I'm getting myself confused on when to use $h = c_p \Delta T$ or $u=c_v \Delta T$, where $c_p$ is the specific heat at constant pressure and $c_v$ is the specific heat at constant volume.
It's in relation to thermodynamic processes such as expanding volumes with pistons and the likes.
Here's what I know (in relation to this):
First law for a closed system (per unit mass)
$$q-w = \Delta u$$
First law for an open system (per unit mass)
$$q-w_s = \Delta (h+\frac12c^2 +gz)$$
Example
Say I've got a piston expanding – causing an ideal gas to expand at constant pressure.
I can say that $\mathrm{d} w = p\mathrm dv$ as well as $\mathrm du = c_v\mathrm dT$ ─ is this correct?
Subbing this in I get
$$\mathrm dq = p\mathrm dv + c_v \mathrm dT,$$
whereas if I decide I want to use
\begin{align}
h & = u+ pv \\
\mathrm dh & = \mathrm du + p \mathrm dv + v \mathrm dp \\
\mathrm du & = \mathrm dh – p \mathrm dv – v \mathrm dp
\end{align}
giving
\begin{align}
\mathrm dq & = p\mathrm dv + \mathrm dh +- p \mathrm dv – v \mathrm dp \\
\mathrm dq & = c_p \mathrm dT – v \mathrm dp .
\end{align}
Which (if any) expression for $\mathrm dq$ is correct? I feel like there's some flaws in my fundamental understanding of whats happening here. Is it to do with open/closed systems?
Best Answer
Fundamentally there's a simple difference: When you are working with a perfect gas at constant volume you can take the variation of inside energy equal to the heat absorbed in the transformation. In this case you must use $C_v$, obviously. In the other side, where the pressure is constant you can't consider the equality defined previously. In fact you must consider that ONLY the heat absorbed is equal to your equation (which is correct in his procedure) with Cp. All of that remembering the values of $C_p$ and $C_v$, also knowing that $C_p - C_v=R$ for an ideal gas.