18. heat, work, & first law of thermodynamics 1. the 1 st law of thermodynamics 2. thermodynamic...
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18. Heat, Work, & First Law of Thermodynamics
1. The 1st Law of Thermodynamics
2. Thermodynamic Processes
3. Specific Heats of an Ideal Gas
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A jet aircraft engine converts the energy of burning fuel into mechanical energy.
How does energy conservation apply in this process?
E combustion = E mech + Q waste
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18.1. The 1st Law of Thermodynamics
Either heating or stirring can raise T of the water.
Joule’s apparatus
1st Law of Thermodynamics:Increase in internal energy = Heat added Work done
U Q W
Thermodynamic state variable = variable independent of history.
e.g., U, T, P, V, …
Not Q, W, …
dU dQ dW
dt dt dt
PE of falling weight
KE of paddle
Heat in water
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Example 18.1. Thermal Pollution
The reactor in a nuclear power plant supplied energy at the rate of 3.0 GW,
boiling water to produce steam that turns a turbine-generator.
The spent steam is then condensed through thermal contact with water taken from a river.
If the power plant produces electrical energy at the rate of 1.0 GW, at what rate is heat
transferred to the river?
dU dQ dW
dt d t d t
3.0 1.0dQ
GW GWdt
2.0dQ
GWdt
From standpoint of power plant:
1st law
( loses heat to river )
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18.2. Thermodynamic Processes
Quasi-static process: Arbitrarily slow process such that system always stays arbitrarily close to thermodynamic equilibrium.
Reversible process:Any changes induced by the process in the universe (system + environment) can be removed by retracing its path.
Reversible processes must be quasi-static.
Irreversible process:Part or whole of process is not reversible.
e.g., any processes involving friction, free expansion of gas ….
Twater = Tgas & rises slowly
system always in thermodynamic equilibrium
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Work & Volume Changes
W F x p A x p V
W d W2
1
V
Vp dV
面積
Work done by gas on piston
extW F x
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GOT IT? 18.1.
Two identical gas-cylinder systems are taken from the same initial state
to the same final state, but by different processes.
Which are the same in both cases:
(a)the work done on or by the gas,
(b)the heat added or removed, or
(c)the change in internal energy?
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Isothermal Processes
Isothermal process: T = constant.
2
1
V
VW p dV 2
1
V
V
n R TdV
V 2
1ln V
Vn R T V
2
1
lnV
W n R TV
3
2U N k T 0U Q W
2
1
lnV
Q W n R TV
Isothermal processes on ideal gas
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Example 18.2. Bubbles!
A scuba diver is 25 m down, where the pressure is 3.5 atm ( 350 kPa ).
The air she exhales forms bubbles 8.0 mm in radius.
How much work does each bubble do as it arises to the surface,
assuming the bubbles remain at 300 K.
PV n R T2
1
lnV
W n R TV
1 1 2 2P V P VT = const
0.94 J
ln 3.5W n R T
2 1
1 2
V P
V P
3.5
1
atm
atm 3.5
1 1 ln 3.5p V 34350,000 0.008 ln 3.5
3Pa m
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Constant-Volume Processes & Specific Heat
Constant-volume process ( isometric, isochoric, isovolumic ) : V = constant
0V 0W p V
U Q
VU Q n C T
CV = molar specific heat at constant volume
Ideal gas: U = U(T) ideal gas VU n C T for all processes
isometric processes
VQ n C T only for isometric processes
1V
V
dQC
n dT
Non-ideal gas:
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Isobaric Processes & Specific Heat
Isobaric Process : constant P
2 1W p V V p V
Q U W U p V
isobaric processesPQ n C T
CP = molar specific heat at constant pressure
P Vn C T n C T p V Ideal gas, isobaric :
Vn C T n R T
P VC C R Ideal gas
Isotherms
1P
P
dQC
n dT
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Adiabatic Processes
Adiabatic process: Q = constant
e.g., insulated system, quick changes like combustion, …
U W
Tactics 18.1. pV const adiabat, ideal gas
1P
V
C
C
1T V const Prob. 66
1 1 2 2
1
p V p VW
Prob. 62
Adiabatic: larger p
cdf
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TACTIC 18.1. Adiabatic Equation
Ideal gas, any process: VdU n C dT
p dV
pV n R T
Adiabatic process: dU dWVn C dT
p dV V d p n R dT
V
p p dV V dpdV
C R
0V VR C p dV C V d p
0p VC p dV C V d p
0dV d p
V p
p
V
C
C
ln lnV p const ln pV pV const
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Conceptual Example 18.1. Ideal-Gas Law vs Adiabatic Equation
The ideal gas law says p V = n R T,
but the adiabatic equation says p V = const.
Which is right ?
Both are right.
The adiabatic equation is a special case where T V +1
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Making the Connection
Suppose you halve the volume of an ideal gas with = 1.4.
What happens to the pressure if the process is
(a) isothermal and
(b) adiabatic?
Ans:
(a) pV = const p = 2 p0 (doubles)
(b) p V = const p = p0 2 = 2.64 p0
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Example 18.3. Diesel Power
Fuel ignites in a diesel engine from the heat of compression (no spark plug needed).
Compression is fast enough to be adiabatic.
If the ignit temperature is 500C, what compression ratio Vmax / Vmin is needed?
Air’s specific heat ratio is = 1.4, & before the compression the air is at 20 C.
1T V const
1 / 1.4 1273 500
273 20
K K
K K
1 / 1
max min
min max
V T
V T
11
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Application: Smog Alert!
Air is poor heat conductor convection is
adiabatic.
If rising air cools slower than surrounding air,
pollution rises high & can be dispersed.
Otherwise, smog.
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GOT IT? 18.2.
Name the basic thermodynamic process involved when each of the following is
done to a piston-cylinder system containing ideal gas,
& tell also whether T, p, V, & U increase or decrease.
(a)the piston is lock in place& a flame is applied to the bottom of the cylinder,
(b)the cylinder is completely insulated & the piston is pushed downward,
(c)the piston is exposed to atmospheric pressure & is free to move, while the
cylinder is cooled by placing it on a block of ice.
(a) isometric; T , p , V =const, U .
(b) adiabatic; T , p , V , U .
(c) isobaric; T , p =const, V , U .
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Ideal Gas Processes
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Cyclic Processes
Cyclic Process : system returns to same thermodynamic state periodically.
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Example 18.4. Finding the Work
An ideal gas with = 1.4 occupies 4.0 L at 300 K & 100 kPa
pressure.
It’s compressed adiabatically to ¼ of original volume,
then cooled at constant V back to 300 K,
& finally allowed to expand isothermally to its original V.
How much work is done on the gas?
1A A B B
AB
p V p VW
741 J
AB (adiabatic):
0BCW BC (isometric):
ln ACA
C
VW n R T
VCA (isothermal):
1.4 1100 4.0 1 4
1.4 1
kPa L
AB A
B
Vp p
V
1
11
A A AAB
B
p V VW
V
ln 4A Ap V 555 J
work done by gas: ABCA AB BC CAW W W W 186 J
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18.3. Specific Heats of an Ideal Gas
3
2ideal gasU N k T 1
V
UC
n T
Ideal gas: 21
2K m v
3
2k T
3
2n R T
3
2R
5
2RP VC C R P
V
C
C
5
3 1.67
Experimental values ( room T ):
For monatomic gases, 5/3, e.g., He, Ne, Ar, ….
For diatomic gases, 7/5 = 1.4, CV = 5R/2, e.g., H2 , O2 , N2 ,
….
For tri-atomic gases, 1.3, CV = 3.4R, e.g., SO2 , NO2 , ….
Degrees of freedom (DoF) = number of independent
coordinates required to describe the system
Single atom: DoF = 3 (transl)
For low T ( vib modes not active ) :
Rigid diatomic molecule : DoF = 5 (3 transl + 2 rot)
Rigid triatomic molecule : DoF = 6 (3 transl + 3 rot)
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The Equipartition Theorem
Equipartition theorem ( kinetic energy version):
For a system in thermodynamic equilibrium, each degree of freedom of a
rigid molecule contributes ½ kT to its average energy.
Equipartition theorem ( general version):
For a system in thermodynamic equilibrium, each degree of freedom described
by a quadratic term in the energy contributes ½ kT to its average energy.
2A
fU n N k T
2V
fC R
2
fn R T
2f
f
12P
fC R
DoF ( f ) CV CP
Monatomic 3 3/2 5/2 5/3
Diatomic 5 5/2 7/2 7/5
Triatomic 6 3 4 4/3
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Example 18.5. Gas Mixture
A gas mixture consists of 2.0 mol of oxygen (O2) & 1.0 mol of Argon (Ar).
Find the volume specific heat of the mixture.
2.2 R
2 2
5
2O OU n R T3
2Ar ArU n R T
2
5 3
2 2mix O ArU n n R T
1 mixmix
UC
n T
5 32.0 1.0
2 22.0 1.0
mol molR
mol mol
2
2
5 32 2O Ar
O Ar
n nR
n n
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Quantum Effects
CV of H2 gas as function of T.
Below 20 K hydrogen is liquid,
above 3200 K it dissociates into individual atoms.
Quantum effect:
Each mechanism has a threshold energy.
Etransl < Erot < Evib
Translation
rotation+Translation
rotation+Translation+vibration
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RepriseQuasi-static process : Arbitrarily slow process such that system always stays arbitrarily close to thermodynamic equilibrium.
Reversible process: Any changes induced by the process in the universe (system + environment) can be removed by retracing its path.
a c : Free expansion with no dissipative work.c b : Adiabatic.
a d : Adiabatic.d b : Free expansion with no dissipative work.
a e : Adiabatic.e b : Adiabatic dissipative work.
Insulated gas
1st law: The net adiabatic work done in all 3 processes are equal (shaded areas are equal).
Dissipative work: Work done on system without changing its configuration, irreversible.