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At constant pressure,∆H=q Where ∆H=enthalpy q=energy transferred by heating ∆H doesn't depend on how we get from one state to another, but q does. How can they be equal in any case? Is it because constant pressure implies just one Pathway from one point to another?

user90596
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  • You should probably learn how the equality $\Delta H = q$ is derived, since it depends on more than just constant pressure. – orthocresol Aug 06 '18 at 14:46
  • I feel that this is not a duplicate because the proposed duplicate's answer does not address this question. – a-cyclohexane-molecule Aug 06 '18 at 17:47
  • So, could somebody answer it then? – user90596 Aug 07 '18 at 02:00
  • If you carried out an adiabatic process in which you suddenly dropped the (external) pressure to a new lower value and then held the (external) pressure at that new value until the gas re-equilibrated at a new volume and temperature, would you consider that a constant pressure process? Or does the external pressure have to be held constant at the original value for the entire expansion (say, while heat is added to the gas to increase its volume)? – Chet Miller Aug 07 '18 at 16:50
  • It should be held constant for the entire expansion – user90596 Aug 07 '18 at 16:53

1 Answers1

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Path functions are so called because they depend on the means by which a system goes from state to state. When we specify a path, the resulting “path function” no longer depends on the actual path taken, and can be considered a state function. For similar examples, consider $C_p$ and $C_V$. In other words:

How can they be equal in any case? Is it because constant pressure implies just one Pathway from one point to another?

Yes.

a-cyclohexane-molecule
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  • @user90596, I'm happy to help, but please note that no one on this site is obligated to answer your question, and in general posting comments like "please answer" is probably far less effective than editing your question statement with further thoughts you've had on the matter. – a-cyclohexane-molecule Aug 07 '18 at 13:45