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Specific Heat Capacity and Heat Capacity at Constant Pressure 📂Thermal Physics

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 11, regarding the constant-volume heat capacity CVC_{V} and constant-pressure heat capacity CpC_{p}.

Cp=CV+R=52R 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, γ:=CpCV\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 CV=∂U∂TC_{V} = \dfrac{\partial U}{\partial T}

    First law of thermodynamics

    dU=δQ+δW d U = \delta Q + \delta W

    According to the first law of thermodynamics, dU(T,V)dU(T,V) is a total differential, and the following holds:

    dU=∂U∂TdT+∂U∂VdV \begin{equation} dU = \dfrac{\partial U}{\partial T} dT + \dfrac{\partial U}{\partial V} dV \label{eq1} \end{equation}

    Similarly, since δW=−pdV\delta W = - p d V holds from the first law of thermodynamics, we obtain:

    dU=δQ+δW=δQ−pdV  ⟹  δQ=dU+pdV d U = \delta Q + \delta W = \delta Q - p dV \\ \implies \delta Q = d U + p d V

    Substituting (eq1)\eqref{eq1} into the above equation gives:

    δQ=dU+pdV=(∂U∂TdT+∂U∂VdV)+pdV=∂U∂TdT+(∂U∂V+p)dV \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 dTdT yields:

    δQdT=∂U∂T+(∂U∂V+p)dVdT \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 dVdT=0\displaystyle {{dV} \over {dT} } = 0, so we get:

    CV=∂Q∂T=δQdT=∂U∂T \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 CpC_{p}

    Upon substituting (eq3)\eqref{eq3} into (eq2)\eqref{eq2}, we obtain the following equation:

    Cp=∂Q∂T=∂U∂T+(∂U∂V+p)dVdT=CV+(∂U∂V+p)dVdT 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 UU

    Average kinetic energy of gas molecules

    <EK>=32kBTr \left< E_{K} \right> = {{3} \over {2}} k_{B} \text{Tr}

    The internal energy UU is obtained by multiplying the average energy by the number of molecules NN. Additionally, since NkB=nRNk_{B} = nR holds, it follows that:

    U=32NkBT=32nRT U = \dfrac{3}{2} N k_{B} T = \dfrac{3}{2} nRT

    However, since we are considering only moles of 11, U=32RTU = \dfrac{3}{2}RT holds. Therefore, we obtain:

    ∂U∂T=CV=32R∂U∂V=0 \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  ⟺  V=RTppV = RT \iff V = \dfrac{RT}{p} holds in the ideal gas equation, we get:

    ∂V∂T=Rp \dfrac{\partial V}{\partial T} = \dfrac{R}{p}

    Substituting these results into (eq3)\eqref{eq3} gives:

    Cp=CV+(∂U∂V+p)dVdT=CV+(0+p)Rp=CV+R=52R 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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