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
Thermodynamic systems
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:- isothermal: T = constant- isochoric: V = constant- isobaric: P = constant- adiabatic: Q = 0
Quasi-static processes
Work in thermodynamics
Expansion: work on piston positive, work on gas negativeCompression: work on piston negative, work on gas positive
Work during a volume change
∫=⇒
==
=
2
1
.
.
V
V
pdVW
pdV
Adxp
dxFdW
Work in pV diagrams
Work done equals area under curve in pV diagram
Careful with the signs…
1st Law of Thermodynamics
WQU −=∆Conservation of energy
Heat is positivewhen it entersthe system
Work is positivewhen it is done bythe system
Heat is negativewhen it leavesthe system
Work is negativewhen it is done onthe system
1st Law of Thermodynamics
pdVdQdU −=Conservation of energy
Heat is positivewhen it entersthe system
Work is positivewhen it is done bythe system
Heat is negativewhen it leavesthe system
Work is negativewhen it is done onthe system
)( 122 VVpW −=
a. isochoricb. isobaric
a. isobaricb. isochoric
)( 121 VVpW −= ∫= f
i
V
VpdVW
isothermal
• The work done by a system depends on the initial and final states and on the path � it is not a state function.
• Amount of heat transferred also depends on the initial, final, and intermediate states � it is not a state function either.
(a) (b) (c)
State Functions
State Functions
12 UUWQU −=−=∆
The internal energy 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!
This pV–diagram shows two ways to take a system from state a (at lower left) to state c (at upper right):
• via stateb (at upper left), or• via stated (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 heat transfer 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
CPS question
Thermodynamic Processes:•Adiabatic: no heat transfer (by insulation or by ve ry fast process)
Q=0 →→→→ U2 – U1 = -W
•Isochoric: constant volume process (no work done)
W=0 →→→→ U2 – U1 = Q
•Isobaric: constant pressure process
p=const. →→→→ W = p (V2 – V1)
•Isothermal: constant temperature process (heat may flow but veryslowly 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
Thermodynamic processes
Isolated systems:
fi UU
U
WQ
=→=∆→==
0
0
Cyclic processes
Pi, f
Vinitial state = final state
WQ
U
=→=∆ 0
The internal energyof an isolated systemsremains constant
Adiabatic processes
WU
Q
−=∆→= 0
Expansion: U decreasesCompression: U increasesEnergy exchange between
“heat” and “work”
First Law for Several Types of Processes
More about cyclic processesP
i, f
V
Work equals the area enclosed by the curves (careful with the sign!!!)
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.
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,
Important formulas
∆U=Q-W (1st law)
(work during a volume change)
pV=nRT (Ideal gas law)
CV=f/2nR (Equipartition theorem)
∫=2
1
V
V
pdVW
Isochoric process: V = constant
V
P
V1,2
1
2 22 nRTVp =
11 nRTVp =
02
1
==→ ∫VV pdVW
TC
TTCQ
V
V
∆=−= )( 12
Heat
reservoir
During an isochoric process, heat enters (leaves) the system and increases (decreases) the internal energy.
(CV: heat capacity at constant volume)
TCQU
WQU
V ∆==∆→−=∆
Ideal gas: isochoric process
Isobaric process: p = constant
V
P
V1
12
22 nRTpV =
11 TNkpV B=
VpVVppdVWV
V∆=−==→ ∫ )( 12
2
1
V2 TC
TTCQ
p
p
∆=
−= )( 12(CP: heat capacity at constant pressure)
VpTC
WQU
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
Ideal gas: isobaric process
Isothermal process: T = constant
V
P
V1
1
2
nRTpV =
1
2ln
2
1
2
1
2
1
V
VnRT
V
dVnRT
dVV
nRTpdVW
V
V
V
V
V
V
=
=
==
∫∫∫
V2
00 =∆⇒=∆ UT
1
2lnV
VnRTWQ ==→
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.
Ideal gas: isothermal process
Heat capacities of an Ideal gasConsider an isobaric process p=constant
TCQ pp ∆=From the 1st Law of Thermodynamics:
pdVdUdTCdQ pp +==but
dTCdU V=pdVdTCdTC Vp +=⇒
nRdTpdVVdppdVnRTpV ==+⇒=From the Ideal gas law:
nRCCnRdTdTCdTC VpVp +=⇒+=⇒
Heat capacities of an Ideal gas
nRf
C
nRCC
V
Vp
2=
+=nR
fCp 2
2+=⇒
f
fnR
fnR
f
C
C
V
p 2
2/
2
2 +=+=⇒
f = #degrees of freedom
For a monoatomic gas f=3
67.13/5;2
5 ===⇒V
pp C
CnRC
Molar heat capacities of various gases at (25 C)
1.058.763.2927.3636.12H2S
1.028.513.4128.3936.90N2O
1.028.453.3928.1736.62CO2
Polyatomic
1.008.402.4920.7429.04CO
1.018.392.5220.9829.37O2
1.018.382.4620.4428.82H2
1.008.322.5020.8029.12N2
Diatomic
0.998.271.5112.5220.79Xe
1.008.341.5012.4520.79Kr
1.008.341.5012.4520.79Ar
0.988.111.5212.6820.79Ne
0.998.271.5112.5220.79He
Monoatomic
(Cp-Cv)/RCp-CvCv/RCvCpGas
V
P
V1
1
2
V2
Adiabatic process: Q = 0),( TVpp = ∫∫ == 2
1
2
1
),(V
V
V
VdVTVppdVW
TNkpV B=
nRdTVdppdV
nRTdpVd
=+→=→ )()(
pdVRdTf
n
pdVdTC
pdVdWdU
V
−=→
−=→−=−=
2
pdVf
VdppdV2−=+
f is the # of degrees of freedom
Ideal gas: adiabatic process
V
P
V1
1
2
V2
),( TVpp =
0)2
1(
2
=++→
−=+
pdVf
Vdp
pdVf
VdppdV
0=+V
dV
p
dp γ
constant0ln
0lnln
0
11
11
11
11
==→=→
=+→
=+ ∫∫
γγγ
γ
γ
γ
VppVVp
pV
V
V
p
p
V
dV
p
dp V
V
p
p
let , and dividing by pV)2
1(f
+=γ
Ideal gas: adiabatic process (contd)
V
P
V1
1
2
V2
constant=γpV
)(1
1
)(
)(
2211
2211
21
VpVp
VpVpR
C
TTnC
TnCUW
V
V
V
−−
=
−=
−=∆−=∆−=
γ
−−
=→ −− 12
11
11
11
)1(
1γγ
γ
γ VVVpW
constant11 == γγ VppV
Ideal gas: adiabatic process (contd)
Using:
V
P
V1
1
2
V2
constant=γpV
γγ2211 VpVp =
222
111
nRTVp
nRTVp
==
2
1
2
1
2
1
T
T
V
V
p
p =→
γ
γ
1
2
2
1
V
V
p
p =→
2
11
1
12
T
T
V
V =→ −
−
γ
γ
or
constant11 == γγ VppV
constant122
111 == −− γγ VTVT
Ideal gas: adiabatic process (contd)
V
P
V1
1
2
V2
constant=γpV constant11 == γγ VppV
constant122
111 == −− γγ VTVT
nRTpV =
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.
Ideal gas: adiabatic process (contd)
−−
= −− 12
11
11 11
)1( γγ
γ
γ VV
VpW
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.
CPS question
Quasi-static process
Character U∆ WQ
adiabatic 0=Q WU −=∆
isothermal T = constant 0=∆U
isochoric
isobaric
V = constant
p = constant
QU =∆ TCQ V ∆= 0=W
VpW ∆=WQU −=∆ TCQ P∆=
1
2lnV
VnRTW =WQ =
)11
()1(
11
11
2
11 −− −−
= γγγ
γ VVVpW0=Q
Summary