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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: BeepoGirl on April 04, 2009, 08:53:20 AM

Title: Physical chem - enthalpy and internal energy
Post by: BeepoGirl 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

Title: Re: Physical chem - enthalpy and internal energy
Post by: Loyal on April 05, 2009, 02:38:17 AM

Your notes are correct.  C_p is larger than C_v.  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.

Title: Re: Physical chem - enthalpy and internal energy
Post by: BeepoGirl 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.

Title: Re: Physical chem - enthalpy and internal energy
Post by: Loyal on April 05, 2009, 04:31:36 PM

Quote from: BeepoGirl 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.

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.

C_p = C_v + (TVB²)/k

All of those numbers are positive numbers so it is impossible for C_p < C_v

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 = C_v
(dU/dT)_p = C_p - P(dV/dT)_p = C_p - 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.

C_p - PVB - C_v < 0

Using the equation
C_p = C_v + (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.