Download “spontaneous” process: (happens without input of work) Life

361 Lec 6
Thur, 8sep16
“spontaneous” process: (happens without input
of work)
Life: enzymes coupling spontaneous reactions with
non-spontaneous (like a heavy weight lifting a lighter
one) through a pulley system
enzymes are the “pulleys” of biochemistry.”
(~ 75,000 enzymes busily doing their thing in the human
body)
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361 Lec 6
Thur, 8sep16
pext = 1 atm
pext = 1 atm
p = 1 atm
IRREVERSIBLE
p = 3 atm
p = changes 3 atm to 1 atm
IRREVERSIBLE
but pext = 1 atm and constant
REVERSIBLE
pext = p during entire
expansion
by removing grains of
sand from piston.
Adding one grain of
sand reverses
direction
pext = 3 atm
p = 3 atm
pext = 1 atm
p = 1 atm
REVERSIBLE
pext = p always
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What are signs of: ∆T,
Adiabatic (q=0) :
Isothermal :
-
q, w,
0
0 +
∆U?
- - 0
∆T,
q, w, ∆U?
+ 0 + +
0 - + 0
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361 Lec 6 Thur, 4sep14
How does the energy of an ideal gas change
with volume at constant temperature?
U is independent of V!
U =U(T)
For ideal gas isothermal process ∆U =
0
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Reversible Isothermal PV work (ideal gas)
pext = p = nRT/V
(balanced forces)
An integral is just a
Sum V2
V2
V2 nRT
w = − ∫ pext dV = − ∫ pdV = − ∫
dV
V1
V1
V1
V
w = −nRT
∫
V2
V1
V2
dV
= −nRT ln
V1
V
Sum of fractional changes
= natural logarithm
ONLY if p = pext = and ideal gas (but is common case)
What are ∆U and q? ∆U =0 (isothermal), therefore q = -w
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Equation of State
example, ideal gas:
pV = nRT
n = mols, p = pressure, V = volume, T = absolute temp.
R = “gas constant” =
8.3145
J
mol-1 K-1
= 0.083145 L bar mol-1 K-1
=0.0821
L atm mol-1 K-1
nRT
= V(n, T, P)
P
nRT
P=
= P(n, T, V)
V
PV
n=
= n(T, P, V)
RT
PV
T=
= T(n, P, V)
nR
Note that V =
or
or
or
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van der Waals Equation of state for real gases
2
n
( p + 2 a )(V - b) = nRT
V
b is the effective volume of the molecule
n/V is mol L-1 i.e., the molar concentration
a is the effective equilibrium constant for association of two
molecules of gas
What does this equation become at low concentration?
p V = nRT
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Some useful state functions are DEFINED from other state
functions. We are immediately concerned with
Enthalpy ≡ H = U + pV (by definition)
Why invent yet another state function?
Because ∆U = q + w. q is easy to measure, but most chemical
reactions have a small work term -pext∆V (due to expansion or
contraction against the atmosphere) I call this “obligatory work”.
We want to tabulate state functions, NOT q and w.
Under the very common conditions: p=pext = constant and
only the obligatory pV work on the atmosphere, the above
definition yields:
∆H = q ( nice!)
Beware, if the reaction is harnessed so that non-obligatory pV work (i.e.,
useful work) is done, then ∆H = q + wuseful
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