Phys 2101
Gabriela González
pV=nRT
If volume is constant (“isochoric” process), W=0, and Tp-1 is constant.
If temperature is constant (“isothermal” process), ΔEint=0, and pV is constant.
If pressure is constant (“isobaric” process), W, Q, and ΔEin are not zero, but TV-1 is
constant.
Adiabatic processes happen when Q = 0: the system is thermally insulated, or
because the process happens very quickly.
Temperature, pressure and volume all change, but the quantities
pVγ and T Vγ-1
are constant,
where γ = CP/CV (=4/3=1.67 for monoatomic gases)
The curve in a pV diagram is
called an “adiabat”
2!
One liter of gas with γ = 1.3 is at 292 K and 1.4 atm pressure. It is
suddenly compressed (adiabatically) to half its original volume.
Find its final pressure and temperature.
Adiabatic process: pVγ and T Vγ-1 are constant:
p2 = p1 (V1/V2)γ = 1.4 atm 21.3 = 3.4 atm
T2 = T1 (V1/V2)γ-1 = 292K 20.3 = 359 K
The gas is now cooled back to 292 K at constant pressure.
What is its final volume?
Constant pressure: p= nRT/V , or T/V, is constant
V3=V2 (T3/T2) = 0.5 l (292 K/ 359 K) = 0.41 l
3!
One mole of an ideal diatomic gas undergoes a transition from
a to c along the diagonal path in the figure. The
temperature of the gas at point a is 1200 K.
(a)  What is the change in internal energy? [-5000 J]
(b)  How much heat is added to the gas? [2000J]
(c)  How much heat is added to the gas
if it goes back to the original state a
through the point b? [-5000J]
4!
pV=nRT
5!
A free expansion process happened when Q = W = 0, for an
isolated system. Since internal energy is constant, temperature
is constant. Pressure and volume change, but the quantity
pV = nRT is constant
The curve in a pV diagram is an isotherm (but no work is done
or heat is exchanged in the process!).
This is a non-reversible process!!
6!
We can easily tell the direction of
the “arrow of time” in irreversible
processes which, if spontaneous,
only happen one way:
•  spilling fluids
•  gas expansion
•  breaking solids
Conservation of energy doesn’t
forbid these processes; there is
another quantity, “entropy”, which
measures the degree of “disorder”:
S = k log W
7!
If an irreversible process occurs
in a closed system, the entropy S
of the system always increases; it
never decreases.
If any process occurs in a closed
system, the entropy of the system
increases for irreversible
processes and remains constant
for reversible processes. It never
decreases!
ΔS ≥ 0
8!
f
dQ
ΔS = S f − Si = ∫
T
i
9!
An 8.0g ice cube at -10oC is put in a Thermos flask containing 100 cm3 of
water at 20oC. The specific heat of ice is 2220 J/kgK, the specific heat of
water is 4190 J/kgK, and the heat of fusion of ice-water is 330 kJ/kg.
What’s the final equilibrium temperature?
Use
ΔQ = 0 = cWmW(Tf-20oC) + cice mice (0oC-(-10oC)) + mice Lf + cW mice (Tf-0oC)
Answer: 12.24oC= 285.39 K
What is the change in entropy of the ice-water system?
Use ΔS= ∫dQ/T = ∫mcdT/T = mc ln(Tf/Ti) for changes in temperature, and
ΔS= ∫dQ/T =Q/T = mLf/T for changes of phase
Answer:
ΔSw= -11.24 J/K
ΔSice = +11.88 J/K
ΔS = 0.64 J/K
10!
Any isothermal process:
 ΔS= ∫dQ/T= (1/T) ∫dQ = ΔQ/T
  ny reversible process:
A
  dQ = dEint
+ dW
 
= nCV dT + p dV
 dQ/T= nCV dT/T + pdV/T
 
= nCV dT/T + pdV/(pV/nR)
 
= nCV dT/T + nRdV/V
Eint = (f/2)nRT = n CV T
W = ∫ p dV
pV = nRT
  ΔS = ∫dQ/T
 
= nCV ln (Tf/Ti) + nR ln (Vf/Vi)
11!
Ideal heat engines use a cycle of reversible thermodynamic processes. A heat
engine transforms energy extracted as heat from thermal reservoirs, into
mechanical work.
Consider a Carnot engine: a cycle with two isothermal processes at a high
temperature TH (a→b) and and a low temperature TL (c →d), and two adiabatic
processes (b→c, d→a).
 
ΔEint = 0 = Q – W → W = Q = |QH|-|QL| > 0
 
ΔS = 0 = |QH|/TH - |QL|/TL
http://galileoandeinstein.physics.virginia.edu/more_stuff/
flashlets/carnot.htm
12!
We use the heat QH to get work W done, so efficiency is
defined as
 
ε = |W|/|QH|
 For a Carnot engine,
 
W = |QH|-|QL|
 
|QH|/TH = |QL|/TL
 so
εC = (|QH|-|QL|)/ |QH| = 1- |QL|/|QH| = 1- TL/TH < 1
Carnot’s theorem: The most efficient cycle with maximum temperature TH
and minimum temperature TL, is the Carnot cycle.
13!
14!
Three ideal Carnot engines operate between (a) 400K and
500K, (b) 500K and 600K, and (c) 400K and 600K.
1. Rank them according to their efficiencies, greatest first.
2.  If they all extract the same amount of energy per cycle from
the high temperature reservoir, rank them according to the
work per cycle done by the engines, greatest first.
15!
Three ideal Carnot engines operate between (a) 400K and
500K, (b) 500K and 600K, and (c) 400K and 600K.
1. Rank them according to their efficiencies, greatest first.
ANS: ε=1-TL/TH : εc > εa > εb
a)  20% (1-4/5=0.2)
b)  16% (1-5/6=0.16)
c)  33% (1-4/6=0.33)
2.  If they all extract the same amount of energy per cycle from
the high temperature reservoir, rank them according to the
work per cycle done by the engines, greatest first.
ANS: ε = |W|/|QH| , so ranking according to W is the same as
according to efficiency (above).
16!