Skip to main content
Chemistry LibreTexts

19.6: The Temperature of a Gas Decreases in a Reversible Adiabatic Expansion

  • Page ID
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)

    We can make the same argument for the heat along C. If we do the three processes A and B+C only to a tiny extent we can write:


    And now we can integrate from \(V_1\) to \(V_2\) over the reversible adiabatic work along B and from \(T_1\) to \(T_2\) for the reversible isochoric heat along C. To separate the variables we do need to bring the temperature to the right side of the equation.:


    The latter expression is valid for a reversible adiabatic expansion of a monatomic ideal gas (say Argon) because we used the \(C_v\) expression for such a system. We can use the gas law \(PV=nRT\) to translate this expression in one that relates pressure and volume see Eq 19.23

    We can mathematically show that the temperature of a gas decreases during an adiabatic expansion. Assuming an ideal gas, the internal energy along an adiabatic path is:

    \[\begin{split} d\bar{U}&= \delta q+\delta w \\ &= 0-Pd\bar{V}\\ &= -Pd\bar{V} \end{split}\]

    The constant volume heat capacity is defined as:

    \[{\bar{C}}_V=\left(\frac{\partial\bar{U}}{\partial T}\right)_V\]

    We can rewrite this for internal energy:


    Combining these two expressions for internal energy, we obtain:


    Using the ideal gas law for pressure of an ideal gas:


    Separating variables:


    This is an expression for an ideal path along a reversible, adiabatic path that relates temperature to volume. To find our path along a PV surface for an ideal gas, we can start in TV surface and convert to a PV surface. Let's go from (\(T_1,V_1\)) to (\(T_2,V_2\)).





    We know that:






    This expression shows that volume and temperature are inversely related. That is, as the volume increase from \(V_1\) to \(V_2\), the temperature must decrease from \(T_1\) to \(T_2\).

    19.6: The Temperature of a Gas Decreases in a Reversible Adiabatic Expansion is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.