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Why can the specific heat at constant Pressure ($C_p$) be used to calculate the change in enthalpy as $\text{d}H = C_p\text{d}T\newcommand{\d}{\text{d}}$ for processes that are not even at constant pressure?

I have $\d H = C_p\d T + V\d p$. Why do we neglect the $V\d p$ term? For gases, $V$ the specific volume, can be very large, so an increase in pressure could have a non-negligible effect on the enthalpy. Is it so? Or not?

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    $\begingroup$ The enthalpy of an ideal gas is a function only of temperature, but not pressure. For a real gas (or other substance), the effect of pressure on enthalpy would also have to be included. $\endgroup$ Jan 27 '21 at 19:31
  • $\begingroup$ But isn't Cp defined for processes at constant pressure? @ChetMiller $\endgroup$
    – Alfred
    Jan 27 '21 at 19:33
  • $\begingroup$ See my answer below. $\endgroup$ Jan 27 '21 at 21:58
  • $\begingroup$ @ChetMiller I updated my question to express my concerns further. Does this explain my question better? $\endgroup$
    – Alfred
    Jan 28 '21 at 2:05
  • $\begingroup$ Your equation for dh is incorrect. It should read $$dh=C_pdT+\left[v-T\left(\frac{\partial v}{\partial T}\right)_P\right]dP$$If you substitute the ideal gas law into the 2nd term, what do you get? $\endgroup$ Jan 28 '21 at 2:18
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Cp of an ideal gas doesn't depend on pressure. Cp of any substance is defined as the partial derivative of enthalpy respect to temperature at constant pressure. But, if the enthalpy of the substance is independent of pressure, then it doesn't matter if the pressure is constant. However, determining Cp by measuring the amount of heat required to change the temperature by a certain amount of the substance at constant pressure is still a standard method of measuring Cp.

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Specific heat capacity $\ce{C_p}$ can be used to calculate the enthalpy change even for the processes that are not at constant pressure as enthalpy is a state function, that is, it doesn't matter by which process you reach the end state, the value of enthalpy change will remain the same. Same is the case when we measure the change in internal energy, it is given by $\Delta U= n \ce{C_v}\Delta T$ where $\ce{C_v}$ is the specific heat capacity at constant volume, but this equation is used for all processes for the same reason as stated above, that is, internal energy is a state function and is independent of the path taken to reach there.

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  • $\begingroup$ But why can Cp still be used to calculate enthalpy? I understand that, but isn't Cp specifically defined at constant pressures? $\endgroup$
    – Alfred
    Jan 27 '21 at 20:25
  • $\begingroup$ Yes, it is defined at constant pressure. Alright, take it this way, suppose the process occurs at constant pressure, then you don't have any problems calculating the enthalpy change, right? Now we know that enthalpy is a state function which means if we go by any other process also enthalpy will come out to be same. But for those processes we cannot evaluate the enthalpy change in a different way, that is why we always use Cp for any process because it doesn't matter by which process we go answer will come out to be same. $\endgroup$ Jan 28 '21 at 6:35
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It follows from the familiar result: $$\Delta H=q_p$$

$$\Delta H=\Delta U+\Delta (pV)$$ Pressure is constant: $$\Delta H =\Delta U+p\Delta V$$ By First Law, $\Delta U=q-p\Delta V$, so, $$\Delta H=q_p$$

By using the fact that $$C_p=\dfrac{q_p}{\Delta T}$$ We get $$\Delta H=C_p\Delta T$$ And take my note: $\Delta H$ is a state function so it doesn't matter how the process is carried out.

Note, that the above answer is for an ideal gas, which the OP used in his question.

Hope this helps. Ask anything if not clear :)

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  • $\begingroup$ This is correct only for an ideal gas. For a real gas, H depends on pressure. $\endgroup$ Jan 28 '21 at 13:22
  • $\begingroup$ @ChetMiller: The OP had an ideal gas in his illustration. $\endgroup$ Jan 28 '21 at 14:34
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    $\begingroup$ It’s important to make it clear the this only applies to an ideal gas. $\endgroup$ Jan 28 '21 at 15:29
  • $\begingroup$ @ChetMiller: done :) $\endgroup$ Jan 28 '21 at 15:41

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