Specific Heat Capacity and Heat Capacity at Constant Pressure
Formulas
The following equation holds for an ideal gas system where the amount of substance is $1$, regarding the constant-volume heat capacity $C_{V}$ and constant-pressure heat capacity $C_{p}$.
$$ C_{p} = C_{V} + R = {{5} \over {2}} R $$
Explanation
Not only do the heat capacities differ depending on whether the process is isochoric or isobaric, but also their relationship fits perfectly in a mathematical sense. Especially, $\gamma := \dfrac{C_{p}}{C_{V}}$ itself may not have significant physical meaning, but it is used importantly in various mathematical contexts.
Proof
Part 1. To show that $C_{V} = \dfrac{\partial U}{\partial T}$
$$ d U = \delta Q + \delta W $$
According to the first law of thermodynamics, $dU(T,V)$ is a total differential, and the following holds:
$$ \begin{equation} dU = \dfrac{\partial U}{\partial T} dT + \dfrac{\partial U}{\partial V} dV \label{eq1} \end{equation} $$
Similarly, since $\delta W = - p d V$ holds from the first law of thermodynamics, we obtain:
$$ d U = \delta Q + \delta W = \delta Q - p dV \\ \implies \delta Q = d U + p d V $$
Substituting $\eqref{eq1}$ into the above equation gives:
$$ \begin{align*} \delta Q =& d U + p d V \\ =& \left( \dfrac{\partial U}{\partial T} dT + \dfrac{\partial U}{\partial V} dV \right) + p d V \\ =& \dfrac{\partial U}{\partial T} dT + \left( \dfrac{\partial U}{\partial V} +p \right) dV \end{align*} $$
Dividing both sides by $dT$ yields:
$$ \begin{equation} \dfrac{\delta Q}{dT} = \dfrac{\partial U}{\partial T} + \left( \dfrac{\partial U}{\partial V} +p \right) \dfrac{dV}{dT} \label{eq2} \end{equation} $$
If the volume is constant then $\displaystyle {{dV} \over {dT} } = 0$, so we get:
$$ \begin{equation} C_{V} = {{\partial Q} \over {\partial T}} = \dfrac{\delta Q}{dT} = \dfrac{\partial U}{\partial T} \label{eq3} \end{equation} $$
Part 2. To derive $C_{p}$
Upon substituting $\eqref{eq3}$ into $\eqref{eq2}$, we obtain the following equation:
$$ C_{p} = {{\partial Q} \over {\partial T}} = \dfrac{\partial U}{\partial T} + \left( \dfrac{\partial U}{\partial V} +p \right) \dfrac{dV}{dT} = C_{V} + \left( \dfrac{\partial U}{\partial V} +p \right) \dfrac{dV}{dT} $$
Part 3. Developing the internal energy $U$
Average kinetic energy of gas molecules
$$ \left< E_{K} \right> = {{3} \over {2}} k_{B} \text{Tr} $$
The internal energy $U$ is obtained by multiplying the average energy by the number of molecules $N$. Additionally, since $Nk_{B} = nR$ holds, it follows that:
$$ U = \dfrac{3}{2} N k_{B} T = \dfrac{3}{2} nRT $$
However, since we are considering only moles of $1$, $U = \dfrac{3}{2}RT$ holds. Therefore, we obtain:
$$ \begin{align*} \dfrac{\partial U}{\partial T} =& C_{V} = {{3} \over {2}} R \\ \dfrac{\partial U}{\partial V} =& 0 \end{align*} $$
Meanwhile, since $pV = RT \iff V = \dfrac{RT}{p}$ holds in the ideal gas equation, we get:
$$ \dfrac{\partial V}{\partial T} = \dfrac{R}{p} $$
Substituting these results into $\eqref{eq3}$ gives:
$$ C_{p} = C_{V} + \left( \dfrac{\partial U}{\partial V} +p \right) \dfrac{dV}{dT} = C_{V} + ( 0 + p ) {{R} \over {p}} = C_{V} + R = { {5} \over {2}} R $$
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