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We want to do a further exercise on "pV work of ideal gases"
discuss - this time it comes to adiabatic expansion work.
We compress or expand a gas
adiabatically, ie thermally
isolated.
There is no possibility for the gas to exchange heat with the environment. Usually
is the temperature
of the gas change in an adiabatic process.
A short Rückglick on the isothermal volume work:
Here, Boyle's law is: pV = const.
p times V is kontant. The implemented work
W = n * R * T * ln (V (I) / V (A))
and the heat Q is-W as delta (L) is zero.
In contrast, the following equations apply when adiabatic process: Here, V p is high time constant kappa
p * V ^ k = const. (One of the three
Poisson Adiabatengleichungen) Kappa is the Poisson adiabatic coefficient -
kappa is a characteristic value for each gas.
One can choose from the numerical value of kappa
Close to the molecular structure of the gas.
Working for the adiabatic process is W = C (V) * delta (T)
and the heat is by definition equal to zero, resulting delta (U) is equal to W.
In the pV diagram is an adiabatic always a steeper curve as a
Isotherm.
One kilogram of water vapor is
of 450 degrees Celsius
adiabatic
from 10 bar to 1 bar
be relaxed.
A process that also
in a steam engine
is used for the production of work: Expansion of 10 bar and 450 ° C at 1 bar.
The process is adiabatic, there is no heat exchange, therefore, is
the temperature
clearly
decrease, because the gas does work when the
Pressure of 10 bar to 1 bar decreases
and the volume increases.
The adiabatic coefficient of water should be assumed to be constant 1.3.
The process should run perfectly reversible;
it is the process variable W (rev)
and a number of state variables
- Delta (S), delta (V), delta (T), delta (p)) -
to be determined.
We outline for all relevant variables
a table.
(1) is the compressed initial state, (2), the final expanded
(I) of the process - the adiabatic expansion of
If ...
Where are the following variables:
Initial temperature: 723 Kelvin
Initial pressure: 10 bar
Initial volume?
(We do not know yet)
Discharge pressure: 1 bar.
For more information:
The process is reversible adiabatic;
the medium is an ideal gas;
and the Poisson adiabatic kappa = 1.3 is also given.
Adiabatic means that heat is zero
(So we can also have a process variable in the table complement)
We
first calculate the amount of substance n
1 kg of water correspond to a molecular weight of
18 g / mol by a molar
55.6 mol.
55.6 mol at 10 bar and 20 ° C after having
ideal gas law
a volume of 334 L
Liter.
Thus, the initial state
completely
described.
We now use one of the three
Poisson Adiabatengleichungen, the initial and final states
Adiabatic connect on a p (A), V (A) kappa = p ^ (E) V (S) ^ kappa.
p (A) and V (A) are given
p (E) is also given, so we can according to the V (E) ^ kappa
cancel
We get V (E) = p ^ kappa (A) / p (D) * V (A) kappa ^
V (S) = p ^ kappa (A) / p (D) * V (A) kappa ^
We draw the kappa-th root and get
V (E) = kappa-th root of (p (A) / p (E)) * V (A)
1.3-th root of
(1000000
by 100000) or
1.3-th root of 10
times resulting in a final volume of 0,334 m³,
the 5.8 times greater than the starting volume.
Note the difference to the isothermal expansion:
This would be a reduction of the pressure by a factor of 10
an increase in the volume by a factor of 10 group.
p (A), V (A) = p (E) V (e)
In the adiabatic expansion in the pressure reduction by a factor of 10 is only an increase of volume by a factor of 5.8,
because the temperature has decreased simultaneously.
The final volume V (E) is almost
2000 liters.
The final temperature T (E) can easily after
ideal gas equation
be calculated,
where we now calculate the final state.
Inserting p (E) and V (E) and
the amount of substance n
providing a final temperature of 425 Kelvin.
Next, we calculate the volume of work,
which the water vapor that:
W = C (V) * delta (T) The temperature change delta (T) we know;
the molar heat C (V) is not given directly, but we can calculate from kappa.
For an ideal gas
C (p) minus C (V) equal to n times R.
The quotient C (p) by C (V) is defined as kappa.
We have two
Equations
and we have
two unknowns (namely, C (p) and C (V)), we solve the Gleichugnssystem to C (V) on
and C obtained (V) = n * R / (kappa - 1)
We calculate the isochoric heat capacity to
C (V) = 1.54 kJ / K
and
W by multiplying the expansion work of C (V) with delta (T) = -298 K
delta (T) = -298 K
We obtain for W is a negative number, that is -459 kJ -
this work is given by the system.
The change in the internal energy delta (U) is easy to calculate,
because after the first main theorem is delta (U) = Q + W and Q = 0, that is
delta (U) = 459 kJ.
The steam
after the process has to a 459 kJ
lower internal energy
than before the process.
This energy difference delta (U) was completely converted into work.
In the PV diagram of the initial state (1) is at 10 bar and 334 liters, and the
Final state
is 1 bar.
This red line
would be an isotherm
when the outdoor temperature (723 K) - a hyperbola -
Since an adiabatic always steeper than
an isotherm is, is the final state (2)
below the red curve. The blue line corresponds to
an isotherm at the final temperature (425 K) - another hyperbola.
The black line represents
the adiabatic curve between (1) and (2).
The work done by the gas which corresponds to the area under the curve in the PV diagram