lect# 7: thermodynamics and entropy reading: zumdahl 10.2, 10.3 outline: isothermal processes (∆t...
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Lect# 7: Thermodynamics and Entropy
• Reading: Zumdahl 10.2, 10.3
• Outline:Isothermal processes (∆T = 0)Isothermal gas expansion and work(w)
Reversible and irreversible processes
Isothermal Processes
• Recall: Isothermal means T = 0.
• Since E = nCvT, then E = 0 for an isothermal process.
• Since E = q + w, then q = -w (isothermal process)
Example: Isothermal Expansion
Consider a mass (M) connected to a ideal gas
confined by a piston. The piston is submerged
in a constant T bath, so T = 0.
Initially, V = V1
P = P1
Pressure of gas is equal to that created by the pull of the hanging mass:
P1 = force/area = M1g/A
A = piston areag = gravitational
acceleration
(9.8 m/s2)
Note: kg m-1 s-2 = 1 Pa
One-Step Expansion: If we change the weight to M1/4, then the pressure becomes
Pext = (M1/4)g/A = P1/4
The mass will be lifted until the internal pressure
equals the external pressure, at which point
Vfinal = 4V1
What work is done in this expansion?
w = -PextV = -P1/4 (4V1 - V1) = -3/4 P1V1
What if we expand in two steps?
In this expansion we go in two steps:
Step 1: M1 to M1/2
Step 2: M1/2 to M1/4
In first step:Pext = P1/2, Vfinal = 2V1
w1 = -PextV = -P1/2 (2V1 - V1) = -1/2 P1V1
In Step 2 (M1/2 to M1/4 ):
Pext = P1/4, Vfinal = 4V1
w2 = -PextV =- P1/4 (4V1 - 2V1) = -1/2 P1V1
wtotal = w1 + w2 = -P1V1/2 - P1V1/2 = -P1V1
Note: wtotal,2 step > wtotal,1 stepMore work was done in the two-step
expansion
Graphically, we can envision this two-step process on a PV diagram:
• Work is given by the area under the “PV” curve.
Infinite Step Expansion
Imagine that we perform a process in which we decrease the weight very slightly (∆M) between an infinite number of (reversible) expansions.
Instead of determining the sum of work performed at each step to get wtotal, we can integrate:
€
w = − PexdVVinitial
V final
∫ = − PdVVinitial
V final
∫ = −nRT
VdV
Vinitial
V final
∫
Graphically, see how much more work is done in the infinite-step expansion (red area)
Two Step
Reversible
If we perform the integration from
V1 to V2:
€
wtotal = − PdVV1
V2
∫ = −nRT
VdV
V1
V2
∫ = −nRT(ln(V )) |V1
V2 = −nRT lnV2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟
€
wrev = −nRT lnV2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟= −qrev
Two Step CompressionNow we will do the opposite, compress
the fully expanded gas:Vinit = 4V1
Pinit = P1/4
Compress in two steps: first put on mass = M1/2, Then, in step two, replace mass M1/2 with a bigger mass M1
In first step:w1 = -PextV = -P1/2 (2V1
- 4V1)
= P1V1
wtotal = w1 + w2 = 2P1V1 (see table 10.3)
In second step:w2 = -PextV = -P1 (V1 -
2V1)
= P1V1
Compression/Expansion
In two-step example:wexpan. = -P1V1
wcomp. = 2P1V1
wtotal = P1V1
qtotal = -P1V1
We have undergone a “cycle” where the system returns to the starting state.
Since isothermal (T = 0) then, E = 0
but, q = -w ≠ 0
Entropy: A Thermodynamic Definition
Let’s consider the four- step cycle illustrated:1: Isothermal expansion
2: Const V cooling3: Isothermal compression
4: Const V heatingVolumeV1 V2
1
2
3
4
Step 1: Isothermal Expansion
at T = Thigh from V1 to V2
T = 0, so E = 0 and q = -w
Do this expansion reversibly,
so that
VolumeV1 V2
1
2
3
4
€
q1 = −w1 = nRThigh lnV2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟
Step 2: Const V cooling to T = Tlow.
V = 0, therefore, w = 0
q2 = E = nCvT = nCv(Tlow-Thigh)
VolumeV1 V2
1
2
3
4
Step 3: Isothermal compression at T = Tlow from V2 to V1.
Since T = 0, then E = 0 and q = -w
Do compression reversibly, then
VolumeV1 V2
1
2
3
4
€
q3 = −w3 = nRTlow lnV1
V2
⎛
⎝ ⎜
⎞
⎠ ⎟
Step 4: Const-V heating to T = Thigh.
V = 0, so, w = 0, and
q4 = E = nCvT = nCv(Thigh-Tlow) = -q2
VolumeV1 V2
1
2
3
4
VolumeV1 V2
1
2
3
4
€
qtotal = q1 + q2 + q3 + q4
€
qtotal = q1 + q3
€
qtotal = nRThigh lnV2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟+ nRTlow ln
V1
V2
⎛
⎝ ⎜
⎞
⎠ ⎟
€
qtotal = nRThigh lnV2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟− nRTlow ln
V2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟
VolumeV1 V2
1
2
3
4
€
qtotal = nRThigh lnV2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟− nRTlow ln
V2
V1
⎛
⎝ ⎜
⎞
⎠ ⎟
€
0 =q1Thigh
+q3
Tlow
The thermodynamic definition of entropy
(finally!)€
S =dqrevTinitial
final
∫ =qrevT
Calculating Entropy: summary
€
S =dqrevTinitial
final
∫
T = 0
€
S =dqrevTinitial
final
∫ = nR lnV finalVinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
V = 0
€
S =dqrevTinitial
final
∫ = nCv lnTfinalTinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
P = 0
€
S =dqrevTinitial
final
∫ = nCp lnTfinalTinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
€
dqrev = nRTdV
V
€
dqrev = nCvdT
€
dqrev = nCPdT
Calculating Entropy: simple example
Example: What is S for the heating of a mole of a monatomic
gas @constant volume from 298 K to 350 K?
€
S =dqrevTinitial
final
∫ = nCv lnTfinalTinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
3/2R
€
S = (1mol) 32R( )ln
350K
298K
⎛
⎝ ⎜
⎞
⎠ ⎟
€
S = 2J K
Connecting with Lecture 6
€
S =dqrevTinitial
final
∫ =qrevT
=nRT
TlnV finalVinitial
⎛
⎝ ⎜
⎞
⎠ ⎟= nR ln
V finalVinitial
⎛
⎝ ⎜
⎞
⎠ ⎟
• From this lecture:
• Exactly the same as derived in the previous lecture!
€
S = Nk lnV finalVinitial
⎛
⎝ ⎜
⎞
⎠ ⎟= nR ln
V finalVinitial
⎛
⎝ ⎜
⎞
⎠ ⎟