Isothermal Process on P-V, T-V, and P-T Diagrams

Isothermal process on p-V, T-V, and p-T diagrams

isothermal ⇒ T = T0 = constant

a = (p1, V1, T0) b = (p2, V2, T0)

pV = nRT0 p T p

a p1 p1 a a b T0 Q T0

b p2 p2 b W

V1 V2 V V1 V2 V T0 T nRT p V 0 T V T p T ( ) = V ( ) = 0 ( ) = multivalued

PHYS 1101, Winter 2009, Prof. Clarke 1 Isochoric process on p-V, T-V, and p-T diagrams

isochoric ⇒ V = V0 = constant

a = (p1, V0, T1) b = (p2, V0, T2)

pV0 = nRT p T p p1 a T1 a p1 a

p2 b T2 b p2 b

V0 V V0 V T2 T1 T nRT p(V) = multivalued T(V) = multivalued p(T) = V0

PHYS 1101, Winter 2009, Prof. Clarke 2 Isobaric process on p-V, T-V, and p-T diagrams

isobaric ⇒ p = p0 = constant

a = (p0, V1, T1) b = (p0, V2, T2)

p0V = nRT p T p

T2 b Q a b a b p0 p0

W T1 a

V1 V2 V V1 V2 V T1 T2 T p V p V p T V 0 p T p ( ) = 0 ( ) = nR ( ) = 0

PHYS 1101, Winter 2009, Prof. Clarke 3 Clicker question 1

Consider the p-V diagram below in which the system evolves from a → b –1 → c. If T0 ~ 240K (and thus RT0 = 2,000 J mol ), how many moles of gas, n, are in the system? p (Nm–2) a) 5 isobar b) 105 a b 105 isotherm c) 50 d) 1,000 T0 isochor e) Not enough information to tell pc c

3 1 Vc V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 4 Clicker question 1

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 5 Clicker question 2

Consider the p-V diagram below in which the system evolves from a → b → c. What is Vc, the volume at state c?

3 p (Nm–2) a) 0.5 m b) 2.0 m3 a b 105 c) 4.0 m3 d) 8.0 m3 T0 e) Not enough information to tell pc c

3 1 Vc V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 6 Clicker question 2

Consider the p-V diagram below in which the system evolves from a → b → c. What is Vc, the volume at state c?

3 p (Nm–2) a) 0.5 m b) 2.0 m3 a b 105 c) 4.0 m3 d) 8.0 m3 T0 e) Not enough information to tell pc c

3 need to know pc 1 Vc V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 7 Clicker question 3

Consider the p-V diagram below in which the system evolves from a → b → c. What is Vc, the volume at state c?

3 p (Nm–2) a) 0.5 m b) 2.0 m3 a b 105 c) 4.0 m3 d) 8.0 m3 T0 e) Not enough information to tell 4 5 ×10 c

3 1 Vc V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 8 Clicker question 3

Consider the p-V diagram below in which the system evolves from a → b → c. What is Vc, the volume at state c?

3 p (Nm–2) a) 0.5 m b) 2.0 m3 a b 105 c) 4.0 m3 d) 8.0 m3 T0 e) Not enough information to tell 4 5 ×10 c p p V = p V ⇒ V = a V = 2 m3 3 c c a a c p a 1 Vc V (m ) c

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 9 Clicker question 4

Consider the p-V diagram below in which the system evolves from a → b → c. What is the net change in internal energy, ΔEint?

p (Nm–2) a) 0 J 4 b) 5.0 ×10 J a b 105 4 c) about 7.0 ×10 J d) 105 J T0 e) Not enough information to tell 4 5 ×10 c

3 1 2 V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 10 Clicker question 4

Consider the p-V diagram below in which the system evolves from a → b → c. What is the net change in internal energy, ΔEint?

p (Nm–2) a) 0 J 4 b) 5.0 ×10 J a b 105 4 c) about 7.0 ×10 J d) 105 J T0 e) Not enough information to tell 4 5 ×10 c ΔE = nC ΔT 3 int V 1 2 V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 11 Clicker question 5

Consider the p-V diagram below in which the system evolves from a → b → c. What is the net work done by the system on its environment, W?

p (Nm–2) a) 0 J 4 b) 5.0 ×10 J a b 105 4 c) about 7.0 ×10 J d) 105 J T0 e) Not enough information to tell 4 5 ×10 c

3 1 2 V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 12 Clicker question 5

Consider the p-V diagram below in which the system evolves from a → b → c. What is the net work done by the system on its environment, W?

p (Nm–2) a) 0 J W 4 b) 5.0 ×10 J a b 105 4 c) about 7.0 ×10 J d) 105 J T0 e) Not enough information to tell 4 5 ×10 c

3 1 2 V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 13 Clicker question 6

Consider the p-V diagram below in which the system evolves from a → b → c. What is the net heat transferred into the system, Q?

4 p (Nm–2) a) –5.0 ×10 J 4 b) 5.0 ×10 J a b 105 c) –105 J d) 105 J T0 e) Not enough information to tell 4 5 ×10 c

3 1 2 V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 14 Clicker question 6

Consider the p-V diagram below in which the system evolves from a → b → c. What is the net heat transferred into the system, Q?

4 p (Nm–2) a) –5.0 ×10 J 4 b) 5.0 ×10 J a b 105 c) –105 J d) 105 J T0 e) Not enough information to tell 4 5 ×10 c Q E W 0 105 J 3 = Δ int + = + 1 2 V (m )

first law of thermodynamics: ΔEint = Q – W ( = nCVΔT ) ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 15 Internal Energy (revisited)

f f E = nC T = nRT = NkT C = C + R int V 2 2 p V 22 n = number of moles; 1 mole = 6.0221 × 10 particles (NA) N = number of particles R = gas constant = 8.3147 J mol–1 K–1 –23 –1 k = Boltzmann’s constant = 1.3807 × 10 J K

type of gas degrees of specific heat at internal specific heat at γ freedom constant energy constant (Cp/CV ) ( f ) volume (CV ) (Eint) pressure (Cp ) monatomic 3 3 R 3 nRT 5 R 5 2 2 2 3 diatomic 5 5 R 5 nRT 7 R 7 2 2 2 5 4 polyatomic ( 3) ~6 3 R 3 nRT 4 R ≥ 3

PHYS 1101, Winter 2009, Prof. Clarke 16 Adiabatic processes

reversible irreversible

a = (p1, V1, T1) a = (p1, V1, T0)

b = (p2, V2, T2) b = (p2, V2, T0) γ pV = constant p1V1 = p2V2

p p

a a p1 adiabat p1 isotherm T1 isotherms T0 T2 b b p2 p2

V1 V2 V V1 V2 V

PHYS 1101, Winter 2009, Prof. Clarke 17 Clicker question 7

Consider the p-V diagram below in which the system evolves reversibly along the adiabat from state a to state b. This gas is…

a) monatomic (γ = 5/3) p (kNm–2) b) diatomic (γ = 7/5) b 16 c) polyatomic (γ = 4/3) d) not enough information to tell adiabat

a 1 3 1 8 V (m ) pV γ = constant

PHYS 1101, Winter 2009, Prof. Clarke 18 Clicker question 7

Consider the p-V diagram below in which the system evolves reversibly along the adiabat from state a to state b. This gas is…

a) monatomic (γ = 5/3) p (kNm–2) b) diatomic (γ = 7/5) b 16 c) polyatomic (γ = 4/3) d) not enough information to tell adiabat γ γ 3γ paVa = 1(8) = 2 = a 1 γ γ 4 3 pbVb = 16(1) = 2 1 8 V (m ) ⇒ γ = 4/3 ⇒ polyatomic pV γ = constant

PHYS 1101, Winter 2009, Prof. Clarke 19 Clicker question 8

Consider the p-V diagram below in which the system evolves reversibly along the adiabat from state a to state b. How much heat is transferred to the system? a) 0 J p (kNm–2) b) 8 kJ b 16 c) 16 kJ d) 128 kJ adiabat e) not enough information to tell

a 1 3 1 8 V (m ) pV γ = constant

PHYS 1101, Winter 2009, Prof. Clarke 20 Clicker question 8

Consider the p-V diagram below in which the system evolves reversibly along the adiabat from state a to state b. How much heat, Q, is transferred to the system? a) 0 J p (kNm–2) b) 8 kJ b 16 c) 16 kJ d) 128 kJ adiabat e) not enough information to tell

1 a By definition, Q = 0 for all adiabatic 1 8 V (m3) processes. pV γ = constant

PHYS 1101, Winter 2009, Prof. Clarke 21 Clicker question 9

Consider the p-V diagram below in which the system evolves reversibly along the adiabat from state a to state b. How much work, W, does the system do on its environment? a) 20 kJ b) –20 kJ p (kNm–2) c) 24 kJ d) –24 kJ b 16 e) 32 kJ f) –32 kJ g) not enough information to tell adiabat

a 1 3 1 8 V (m )

Eint = nCVT = n3RT (for a polyatomic gas)

first law: ΔEint = Q – W ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 22 Clicker question 9

ΔEint = 0 – W = 3Δ(nRT) = 3Δ(pV) 1 W a = 3(16 – 8) = 24 ⇒ W = –24 kJ 3 1 8 V (m )

Eint = nCVT = n3RT (for a polyatomic gas)

first law: ΔEint = Q – W ideal gas law: pV = nRT PHYS 1101, Winter 2009, Prof. Clarke 23 Summary of Processes

a = (p , V , T , S ) S 1 1 2 3 S isentrop: Δ = 0 p S 1 (reversible adiabat: Q = 0) 3 b = (p1, V2, T1, S1)

S2 isobar: Δp = 0 c = (p2, V2, T2, S2) S4 isochor: ΔV = 0 a d = (p4, V2, T3, S3) p b 1 isotherm: ΔT = 0 e = (p3, V1, T3, S4)

T1 c p2 e p3 T2 p d 4 T3

V1 V2 V free expansion (irreversible adiabat: Q = 0)

PHYS 1101, Winter 2009, Prof. Clarke 24 Summary of Processes

ΔE = process W Q int ΔS nCVΔT pC pC V p V – V p V – V V V – V nC ln 2 isobar ( 2 1) ( 2 1) ( 2 1) p ( ) R R V1 VC VC p V p – p V p – p nC ln 2 isochor 0 ( 2 1) ( 2 1) V ( ) R R p1

V2 V2 V2 isotherm nRT ln( ) nRT ln( ) 0 nR ln ( ) V1 V1 V1 p V – p V p V – p V isentrop 1 1 2 2 0 2 2 1 1 0 γ – 1 γ – 1

free V2 0 0 0 nR ln ( ) expansion V1

PHYS 1101, Winter 2009, Prof. Clarke 25 All processes on p-V, T-V, and T-S diagrams

p T T p a p p

a a T T

V T S V V S S

V V S

An isobar (p), isotherm (T), isentrop (S), and isochor (V) emanating from the same initial state (a) as manifest on a p-V, T-V, and a T-S diagram.

PHYS 1101, Winter 2009, Prof. Clarke 26 Clicker question 10

Consider the p-V diagram below in which n = 1 mole of gas evolves reversibly from state a to state b along the path shown. What is the net change in entropy? (Note, e = 2.71828 = Euler’s number, and thus ln(e) = 1.) p a) Cp b) Cp e

c) CV d) CV e a b e) R f) R e g) no where near enough information!!

V nC ln 2 (isobar) V ΔS = p ( ) 1 e V1 p S = nC ln 2 (isochor) Δ V ( p ) some formulae, in case they help… 1 V nR ln 2 (isotherm) ΔS = ( ) V1 PHYS 1101, Winter 2009, Prof. Clarke 27 Clicker question 10

Consider the p-V diagram below in which n = 1 mole of gas evolves revers -ibly from state a to state b along the path shown. What is the net change in entropy? (Note, e = 2.71828 = Euler’s number, and thus ln(e) = 1.) p a) Cp b) Cp e

c) CV d) CV e a b e) R f) R e g) Yes there is!!

V nC ln 2 (isobar) V ΔS = p ( ) 1 e V1 p Since S is a state variable, it doesn’t S = nC ln 2 (isochor) Δ V ( p ) matter which path from a to b you 1 V nR ln 2 (isotherm) choose. Thus, choose the isobar. ΔS = ( ) V1 PHYS 1101, Winter 2009, Prof. Clarke 28 Clicker question 11

The evolution of a system from state a to state b is shown on both the p-V and T-S diagrams below. What is the change in internal energy?

p (Nm–2) T (K) a 500 ΔE = Q – W b a int 2,000 = nCV ΔT b 250 pV = nRT

3 –1 1 2 V (m ) 10 30 S (JK )

a) 2000 J b) –2000 J c) 5000 J d) –5000 J e) 7000 J f) –7000 J

PHYS 1101, Winter 2009, Prof. Clarke 29 Clicker question 11

The evolution of a system from state a to state b is shown on both the p-V and T-S diagrams below. What is the change in internal energy?

p (Nm–2) T (K) Q ~ –7000 J a 500 b a W = –2000 J 2,000 ΔEint = Q – W b 250 ~ –5000 J W = –2000 J Q ~> –7500 J

3 –1 1 2 V (m ) 10 30 S (JK )

a) 2000 J b) –2000 J c) 5000 J d) –5000 J e) 7000 J f) –7000 J

PHYS 1101, Winter 2009, Prof. Clarke 30 Clicker question 12

The evolution of a system from state a to state b is shown on both the p-V and T-S diagrams below. About how many moles of gas are in the system? (Take R = 8.) p (Nm–2) T (K) a 500 ΔE = Q – W b a int 2,000 = nCV ΔT b 250 pV = nRT

3 –1 1 2 V (m ) 10 30 S (JK )

a) 0.5 b) 1 c) 1.5 d) 2 e) not enough information to tell

PHYS 1101, Winter 2009, Prof. Clarke 31 Clicker question 12

The evolution of a system from state a to state b is shown on both the p-V and T-S diagrams below. About how many moles of gas are in the system? (Take R = 8.) p (Nm–2) T (K) pV = nRT a 500 b a at state b:

2,000 pV = 2000

b 250 RT ~ 2000 ⇒ n ~ 1

3 –1 1 2 V (m ) 10 30 S (JK )

a) 0.5 b) 1 c) 1.5 d) 2 e) not enough information to tell

PHYS 1101, Winter 2009, Prof. Clarke 32 Clicker question 13

The evolution of a system from state a to state b is shown on both the p-V

and T-S diagrams below. With n = 1 and ΔEint = –5000 J, this gas is:

p (Nm–2) T (K) a 500 ΔEint = nCV ΔT b a

2,000 CV ~ 12 (monatomic) b 250 CV ~ 21 (diatomic)

CV ~ 25 (polyatomic)

3 –1 1 2 V (m ) 10 30 S (JK )

a) monatomic b) diatomic c) polyatomic d) not enough information to tell

PHYS 1101, Winter 2009, Prof. Clarke 33 Clicker question 13

The evolution of a system from state a to state b is shown on both the p-V

and T-S diagrams below. With n = 1 and ΔEint = –5000 J, this gas is:

p (Nm–2) T (K) a 500 ΔEint = nCV ΔT b a

2,000 CV ~ 12 (monatomic) b 250 CV ~ 21 (diatomic)

CV ~ 25 (polyatomic)

3 –1 1 2 V (m ) 10 30 S (JK )

a) monatomic b) diatomic c) polyatomic d) not enough information to tell

PHYS 1101, Winter 2009, Prof. Clarke 34 Three pictorial representations of an engine

a→c Qin p a Qin TH

W > 0

Qout c expansion stroke V c→a In a complete thermodynamical cycle, gas T expands at high pressure C Qout and compresses at low Q = Q pressure allowing work, H in Q = Q W, to be extracted in C out W = W each cycle. out compression PHYS 1101, Winter 2009, Prof. Clarke stroke 35 Maximum efficiency and the Carnot cycle

p S T T 1 Q S Qin in a 2 a a b Qin adiabats TH TH isotherms b W > 0 W > 0 TC TH TC Q TC out c d c d c Qout Qout S S S S S S 1 2 1 2 V non-optimal thermo optimal thermodyn the Carnot engine cycle -dynamical cycle for -amical cycle for an on a p-V diagram an engine engine (Carnot cycle)

Note that for an engine, the thermodynamical cycle is always clockwise.

PHYS 1101, Winter 2009, Prof. Clarke 36 Refrigerators (heat pumps)

p p a Qout S1 S a 2 Qout adiabats W < 0 isotherms Qin c b TC TH V d c For a refrigerator, the Qin

cycle is always V counterclockwise. As for an engine, the most Expansion happens at optimal thermodynamical low pressure, cycle for a refrigerator is the compression at high Carnot cycle traversed in pressure and this takes Q = Q C in the counterclockwise work. Heat is drawn in at Q = Q H out direction (opposite to the T and expelled at T . W = –W C H in engine). PHYS 1101, Winter 2009, Prof. Clarke 37