Derivation of Clausius Clapyron Equation
David Farris
the Clausius-Clapyron Equation (log (P2/P1)=(heat of
vaporization)*((T2-T1)/T2T1)/R) for liquids, specifically how one gets
from an exponential curve for a graph of temperature v. pressure to get
a linear curve with negative slope in a graph of the reciprocal of
temperature against the log of pressure. I haven't found the derivation in
any books that I found or that she reccomended. COuld someone shed some
light on this? Thanks.
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David Farris | Ash nazg durbatuluk, ash nazg gimbatul,
dfa...@netcom.com | ash nazg thrakatuluk agh burzum-ishi krimpatul
Michael Atwell
The Gibbs-Helmholtz equation is
(d(delta G/T)/dt)=-(delta H/T^2)
The equilibrium constant can be written as:
ln Kp = - delta G/RT
Differentiate this equation, with respect to t,
d ln Kp/dt = - (1/R)*(d (delta G/t)/dt)
Substitute the Gibbs-Helmholtz equation into the equation,
d ln Kp/dt = delta H/RT^2
which is,
d ln p/dt = delta H of vaporization/RT^2
Integrate with respect to T, from T1 to T2,
ln (p2/p1) = -(delta H of vap/R) * (1/T2 - 1/T1)
Algebraic rearrangement,
ln (p2/p1) = -(delta H of vap/R) * (T1/T2T1 - T2/T2T1)
ln (p2/p1) = (delta H of vap/R) * ((T2 - T1)/(T2T1))
Which is the Clausius-Clayperon equation.
I realize this is hard to read, and may not help. If it doesn't help,
any Physical Chemistry text worth its salt should have this derivation.
Mike
John Milligan
dfa...@netcom.com (David Farris) wrote:
>My high school chemistry teacher was unable to explain the derivation
of
>the Clausius-Clapyron Equation (log (P2/P1)=(heat of
>vaporization)*((T2-T1)/T2T1)/R) for liquids, specifically how one gets
>from an exponential curve for a graph of temperature v. pressure to get
>a linear curve with negative slope in a graph of the reciprocal of
>temperature against the log of pressure. I haven't found the
derivation in
>any books that I found or that she reccomended. COuld someone shed
some
>light on this? Thanks.
Assuming that you started with
P = AExp[-dH/RT]
taking log (that is the natural log) of both sides gives
Log[P] = Log[A]-dH/RT
If we have two sets of data P1, T1 and P2, T2 we have two equations for
the same substance
Log[P1] = Log[A]-dH/R(T1)
and
Log[P2] = Log[A]-dH/R(T2)
Subtracting one from the other we have
Log[P1]-Log[P2] = Log[A]-dH/R(T1) - (Log[A]-dH/R(T2))
the left hand side gives
Log[P1/P2]
due to the properties of logarithms. The right hand side gives
dH/R(T2)-dH/R(T1)
since A is the same for the same substance, Log[A] drops out when
we subtract the two equations. This expression can be rearranged to
(dH/R)*((1/T2)-(1/T1))
since dH and R are the same for the same substance. This gives the
familiar Clausius-Clapeyron equation.
Log[P1/P2] = (dH/R)*((1/T2)-(1/T1)).
I am suprised that your teacher did not know this. This derivation is
in most college level general chemistry books that I have seen.
________________________________________________
Dr. John Milligan "Spoon !!!!!!"
LAVC -The Tick
CSULA
abr...@urvax.urich.edu
dfa...@netcom.com (David Farris) wrote:
>My high school chemistry teacher was unable to explain the derivation
of
>the Clausius-Clapyron Equation (log (P2/P1)=(heat of
>vaporization)*((T2-T1)/T2T1)/R) for liquids, specifically how one gets
>from an exponential curve for a graph of temperature v. pressure to
get
>a linear curve with negative slope in a graph of the reciprocal of
>temperature against the log of pressure. I haven't found the
derivation in
>any books that I found or that she reccomended. COuld someone shed
some
>light on this? Thanks.
the Clapeyron equation, dp/dT = (Delta S)/(Delta V), where delta S is
the difference in molar entropies between the gas phase and liquid
phase, and delta V is the difference in molar volumes. The next step
is to note that since the process of vaporization is isothermal, that
we can replace Delta S by Delta H(vap) over T. This yields dp/dt =
(Delta H)/(T delta V). The next step is an approximation. Note that
the molar volume of a gas under conditions far from the critical point
will be far larger than the molar volume of the corresponding liquid.
A good example is water at STP. Treated as an ideal gas (a fair
approximation) the vapor has a molar volume of 22.14L while the liquid
has a molar volume of .018 L. This large difference means that delta V
is approximately equal to V(gas). We approximate the vapor as an ideal
gas, and replace V molar by RT/p. Collecting variables yields dp/p =
(delta H)/ (R T squared) dT. Integrating both sides yields the
Clausius Clapeyron equation.
John Milligan
mat...@emerald.tufts.edu (Michael Atwell) wrote:
>d ln Kp/dt = - (1/R)*(d (delta G/t)/dt)
>
>Substitute the Gibbs-Helmholtz equation into the equation,
>
>d ln Kp/dt = delta H/RT^2
What happened to the delta S that should be here upon substitution???
since delta G = delta H - T delta S when you do the above step you
should have a delta S in your equation.
Or have you skipped a step or two (or more) and I am just missing it??
Michael Atwell
: In article <4au14g$6...@d2.tufts.edu>,
: mat...@emerald.tufts.edu (Michael Atwell) wrote:
: >d ln Kp/dt = - (1/R)*(d (delta G/t)/dt)
: >
: >Substitute the Gibbs-Helmholtz equation into the equation,
: >
: >d ln Kp/dt = delta H/RT^2
: What happened to the delta S that should be here upon substitution???
: since delta G = delta H - T delta S when you do the above step you
: should have a delta S in your equation.
: Or have you skipped a step or two (or more) and I am just missing it??
I was merely substituing the Gibbs-Hemholtz equation that I referred to
at the beginning of my entry. When deriving the Gibbs-Hemholtz equation,
S cancels. It is derived this way:
(dG/dT) = -S (at constant p)
Substitute for S,
(dG/dT) = (G-H)/T
By rule of differentiation,
(d(G/T)/dT) = (1/T)*(dG/dT) - (1/T^2)*G
Substitute in the original expression, (dG/dT) = -S,
(d(G/T)/dT) = -(TS - G)/T^2
Which is,
(d(G/T)/dT) = - H/T^2
which is the Gibbs-Hemholtz eq., which is what I used in substition.
That is why no S is present after substitution.
All d's are partial derivatives, with p held constant.
Mike
Ernst U. Wallenborn
> In article <4au14g$6...@d2.tufts.edu>,
> mat...@emerald.tufts.edu (Michael Atwell) wrote:
> >d ln Kp/dt = - (1/R)*(d (delta G/t)/dt)
> >
> >Substitute the Gibbs-Helmholtz equation into the equation,
> >
> >d ln Kp/dt = delta H/RT^2
>
> What happened to the delta S that should be here upon substitution???
> since delta G = delta H - T delta S when you do the above step you
> should have a delta S in your equation.
>
> Or have you skipped a step or two (or more) and I am just missing it??
>
> Dr. John Milligan "Spoon !!!!!!"
> LAVC -The Tick
> CSULA
>
The idea is (D denoting Delta):
K = exp{-DG/RT}
ln K = -DG/RT = -DH/RT + TDS/RT = -DH/RT + DS/R
DS/R is independent of T, hence:
d(ln K)/dT = d/dT (-DH/RT + DS/R) = DH/RT^2
Hope that makes it clear.
---
-ernst wallenborn.
i'm not a bug.
i'm an undocumented feature.
John Milligan
wa...@phys.chem.ethz.ch (Ernst U. Wallenborn) wrote:
>
>The idea is (D denoting Delta):
>
>K = exp{-DG/RT}
>
>ln K = -DG/RT = -DH/RT + TDS/RT = -DH/RT + DS/R
>
>DS/R is independent of T, hence:
>
>d(ln K)/dT = d/dT (-DH/RT + DS/R) = DH/RT^2
That makes it more clear. It should have been explained in the original
posting. Thanks.