Proof that the heat capacity of a system at constant volume is greater than zero at the equilibrium - Physics Stack Exchange most recent 30 from physics.stackexchange.com 2019-08-19T11:57:55Z https://physics.stackexchange.com/feeds/question/334125 http://www.creativecommons.org/licenses/by-sa/3.0/rdf https://physics.stackexchange.com/q/334125 1 Proof that the heat capacity of a system at constant volume is greater than zero at the equilibrium Gabu https://physics.stackexchange.com/users/145303 2017-05-19T02:22:13Z 2017-05-19T07:20:41Z <p>I'm trying to prove that the heat capacity at constant volume, given by</p> <p>\begin{equation} C_{_V} = \left(\frac{\partial U}{\partial T}\right)_{V}, \end{equation}</p> <p>where $U$ is the energy of the system and $T$ it's temperature, is greater than zero when the system is at equilibrium.</p> <p>To do so I started considering</p> <p>\begin{equation} \left(\frac{\partial^{2} S}{\partial U^{2}}\right)_{_V} = -\frac{1}{T^{2}}\left(\frac{\partial T}{\partial U}\right)_{_V} = -\frac{1}{C_{_V}T^{2}} \end{equation}</p> <p>and so, I have to prove that</p> <p>\begin{equation} \left(\frac{\partial^{2} S}{\partial U^{2}}\right)_{_V} &lt;0 \end{equation}</p> <p>at equilibrium. Now, at equilibrium, the entropy is maximum for any given value of the energy $U$, which can be stated as</p> <p>\begin{equation} \left(\frac{\partial S}{\partial X}\right)_{_U} = 0 \,\,\,\,\,\,\,\,\,\,\mathrm{and}\,\,\,\,\,\,\,\,\,\, \left(\frac{\partial^{2} S}{\partial X^{2}}\right)_{_U}&lt;0 \end{equation}</p> <p>where $X$ is any extensive parameter of the system. This means that, for a given value of $U$, at equilibrium, the entropy is maximum with respect to this paramenter. Since $U$ is also an extensive paramenter of the system can I write the maximum entropy principle like</p> <p>\begin{equation} \left(\frac{\partial^{2} S}{\partial U^{2}}\right)_{_X}&lt;0 \end{equation} ?</p> <p>Thank you.</p>