# Please explain the solution if it's correct Here pressure isn't constant, so how $\Delta H=C_p \Delta T$?

• $C_p$ is not well-named; it frequently applies when the pressure changes. See, e.g., my answer here: physics.stackexchange.com/q/203605 Sep 30, 2016 at 16:42
• @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??? Sep 30, 2016 at 16:52
• 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. Sep 30, 2016 at 17:15
• For an ideal gas, U and H are functions only of T, and not V and P, respectively. Sep 30, 2016 at 19:38

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