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 D T = 0.
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Isothermal processes (∆T = 0)
Isothermal gas expansion and work(w)
Reversible and irreversible processes
q = -w (isothermal process)
Consider a mass (M) connected to a ideal gas
confined by a piston. The piston is submerged
in a constant T bath, so DT = 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 area
g = gravitational
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 = -PextDV = -P1/4 (4V1 - V1) = -3/4 P1V1
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 = -PextDV = -P1/2 (2V1 - V1) = -1/2 P1V1
In Step 2 (M1/2 to M1/4 ):
Pext = P1/4, Vfinal = 4V1
w2 = -PextDV =- P1/4 (4V1 - 2V1) = -1/2 P1V1
wtotal = w1 + w2 = -P1V1/2 - P1V1/2 = -P1V1
Note: wtotal,2 step > wtotal,1 step
More 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.
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:
Graphically, see how much more work is done in the infinite-step expansion (red area)
If we perform the integration from
V1 to V2:
Now 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 = -PextDV = -P1/2 (2V1 - 4V1)
In second step:
w2 = -PextDV = -P1 (V1 - 2V1)
wtotal = w1 + w2 = 2P1V1 (see table 10.3)
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 (DT = 0)
then, DE = 0
but, q = -w ≠ 0
Let’s consider the four- step cycle illustrated:
1: Isothermal expansion
2: Const V cooling
3: Isothermal compression
4: Const V heating
Step 1: Isothermal Expansion
at T = Thigh from V1 to V2
DT = 0, so DE = 0 and q = -w
Do this expansion reversibly,
Step 2: Const V cooling to
T = Tlow.
DV = 0, therefore, w = 0
q2 = DE = nCvDT
Step 3: Isothermal compression at T = Tlow from V2 to V1.
Since DT = 0, then DE = 0
and q = -w
Do compression reversibly, then
Step 4: Const-V heating to
T = Thigh.
DV = 0, so, w = 0, and
q4 = DE = nCvDT
= nCv(Thigh-Tlow) = -q2
The thermodynamic definition of entropy
DT = 0
DV = 0
DP = 0
Example: What is DS for the heating of a mole of a monatomic
gas @constant volume from 298 K to 350 K?
Connecting with Lecture 6
• From this lecture:
• Exactly the same as derived in the previous lecture!