isolated systems Energy exchange between remains constant “heat” and “work” More about cyclic processes

P i, f

V

Work equals the area enclosed by the curves (careful with the sign!!!) CPS question An ideal gas is taken around the cycle shown in this pV –diagram, from a to b to c and back to a. Process b → c is isothermal. For this complete cycle,

A. Q > 0, W > 0, and ∆U = 0. B. Q > 0, W > 0, and ∆U > 0. C. Q = 0, W > 0, and ∆U < 0. D. Q = 0, W < 0, and ∆U > 0. E. Q > 0, W = 0, and ∆U > 0. Important formulas

∆U=Q-W (1 st law)

V2 W = ∫ pdV (work during a volume change) V1 pV=nRT ()

CV=f /2 nR () Ideal gas:

P Isochoric process: V = constant = 2 p2V nRT 2 V → W = 2 pdV = 0 ∫V 1 1 Q = C (T −T ) pV = nRT V 2 1 1 1 (C V: = ∆ at constant volume) CV T V1,2 V ∆U = Q −W → ∆ = = ∆ U Q CV T reservoir

During an isochoric process, heat enters Heat (leaves) the system and increases (decreases) the internal energy. Ideal gas: P = Isobaric process: p = constant pV 2 nRT 2 2 → = V2 = − = ∆ 1 W pdV p(V2 V1) p V ∫V 1

pV = Nk T = − 1 B 1 Q C p (T2 T1) V V = ∆ (C P: heat capacity 1 2 V C p T at constant pressure)

→ ∆U = Q −W reservoir = ∆ − ∆ CP T p V

Heat During an isobaric expansion process, heat enters the system. Part of the heat is Work used by the system to do work on the environment ; the rest of the heat is used to increase the internal energy. Ideal gas: P 1 pV = nRT Isothermal process: T = constant

2 ∆T = 0 ⇒ ∆U = 0

V2 V2 nRT W = pdV = dV ∫V ∫V 1 1 V1 V2 V V

V2 dV = nRT Expansion: ∫V 1 heat enters the system V all of the heat is used by the system V2 to do work on the environment. = nRT ln V1 Compression: V the work done on the system increases → = = 2 its internal energy, Q W nRT ln all of the energy leaves the system at V1 the same as the heat is removed. Heat capacities of an Ideal gas

Consider an isobaric process p=constant = ∆ Qp Cp T From the 1 st Law of Thermodynamics: = = + dQ p CpdT dU pdV but ⇒ C dT = C dT + pdV = p V dU CV dT

From the Ideal gas law: pV = nRT ⇒ pdV +Vdp = pdV = nRdT

⇒ = + ⇒ = + CpdT CV dT nRdT Cp CV nR Heat capacities of an Ideal gas C = C + nR p V + ⇒ = f 2 f = #degrees f Cp nR of freedom C = nR 2 V 2 C f + 2 f f + 2 ⇒ p = nR / nR = CV 2 2 f

For a monoatomic gas f=3 C ⇒ = 5 p = = Cp nR ; 3/5 67.1 2 CV Molar heat capacities of various gases at (25 C)

Gas Cp Cv Cv/R Cp-Cv (Cp-Cv)/R Monoatomic He 20.79 12.52 1.51 8.27 0.99 Ne 20.79 12.68 1.52 8.11 0.98 Ar 20.79 12.45 1.50 8.34 1.00 Kr 20.79 12.45 1.50 8.34 1.00 Xe 20.79 12.52 1.51 8.27 0.99 Diatomic N2 29.12 20.80 2.50 8.32 1.00 H2 28.82 20.44 2.46 8.38 1.01 O2 29.37 20.98 2.52 8.39 1.01 CO 29.04 20.74 2.49 8.40 1.00 Polyatomic CO2 36.62 28.17 3.39 8.45 1.02 N2O 36.90 28.39 3.41 8.51 1.02 H2S 36.12 27.36 3.29 8.76 1.05 Ideal gas:

Adiabatic process: Q = 0 P = 2 p p(V ,T ) V V W = ∫ 2 pdV = ∫ 2 p(V ,T )dV V1 V1 1

V pV = Nk T 2 V1 V B → d( pV ) = d(nRT ) → pdV +Vdp = nRdT 2 = − = − pdV +Vdp = − pdV dU dW pdV f → = − CV dT pdV f → n RdT = − pdV 2

f is the # of degrees of freedom Ideal gas: adiabatic process (contd) 2 pdV +Vdp = − pdV P f 2 p = p(V ,T ) 2 →Vdp + 1( + ) pdV = 0 f 1 2 let γ = 1( + ), and dividing by pV f V V 2 1 V dp dV + γ = 0 p V

p dp V dV + γ = 0 ∫p ∫V 1 p 1 V p V → ln + γ ln = 0 p1 V1 γ → pV = → γ = γ = ln γ 0 pV p1V1 constant p1V1 Ideal gas: adiabatic process (contd) = −∆ = − ∆ P γ W U nC V T pV = constant 2 = − nC V (T1 T2 )

= CV − 1 ( p1V1 p2V2 ) R V V 1 2 1 V = ( pV − p V ) γ −1 1 1 2 2

Using: γ = γ = pV p1V1 constant

γ 1  1 1  → W = pV  −  1 1 γ −  γ −1 γ −1  ( )1 V1 V2  Ideal gas: adiabatic process (contd)

γ = γ = P γ = pV p1V1 constant 2 pV constant

1 pV = nRT 1 1 1 → p1 V1 = T1 = p2V2 nRT 2 p2 V2 T2 V2 V 1 V γ γ γ p V = → 1 = 2 p1V1 p2V2 γ p2 V1 γ −1 → V2 = T1 γ −1 V1 T2 γ −1 = γ −1 = or T1V1 T2 V2 constant Ideal gas: adiabatic process (contd)

γ γ P γ = pV = pV = constant 2 pV constant 1 1 γ −1 = γ −1 = T1V1 T2 V2 constant 1 pV = nRT V V γ 2 1 V pV  1 1  W = 1 1  −  γ −  γ −1 γ −1  ( )1 V1 V2 

During an adiabatic expansion process, the reduction of the internal energy is used by the system to do work on the environment.

During an adiabatic compression process, the environment does work on the system and increases the internal energy. CPS question When an ideal gas is allowed to expand isothermally from

volume V1 to a larger volume V2, the gas does an amount of work equal to W12 . If the same ideal gas is allowed to expand adiabatically from

volume V1 to a larger volume V2, the gas does an amount of work that is

A. equal to W12 .

B. less than W12 .

C. greater than W12 .

D. either A., B., or C., depending on the ratio of V2 to V1. Summary

Quasi-static Character ∆ process U Q W ∆ = = ∆ = isochoric V = constant U Q Q CV T W 0 ∆ = − = ∆ W = p∆V isobaric p = constant U Q W Q CP T V isothermal T = constant ∆U = 0 Q = W W = nRT ln 2 V1 γ 1 1 1 W = pV ( − ) adiabatic Q = 0 ∆U = −W Q = 0 1 1 γ − γ −1 γ −1 ( )1 V2 V1