Why Is Dry Ice's Energy Change Considered Internal Rather Than Enthalpy?

[metallica007]

Hi everyone
I need your help in the following problem:
Some dry ice (solid carbon dioxide) at -78oC is placed in a steel jar, which is subsequently sealed. The jar is immersed in a large beaker of water at room temperature until the dry ice has completely vaporized.
If the measured energy change was 580 J, was this an enthalpy change or an internal energy change?
Answer: internal energy change
My question is, why the answer is internal energy instead of enthalpy change? could be you please explain it to me

 

[Point Conception]

delta H = delta E + delta (PV)
In this case with no change in volume with the sealed jar and therefor no work done on the
surroundings. Then the change in enthalpy = the change in internal energy

 

[DrDu]

delta H = delta E + delta (PV)
In this case with no change in volume with the sealed jar and therefor no work done on the
surroundings. Then the change in enthalpy = the change in internal energy

The pressure increases in that process while V remains constant. Hence delta PV=V delta P and this is different from 0. Hence delta E is not equal to delta H.

 

[Point Conception]

The pressure increases in that process while V remains constant. Hence delta PV=V delta P and this is different from 0. Hence delta E is not equal to delta H.

I can see that V delta P is not 0 above.
But with: delta U = delta H + delta work
And with: work = -p delta V
Therefore it still looks like delta U = delta H
in the context of lines 2 & 3

From this text: Physical Chemistry, GI Brown
change in energy = heat absorbed or evolved (q) - work done in expansion or contraction
(p delta V)
Internal energy change, delta U For a reaction carried out at a constant volume ( in a sealed container ) and if no other work is done , the change in energy called internal energy, delta U is equal to q, ( at constant volume )

 

[DrDu]

But with: delta U = delta H + delta work

That's not correct, or, to be more explicit, it only holds true for isobaric processes, where the concept of Enthalpy is most useful.

 

[metallica007]

Thx a lot everyone

 

[Point Conception]

http://www.patana.ac.th/secondary/science/anrophysics/ntopic10/commentary.htm

Scroll down to Isometric Processes:
This is a change that happens at constant volume
The work done in an isometric process is zero.
Therefore any change in internal energy of a gas in an isometric change is wholy due
to heat exchange between the system and the surroundings. delta U = Q

 

[Zeppos10]

The discussion is about the difference between U and H:
dU=q+w is an energy-balance for the system: if the volume of the system changes, some work needs to be done to displace against the external/environmental pressure p(env): This can also be taken as an energy-term: p(env)V is called 'flow-energy' /'pressure-energy/ pV-energy/ energy of displacement. Hence you can define H=U+p(env)V as the sum of 2 forms of energy. The energy balance now becomes dH= q +w' , where w' = useful work = - shaft work, and if no usefull work is done dH=q. In (reversible) thermodynamics no distinction is made between internal and external pressure (because they are assumed equal), which causes a lot of confusion. In the above: d(p(env)V)=0 hence dU=dH. See
https://www.physicsforums.com/showthread.php?t=88987&referrerid=219693 [post #7]

 

[DrDu]

Hi everyone
I need your help in the following problem:

If the measured energy change was 580 J, was this an enthalpy change or an internal energy change?
Answer: internal energy change
My question is, why the answer is internal energy instead of enthalpy change? could be you please explain it to me

The main problem is that the question isn't well posed. If they had said "the measured heat exchanged" instead of "energy change".
But the measured energy change? What do I know how and whether they measured internal energy or enthalpy? Both potentials are well defined but only the internal energy change is equal to the heat generated in the process at hand.