Jaimes Soto Braulio
Proceso Isobárico.
Isobars process is that thermodynamical process that occurs at constant pressure. There,
the heat transferred at constant pressure is related to other variables by:
\ triangle Q = \ triangle U + P \ triangle V,
Where:
Q \! = Heat transferred.
U \! = Internal Energy.
P \! = Pressure.
V \! = Volume.
On a P-V diagram, a process isobars shown as a horizontal line.
Volume vs pressure graph, the isobars in the pressure is constant. Work (W) is the
integral of the pressure in the volume
You might say otherwise as:
Evolution of a thermodynamic system at constant pressure. The water boiled in an open
container into the atmosphere is an example of process isobars.
When a system undergoes a thermodynamic process isobars from state defined by the
variables p and V1, the state defined by p and V2, the work done is given by W = p (V2
- V1). The work done by the system is positive when the increase in volume is positive,
is working on the thermodynamic system if the increase in volume is negative.
The heat produced or absorbed when a system undergoes a thermodynamic process
isobars is equal to the change in enthalpy of the process.
Problem:
A mass of 0.6 [g] of air are in a cylinder with a piston
3 pressure [bar], a temperature of 176 [oC] and an initial volume of
260
[cm3].
For
each
point
calculate
the
final
state,
the
work, the heat involved in the process and the type of process
that question.
e. Is compressed at constant pressure and temperature is tripled.
Solution:
Tf=?
Vf=?
W=?
Q=?
Formulas
W= -nR(Tf-T0)= -p(Vf-V0)
Tf/T0=Vf/V0
W= U-Q
Vf=V0(Tf/T0)=260(52800/176)=780 [cm^3]
W=-300000(780-260)(1m^3/1x10^6 m^3)= -156 [J]
W= U-Q = W= -Q , Q= -W
Porque se desprecia la energía interna.
Q= 156 [J]
La temperatura final resulta ser 528[°C] porque la temperatura inicial aumenta el triple:
176(3)= 528[°C]