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Here pressure isn't constant, so how $\Delta H=C_p \Delta T$?

  • $\begingroup$ $C_p$ is not well-named; it frequently applies when the pressure changes. See, e.g., my answer here: physics.stackexchange.com/q/203605 $\endgroup$ Sep 30, 2016 at 16:42
  • $\begingroup$ @Jahan Claes Ohh.. So for ideal gas, Cp doesn't depend upon p..OK But here the gas isn't necessarily ideal... Then also Cp doesn't depend upon p??? $\endgroup$
    – Kshitij
    Sep 30, 2016 at 16:52
  • $\begingroup$ The solution is using the ideal gas formula for $C_p$ and $C_v$, so they're assuming an ideal gas. If it wasn't an ideal gas, there isn't much you can say, since you don't have explicit formulas for anything. $\endgroup$ Sep 30, 2016 at 17:15
  • $\begingroup$ For an ideal gas, U and H are functions only of T, and not V and P, respectively. $\endgroup$ Sep 30, 2016 at 19:38

1 Answer 1


Enthalpy $H=U+PV$ and for an ideal gas enthalpy does not depend on pressure because $PV = RT$ for one mole ($n=1)$

So you have $\Delta U = C_v \Delta T$ and $\Delta H = \Delta U + \Delta PV = C_v \Delta T + R \Delta T = C_p \Delta T$.


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