enter image description here

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 $\endgroup$ Commented 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
    Commented 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. $\endgroup$ Commented 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. $\endgroup$ Commented 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 $\Delta H = \Delta U + \Delta PV = C_v \Delta T + R \Delta T = C_p \Delta T$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.