Thermodynamic systems Isolated systems can exchange neither energy - - PowerPoint PPT Presentation
Thermodynamic systems Isolated systems can exchange neither energy nor matter with the environment. reservoir reservoir Heat Heat Work Work Open systems can exchange Closed systems exchange energy both matter and energy with but not
Isolated systems can exchange neither energy nor matter with the environment. Closed systems exchange energy but not matter with the environment. Heat Work reservoir Open systems can exchange both matter and energy with the environment. Heat Work reservoir
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. Thermal equilibrium means that It makes sense to define a temperature. Examples of quasi-static processes:
T = constant
V = constant
P = constant
Q = 0
Expansion: work on piston positive, work on gas negative Compression: work on piston negative, work on gas positive
2 1
V V
Work done equals area under curve in pV diagram
Careful with the signs…
Heat is positive when it enters the system Work is positive when it is done by the system Heat is negative when it leaves the system Work is negative when it is done on the system
Heat is positive when it enters the system Work is positive when it is done by the system Heat is negative when it leaves the system Work is negative when it is done on the system
) (
1 2 2
V V p W − =
) (
1 2 1
V V p W − =
∫
=
f i
V V pdV
W
isothermal
states it is not a state function either. (a) (b) (c)
1 2
The internal energy U is a state function: the energy gain (loss) only depends on the initial and final states, and not
Even though Q and W depend on the path, ∆U does not!
This pV–diagram shows two ways to take a system from state a (at lower left) to state c (at upper right):
For which path is W > 0?
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 heat transfer the greatest?
CPS question
Thermodynamic Processes:
Q=0 → → → → U2 – U1 = -W
W=0 → → → → U2 – U1 = Q
p=const. →
→ → → W = p (V2 – V1)
slowly so that thermal equilibrium is not disturbed) ∆ ∆ ∆ ∆U=0,Q =-W only for ideal gas. Generally ∆
∆ ∆ ∆U,Q, W not zero
any energy entering as heat must leave as work
Isolated systems:
f i
U U U W Q = → = ∆ → = =
Cyclic processes P
i, f
V initial state = final state
W Q U = → = ∆
The internal energy
remains constant
Adiabatic processes
W U Q − = ∆ → = 0
Expansion: U decreases Compression: U increases Energy exchange between “heat” and “work”
First Law for Several Types of 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,
∆U=Q-W (1st law) (work during a volume change) pV=nRT (Ideal gas law) CV=f/2nR (Equipartition theorem)
=
2 1
V V
pdV W
Isochoric process: V = constant
V P V1,2 1 2
2 2
nRT V p =
1 1
nRT V p =
2 1
= = →
V V pdV
W
T C T T C Q
V V
∆ = − = ) (
1 2
Heat
reservoir During an isochoric process, heat enters (leaves) the system and increases (decreases) the internal energy.
(CV: heat capacity at constant volume)
T C Q U W Q U
V∆
= = ∆ → − = ∆
Isobaric process: p = constant
V P V1 1 2
2 2
nRT pV =
1 1
T Nk pV
B
=
V p V V p pdV W
V V
∆ = − = = →
∫
) (
1 2
2 1
V2
T C T T C Q
p p
∆ = − = ) (
1 2
(CP: heat capacity at constant pressure)
V p T C W Q U
P
∆ − ∆ = − = ∆
→
During an isobaric expansion process, heat enters the system. Part of the heat is used by the system to do work on the environment; the rest of the heat is used to increase the internal energy. Heat Work reservoir
Isothermal process: T = constant
V P V1 1 2
nRT pV =
1 2
ln
2 1 2 1 2 1
V V nRT V dV nRT dV V nRT pdV W
V V V V V V
= = = =
∫ ∫ ∫
V2
= ∆ ⇒ = ∆ U T
1 2
ln V V nRT W Q = = →
Expansion: heat enters the system all of the heat is used by the system to do work on the environment. Compression: the work done on the system increases its internal energy, all of the energy leaves the system at the same time as the heat is removed.
Consider an isobaric process p=constant
T C Q
p p
∆ =
From the 1st Law of Thermodynamics:
pdV dU dT C dQ
p p
+ = =
but
dT C dU
V
= pdV dT C dT C
V p
+ = ⇒ nRdT pdV Vdp pdV nRT pV = = + ⇒ =
From the Ideal gas law:
nR C C nRdT dT C dT C
V p V p
+ = ⇒ + = ⇒
nR f C nR C C
V V p
2 = + = nR f Cp 2 2 + = ⇒ f f nR f nR f C C
V p
2 2 / 2 2 + = + = ⇒
f = #degrees
For a monoatomic gas f=3
67 . 1 3 / 5 ; 2 5 = = = ⇒
V p p
C C nR C
Molar heat capacities of various gases at (25 C)
1.05 8.76 3.29 27.36 36.12 H2S 1.02 8.51 3.41 28.39 36.90 N2O 1.02 8.45 3.39 28.17 36.62 CO2 Polyatomic 1.00 8.40 2.49 20.74 29.04 CO 1.01 8.39 2.52 20.98 29.37 O2 1.01 8.38 2.46 20.44 28.82 H2 1.00 8.32 2.50 20.80 29.12 N2 Diatomic 0.99 8.27 1.51 12.52 20.79 Xe 1.00 8.34 1.50 12.45 20.79 Kr 1.00 8.34 1.50 12.45 20.79 Ar 0.98 8.11 1.52 12.68 20.79 Ne 0.99 8.27 1.51 12.52 20.79 He Monoatomic (Cp-Cv)/R Cp-Cv Cv/R Cv Cp Gas
V P V1 1 2 V2
Adiabatic process: Q = 0
) , ( T V p p =
∫ ∫
= =
2 1 2 1
) , (
V V V V
dV T V p pdV W
T Nk pV
B
=
nRdT Vdp pdV nRT d pV d = + → = → ) ( ) (
pdV RdT f n pdV dT C pdV dW dU
V
− = → − = → − = − = 2
pdV f Vdp pdV 2 − = + f is the # of degrees of freedom
V P V1 1 2 V2
) , ( T V p p =
) 2 1 ( 2 = + + → − = + pdV f Vdp pdV f Vdp pdV
= + V dV p dp γ
constant ln ln ln
1 1 1 1 1 1
1 1
= = → = → = + → = + ∫
∫
γ γ γ γ
γ γ V p pV V p pV V V p p V dV p dp
V V p p
let , and dividing by pV ) 2 1 ( f + = γ
V P V1 1 2 V2
constant =
γ
pV
) ( 1 1 ) ( ) (
2 2 1 1 2 2 1 1 2 1
V p V p V p V p R C T T nC T nC U W
V V V
− − = − = − = ∆ − = ∆ − = γ
− − = →
− − 1 2 1 1 1 1
1 1 ) 1 ( 1
γ γ γ
γ V V V p W
constant
1 1
= =
γ γ
V p pV
Using:
V P V1 1 2 V2
constant =
γ
pV
γ γ 2 2 1 1
V p V p =
2 2 2 1 1 1
nRT V p nRT V p = =
2 1 2 1 2 1
T T V V p p = →
γ γ 1 2 2 1
V V p p = →
2 1 1 1 1 2
T T V V = →
− − γ γ
constant
1 1
= =
γ γ
V p pV
constant
1 2 2 1 1 1
= =
− − γ γ
V T V T
V P V1 1 2 V2
constant =
γ
pV
constant
1 1
= =
γ γ
V p pV
constant
1 2 2 1 1 1
= =
− − γ γ
V T V T
nRT pV =
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.
− − =
− − 1 2 1 1 1 1
1 1 ) 1 (
γ γ γ
γ V V V p W
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
CPS question
Quasi-static process Character
U ∆
W
Q
adiabatic
= Q
W U − = ∆
isothermal T = constant
= ∆U
isochoric isobaric V = constant p = constant
Q U = ∆
T C Q
V∆
=
= W
V p W ∆ =
W Q U − = ∆
T C Q
P∆
=
1 2
ln V V nRT W =
W Q =
) 1 1 ( ) 1 ( 1
1 1 1 2 1 1 − − −
− =
γ γ γ
γ V V V p W
= Q