Download Example 15-4 Two Thermodynamic Processes: Isochoric and Isobaric

Example 15-4 Two Thermodynamic Processes: Isochoric and Isobaric
Set Up
We’ll use the ideal gas law to determine the
final volume of the gas. For the isochoric
(constant-volume) process A S B, the gas does
no work, so the quantity of heat Q that enters
the gas is equal to the internal energy change
U of the gas. (The gas is cooled in this process, so Q 6 0 and U 6 0.) We’ll determine
U from the temperature change. For the isobaric (constant-pressure) process B S C, the
work W that the gas does is given by Equation
15-4; this is positive because the gas expands.
The internal energy change U is again given
by the temperature change. We’ll then determine Q for this process from the first law of
thermodynamics.
(a) Use the ideal gas law to find the final
volume Vf.
A
3.03 × 105
The light blue curve is an isotherm.
1.01 × 105
B
1.20 × 10
C
–3
Vf = ?
Volume (m3)
Process B → C isobaric
(constant pressure)
Figure 15-9 ​Isochoric cooling and isobaric expansion In
this two-step process for an ideal gas, how much work is done,
how much heat flows into the gas, and by how much does the
internal energy change?
Ideal diatomic gas:
5
U = nRT
2
Ideal gas law:
pV = nRT
Process A
(14-7)
system =
gas
(15-9)
C
The gas pushes upward
and does work on the
piston: W > 0.
(15-4)
Isochoric (constant-volume)
process:
Heat flows out of
the gas: Q < 0.
Process B
Work done in an isobaric
(constant-pressure) process:
W = p(Vf 2 Vi)
B
The volume of the gas
is constant: W = 0.
First law of thermodynamics:
(15-2)
U = Q - W
U = Q
Solve
Process A → B isochoric
(constant volume)
Pressure (Pa)
The pV diagram in Figure 15-9 shows two thermodynamic
processes that occur in an ideal diatomic gas, for which the
­internal energy is U = 15>22nRT. The gas is in a cylinder
with a moveable piston (see Figure 15-5) and is initially in the
state labeled A on the diagram, at pressure pi = 3.03 * 105 Pa
and volume Vi = 1.20 * 1023 m3. The piston is first locked in
place so that the volume of the gas cannot change, and the
cylinder is cooled so that the temperature and pressure of the
gas both decrease. When the gas is in state B, the pressure of
the gas is pf = 1.01 * 105 Pa = 1.00 atm, the same as the air
pressure outside the cylinder. The piston is then unlocked so
it is free to move, and the cylinder is slowly heated so that
the gas expands at constant pressure. The temperature of the
gas increases until in the final state (C in Figure 15-9), when
the gas is at the same temperature as in the initial state A.
(a) What is the final volume of the gas? (b) Find the values
of W, U, and Q for the isochoric process A S B. (c) Find
the values of W, U, and Q for the isobaric process B S C.
(d) Find the values of W, U, and Q for the net process
A S B S C.
Heat flows into
the gas to make
it expand: Q > 0.
system =
gas
The initial pressure and volume of the gas are
pi = 3.03 * 105 Pa
Vi = 1.20 * 1023 m3
From the ideal gas law, Equation 14-7,
piVi = nRTi
where Ti is the initial temperature of the gas in state A.
The final pressure of the gas is
pf = 1.01 * 105 Pa
The ideal gas law tells us that
pfVf = nRTf
where Tf is the final temperature of the gas in state C. The initial and
final temperatures are the same (Ti = Tf) and so
piVi = pfVf
Vf =
13.03 * 105 Pa2 11.20 * 10-3 m3 2
piVi
=
= 3.60 * 10-3 m3
pf
1.01 * 105 Pa
The final pressure is one-third of the initial pressure, and the final
volume is three times the initial volume.
(b) For the isochoric process A S B the work
done is WA S B = 0. Find the change in internal
energy U and quantity of heat Q for this
process.
The change in internal energy U is the difference between the value of
U in state B and the value of U in state A:
5
5
UA S B = UB - UA = nRTB - nRTi
2
2
We don’t know the number of moles of gas n, nor do we know the
temperatures Ti (in the initial state A, for which the pressure is pi and
the volume is Vi) and TB (in state B, for which the pressure is pf and
the volume is Vi). But from the ideal gas law pV = nRT, so
piVi = nRTi and pfVi = nRTB
So the change in internal energy is
5
5
5
UA S B = pfVi - piVi = 1pf - pi 2Vi
2
2
2
5
5
= 11.01 * 10 Pa - 3.03 * 105 Pa2 11.20 * 10-3 m3 2
2
= -606 J
(Remember that 1 Pa = 1 N>m2 and 1 N # m = 1 J.) From
Equation 15-9 for an isochoric process,
QA S B = UA S B = 2606 J
Heat flows out of the gas and the internal energy of the gas
decreases.
(c) Find W, U, and Q for the isobaric process
B S C.
The work done in an isobaric process is given by Equation 15-4. The
constant pressure is pf = 1.01 * 105 Pa, and the volume increases
from Vi = 1.20 * 1023 m3 to Vf = 3.60 * 1023 m3:
WB S C = pf (Vf 2 Vi)
= (1.01 * 105 Pa) (3.60 * 1023 m3 2 1.20 * 1023 m3)
= +242 J
The gas does positive work as it expands. As in part (b), we calculate
the internal energy change with the aid of the ideal gas law:
5
5
nRTf - nRTB
2
2
5
5
5
= pfVf - pfVi = pf 1Vf - Vi 2
2
2
2
UB S C = UC - UB =
This is just 5>2 times the above expression for WB S C, so
5
5
WB S C = 1 +242 J2 = +606 J
2
2
From the first law of thermodynamics, the heat that flows into the gas
in this process is
UB S C =
QB S C = UB S C + WB S C = (+606 J) + (+242 J)
= +848 J
(d) Find the values of W, U, and Q for the
combined process A S B S C.
The total amount of work done by the gas is
WA S B + WB S C = 0 + (+242 J) = +242 J
The total change in internal energy of the gas is
UA S B + UB S C = (2606 J) + (+606 J) = 0
The total quantity of heat that flows into the gas is
QA S B + QB S C = 2606 J + (+848 J) = +242 J
Reflect
For the combined process, there is zero net change in the internal energy U of the ideal gas. That’s because U depends
only on the number of moles of gas and the temperature, which have the same values in the final state C as in the initial
state A. In the combined process 242 J of heat flows into the gas, and the gas uses this energy to do 242 J of work.