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Topic: ethalpy and internal energy  (Read 3121 times)

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Offline simpleton

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ethalpy and internal energy
« on: August 19, 2009, 04:52:02 AM »
Dear all,

I have some doubts regarding biophysical chemistry and would like to seek your advice.

(1) If an experiment is conducted in a closed system, we will regard the work as irreversible? Since (e.g. exothermic process) heat is transferred out of the system and so we could not use the heat lost to reverse the process?

(2) The theory regarding enthaply. We can only calculate for delta-H when the process is conducted under a constant pressure? My understanding is that because under constant pressure, in order to maintain a constant pressure, the heat will be consumed for expansion or compression hence we need to factor in the heat consumed. In order to get the true heat changes for experiment conducted under constant pressure, we must use delta-H and not delta-U?

(3) If the process is not conducted under constant pressure, then we will calculate for delta-U to get the true heat changes within the system?

Really appreciate all the helps. Thank you very much in advance.  :)

Offline zxt

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Re: ethalpy and internal energy
« Reply #1 on: August 19, 2009, 09:02:29 AM »
Dear all,

I have some doubts regarding biophysical chemistry and would like to seek your advice.

(1) If an experiment is conducted in a closed system, we will regard the work as irreversible? Since (e.g. exothermic process) heat is transferred out of the system and so we could not use the heat lost to reverse the process?

(2) The theory regarding enthaply. We can only calculate for delta-H when the process is conducted under a constant pressure? My understanding is that because under constant pressure, in order to maintain a constant pressure, the heat will be consumed for expansion or compression hence we need to factor in the heat consumed. In order to get the true heat changes for experiment conducted under constant pressure, we must use delta-H and not delta-U?

(3) If the process is not conducted under constant pressure, then we will calculate for delta-U to get the true heat changes within the system?

Really appreciate all the helps. Thank you very much in advance.  :)

First, according to Carnot cycle, work can totally transfer into heat but can not heat totally tansfer into work. So the process is irreversible.

Second,  :delta:H= :delta:U+ :delta:(pv), because absolute value of U cannot be calculated and measured now, so is H. If P is constant,  :delta:H= :delta:U+ :delta:(pv), or V(System Volume) is constant,  :delta:H= :delta:U=Q.

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