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Topic: Physical chem - enthalpy and internal energy  (Read 7760 times)

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Offline BeepoGirl

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Physical chem - enthalpy and internal energy
« on: April 04, 2009, 08:53:20 AM »
According to my lecture notes, the magnitude of change in enthalpy at constant pressure is always more than the magnitude of change in internal energy at constant volume, with the same amount of heat supplied.

To my mind that's logically the wrong way round, is this true and is it only for a certain set of conditions?

Thanks

Offline Loyal

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Re: Physical chem - enthalpy and internal energy
« Reply #1 on: April 05, 2009, 02:38:17 AM »
Your notes are correct.  Cp is larger than Cv.  What was your thought process when you were arriving at your answer?  Maybe if I understand where you are coming from I might be able to remove some of the confusion.
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Offline BeepoGirl

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Re: Physical chem - enthalpy and internal energy
« Reply #2 on: April 05, 2009, 10:14:10 AM »
As internal energy at constant volume is equal to heat supplied, but internal energy at constant pressure also involves doing expansion work so it is an underestimate of the heat supplied.

Offline Loyal

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Re: Physical chem - enthalpy and internal energy
« Reply #3 on: April 05, 2009, 04:31:36 PM »
As internal energy at constant volume is equal to heat supplied, but internal energy at constant pressure also involves doing expansion work so it is an underestimate of the heat supplied.

If we are talking enthalpy at constant P and internal energy at constant V then it is easily proved by the relationship between their heat capacities.

Cp = Cv + (TVB2)/k

All of those numbers are positive numbers so it is impossible for Cp < Cv

However if I am reading you correctly we have internal energy at constant V and then at constant P?  I would still think it would be larger given the relationship between the two. 

This can be proven mathematically.

If that is the case then we have the value of dU = (dU/dT)v dT and  dU = (dU/dT)p dT at constant V and P respectively. (Those are suppose to be partials, but I can't find the symbol for it)

(dU/dT)V = Cv
(dU/dT)p = Cp - P(dV/dT)p = Cp - PVB

For (dU/dT)p to be less than (dU/dT)v the difference (dU/dT)p - (dU/dT)v must be negative. So we have the equation.

Cp - PVB - Cv < 0

Using the equation
Cp = Cv + (TVB2)/k

- PVB + (TVB2)/k < 0

or

PVB > (TVB2)/k

after some refinement we get
P > T(B/k)

or

P > T(dP/dT)V

This is the condition that must be met for (dU/dT)v to be greater.  However by simple evaluation we can see that this isn't true for many cases.

For an ideal gas this can never be true since
T(dP/dT)V = nRT/V = P.   
For a Van der Waals gas this again can not be true since T(dP/dT)V = nRT/(V-nb) and this would require -(an2)/V2 to be greater than zero which is impossible since all of those numbers are possitive.

Yet an interesting case occurs for a solid.  Say for instance Al, under the assumption that B and k remain constant you get

P > T(B/k)
P > T(69.3*10-6/1.33*10-6)
P > T*52.1 (bar/K)

However this would mean it would have to be under extreme conditions and under those conditions B and k would no longer be constant or those values.  So the assumption does not hold.

So from this proof, I would say it is safe to assume that (dU/dT)P is always greater than (dU/dT)V for nearly all reasonable conditions.  Thus
for the same temperature range  :delta: U at constant P is greater than  :delta: U at constant V.

Chemistry Student(Senior) at WSU

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