Heat transfer thermodynamics and thermodynamic system
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Jul 22, 2024
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Language: en
Added: Jul 22, 2024
Slides: 31 pages
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THERMODYNAMICS
16.1 Work
Heat
Form of energy which is transferred from one body to
another body at lower temperature, by virtue of the
temperature difference between the bodies.
Internal energy
Total energy content of a system. It is the sum of all
forms of energy possessed by the atoms and
molecules of the system.
16.1 Work
System
The collection of matter within prescribed and identifiable boundaries.
Surroundings
Everything else in the environment.
Thermodynamics system
System that can interact with its surroundings or environment in at least two
ways, one of which is heat transfer.
16.1 Work
Work
Assume a system in a piston that contains fluid, the work done by the system
as its volume changes.
pA
dx
A
16.1 Work
When the piston moves out an infinitesimal distance
dx, the work dW done by this force is
dw = Fdx = pAdx
A : cross-sectional area of the cylinder
p : pressure exerted by the system at the
surface of the piston.
F : total force exerted by the system on
the piston
16.1 Work
But dV=Adx, therefore
dW=pdV.
The work done in a volume
change from V
1to V
2,AdxdV 2
1
1
2
P
P
nRT
V
V
nRTW lnln
Graph p versus V
pp
(a) p vs V at changes pressure
V
(a) p vs V at constant pressure
V
16.2 First Law of Thermodynamics
The internal energy of a
system changes from an initial
value U1 to a final value U2
due to heat, Q and work, W:WQUUU
12
Cyclic process
The change in internal energy depends on the gas
temperature.
It also depends on the initial and final state.
Cyclic process :
Initial state = Final state
U1 = U2
Q = W
Example:
Two moles of the monatomic argon gas expand isothermally at 298 K, from an
initial volume of 0.025 m3 to a final volume of 0.050 m3 (Argon is an ideal
gas), find (a) work done by the gas
(b) the heat supplied to the gas.
(a) W=+3400 J,
(b) Q =+3400 J
16.3 Thermodynamic
Processes
Isothermal process
Isochoric process
Isobaric process
Adiabatic process
Graph Isothermal process
The work done by the system is the area under the curve of a pV-diagram
as a graph shown below..
p
V
V
i V
f
Work,W
p
i
p
f
Isothermal process2
1
1
2
lnln
P
P
nRT
V
V
nRTW
Isochoric process
a constant-volume process, where there is no work done in the system, W = 0.
The first law of thermodynamic will be,
U2-U1 = ΔU = Q.
Graph Isochoric process
The graph pV of an isochoric process is shown below.
p
v
Isothermal process
a constant-temperature process.
Heat transfer and work in the system must not
change and the internal energy will be zero,
ΔU=0.
The changes in pressure,p and volume,V will
take place.
Isobaric process
Isobaric process is a constant-pressure process.
W = p(V2 –V1)
The first law of thermodynamic will be,
Q= ΔU +W = ΔU + p( V2-V1)
Graph Isobaric process
p
V
Isobaric process carried out at a constant pressure
Adiabatic process
a process with no heat
transfer to or from a system,
Q=0.
ΔU = -W
Example
The temperature of three moles of a monatomic
ideal gas is reduced from Ti = 540 K to Tf = 350 K
by two different methods. In the first method
5500 J of heat flows into the gas, while in the
second method, 1500 J of heat flows into it. In
each case, find (a) the change in the internal
energy of the gas and (b) the work done by the
gas.
16.4 Molar Specific Heat Capacities At Constant Pressure
And Volume
Molar specific heat capacities of constant pressure, Cp
When an n mole of gas undergoes a heating process,
the equation that relates Q and the temperature changes, dT
is written as ;
Q = nCdT ……………..(1)
Whereas Q = heat
n = number of mole
C = molar specific heat capacities
dT = temperature changes
When the gas is heated at constant pressure, the (1) equation will be;
Q = nCpdT
Or Cp =
Where Cp = molar specific heat capacities at constant
pressure.ndT
Q
Molar specific heat capacities of constant volume, Cv
Hence, when an n mole of gas is heated at constant volume, the (1)
equation will be;
Q= nCvdT
Or
Where Cv = molar specific heat capacities at constant volume
The universal gas constant, R
the substraction between Cp and Cv that is;
Cp-Cv = R
Where R = 8314 J mol-1 K
Example
A certain perfect gas has specific heat capacities as follows:
Cp = 0.84 kJ / mol K and
Cv = 0.657 kJ / mol K
Calculate the gas constant.
16.5 Specific Heat At Constant Pressure And
Volume
Example
A 2.00 mol sample of an ideal gas with γ = 1.40 expands slowly and
adiabatically from a pressure of 5.00 atm and a volume of 12.0 L to a final
volume of 30.0 L.
(a) What is the final pressure of the gas?
(b) What are the initial and final temperature?