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kelvin490
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We know that for constant pressure thermodynamic processes, dH=dqp. My question is, does it implies that only reversible work is possible in this processes so that dw=0 because dv is zero? In addition, does qv necessarily be reversible heat transfer in this case? What if the heat transfer is irreversible?
Similar question for dU=dqv, does the process need to be reversible?
Another question is, do the above relations have anything to do with whether or not the system is an ideal gas? I have heard from a lecture that du≠dqv in Joule's free expansion for non-ideal gas. Consider reversibility and whether it's ideal gas there are four combinations of situations (ideal gas reversible,ideal gas irreversible, non-ideal gas reversible, non-ideal gas irreversible). I get confused with how these factors affect the thermodynamic relations.
Similar question for dU=dqv, does the process need to be reversible?
Another question is, do the above relations have anything to do with whether or not the system is an ideal gas? I have heard from a lecture that du≠dqv in Joule's free expansion for non-ideal gas. Consider reversibility and whether it's ideal gas there are four combinations of situations (ideal gas reversible,ideal gas irreversible, non-ideal gas reversible, non-ideal gas irreversible). I get confused with how these factors affect the thermodynamic relations.
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