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Topic: About thermodynamics!  (Read 11323 times)

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Offline Winga

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About thermodynamics!
« on: November 10, 2004, 08:34:04 AM »
E (or U) = q + w
dE = q - PdV

1. Why we don't use dq instead of q?

2. dE = q - PdV, is it modified from this dE = q - (VdP + PdV), in constant pressure, dp = 0 ?

3. Why w always = - PdV?
When the system expands in volume, the direction of expansion is opposite to the direction of external pressure, so w is -ve.
When the system reduces in volume, both directions (w & p) are the same, but PdV is still -ve. So, why we need to add a -ve sign to PdV?

Demotivator

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Re:About thermodynamics!
« Reply #1 on: November 10, 2004, 09:00:53 AM »
1) Technically, dE = dq - PdV.
   I suppose it's just carelessness or oversight.

2) yes

3) The system either does work or work is done on the system. When volume expands, the system does work in pushing against it's surroundings. That means the energy of the system will decrease and therefore work is negative. -PdV is negative since final vol > initial.
When volume contracts, the surroundings do work on the system thus raising the energy of the system. In this case work must be positive. -PdV is positive since final vol < initial.

Offline Winga

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Re:About thermodynamics!
« Reply #2 on: November 10, 2004, 09:45:02 AM »
In Q.3, adding a -ve sign is to show that change in volume cannot be -ve, right?

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Re:About thermodynamics!
« Reply #3 on: November 10, 2004, 10:18:41 AM »
No.
Change in volume can be negative. When compressed, final volume is less than initial volume so change in volume is negative.
The neg sign is to make the WORK positive when vol change is negative.

Offline maxyoung

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Re:About thermodynamics!
« Reply #4 on: November 14, 2004, 05:57:37 AM »
what is the meaning of dq? q2-q1??
q is not a state function, so dq has no meaning.

w is not a state function either. So dE = q + w
and w = -pdv
not w = -d(pv) = -vdp-pdv and dp=0
 w = -pdv applies to all conditions not only constant pressure process.


Demotivator

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Re:About thermodynamics!
« Reply #5 on: November 14, 2004, 05:03:04 PM »
1) dq and dw  are legitimate differentials.
   Although dq and dw  are  not state functions, the sum of these two inexact differentials is the exact differential, dE.
 The integral(dq) is q .

2) you're right,   dw = -pdv

Offline maxyoung

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Re:About thermodynamics!
« Reply #6 on: November 15, 2004, 05:58:23 AM »
Mathematically, dq is right, but in the sense of chemistry it has no meaning.
am i right?

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Re:About thermodynamics!
« Reply #7 on: November 15, 2004, 11:20:44 AM »
Differentials are infinitesimal increments so in that sense they have "non quantifiable meaning" until integrated.
A ratio of differentials (derivative), however, has meaning as the instantaneous rate of change, as in dq/dT (heat capacity).

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