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Enthalpy for Constant Pressure and Constant Volume Systems?

  1. Feb 11, 2015 #1
    I understand why entropy (a state function) is very useful in chemistry. Since most chemistry systems are at constant pressure, ##\Delta H=Q_v## from the derivation below:

    ##H=U+PV##
    ##dH=dU+dPV+PdV##
    ##dU=\delta Q-\delta W##
    ## dU=\delta Q-PdV##
    ## dH=\delta Q+dPV##
    ##dPV=0##
    ## dH=\delta Q_v##

    However, for constant volume work is not done, so that seems to leave this pesky ##dPV## term. That is:

    ##dH=\delta Q_p+dPV## What is the meaning of this term. ##PdV## is obviously microscopic work, but what the other term resulting from the product rule means physically escapes me.

    Thanks,
    Chris
     
  2. jcsd
  3. Feb 11, 2015 #2
    Use the following logic,
    $$ \text{d}H = \delta\text{Q} + v\,\text{d}P $$
    knowing that,
    $$ \text{d}S = \frac{\delta\text{Q}}{T} $$
    rearrange to get the Gibbs relation
    $$ \text{d}H = T\text{d}S+ v\,\text{d}P $$
    Assume boldly $$\text{d}S=0$$ And you get reversibly adiabatic shaft work.
    $$ \text{d}H = v\,\text{d}P $$
     
  4. Feb 11, 2015 #3
    I follow the math, but can you give me a physical example of a system that reversibly does shaft work.

    Thanks,
    Chris
     
  5. Feb 12, 2015 #4
    Well before I answer this we must be very careful in this assumption $$\text{d}S=0$$ Many physical systems such as turbines, compressors and pumps are very well insulated therefore it is possible to consider any heat transfer negligible and ultimately they may be modeled as isentropic.

    However in reality, due to many factors there are heat losses in the system therefore the assumption $$\text{d}S=0$$ does not hold. The concept is still usable in modeling, but you would have to account for the heat transfer somehow.
     
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