give Pressure-temperature graph and volume-temperature graph for isobaric process..!!
Answers
Answer:
From the equation state for an ideal gas PV=nRT,
We have V=(
P
0
nR
)T or VαT
At V=V
0
, we get T
1
=
nR
P
0
V
0
And at V=2V
0
, we get
T
2
=
nR
2P
0
V
0
For the graph of P versus T the variation is a straight line normal to the pressure axis, the temperature varies from T
1
to T
2
as shown in the figure.
For the graph of V versus T, the equation V=(
P
0
nR
)T or V=kT shows that the volume varies directly with the temperature (Charlie's law). So, the graph is a straight line inclined to the (V-T) axes, and passes through the origin (when produced) as shown in figure.
solution
Explanation:
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Recall from the previous section…
∆U = Q + W
Q > 0 system absorbs heat from the environment
Q < 0 system releases heat to the environment
W > 0 work done on the system by the environment
W < 0 work done by the system on the environment
A system can be described by three thermodynamic variables — pressure, volume, and temperature. Well, maybe it's only two variables. With everything tied together by the ideal gas law, one variable can always be described as dependent on the other two.
⎧
⎪
⎪
⎨
⎪
⎪
⎩ P = nRT
V
PV = nRT ⇒ V = nRT
P
T = PV
nR
Temperature is the slave of pressure and volume on a pressure-volume graph (PV graph).
Function of State
∆U =
3
2
nR∆T
Function of Path: Work
W = ∫ F · ds = ∫ P dV
W = − area on PV graph
Function of Path: Heat
Q = ∆U + W = nc∆T
cP = specific heat at constant pressure
cV = specific heat at constant volume
