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5.5: The Joule Experiment

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  • Going back to the expression for changes in internal energy that stems from assuming that \(U\) is a function of \(V\) and \(T\) (or \(U(V, T)\) for short)

    \[ dU = \left( \dfrac{\partial U}{\partial V} \right)_TdV+ \left( \dfrac{\partial U}{\partial T} \right)_V dT\]

    one quickly recognizes one of the terms as the constant volume heat capacity, \(C_V\). And so the expression can be re-written

    \[ dU = \left( \dfrac{\partial U}{\partial V} \right)_T dV + C_V dT\]

    But what about the first term? The partial derivative is a coefficient called the “internal pressure”, and given the symbol \(\pi_T\).

    \[ \pi_T = \left( \dfrac{\partial U}{\partial V} \right)_T \]

    James Prescott Joule (1818-1889) recognized that \(\pi_T\) should have units of pressure (Energy/volume = pressure) and designed an experiment to measure it.

    He immersed two copper spheres, A and B, connected by a stopcock. Sphere A is filled with a sample of gas while sphere B was evacuated. The idea was that when the stopcock was opened, the gas in sphere A would expand (\(\Delta V > 0\)) against the vacuum in sphere B (doing no work since \(p_{ext} = 0\). The change in the internal energy could be expressed

    \[ dU = \pi_T dV + C_V dT\]

    But also, from the first law of thermodynamics

    \[ dU = dq + dw\]

    Equating the two

    \[ \pi_T dV + C_V dT = dq + dw\]

    and since \(dw = 0\)

    \[ \pi_T dV + C_V dT = dq \]

    Joule concluded that \(dq = 0\) (and \(dT = 0\) as well) since he did not observe a temperature change in the water bath which could only have been caused by the metal spheres either absorbing or emitting heat. And because \(dV > 0\) for the gas that underwent the expansion into an open space, \(\pi_T\) must also be zero! In truth, the gas did undergo a temperature change, but it was too small to be detected within his experimental precision. Later, we (once we develop the Maxwell Relations) will show that

    \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \left( \dfrac{\partial p}{\partial T} \right)_V -p \label{eq3}\]

    Application to an Ideal Gas

    For an ideal gas \(p = RT/V\), so it is easy to show that

    \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{R}{V} \label{eq4}\]

    so combining Equations \ref{eq3} and \ref{eq4} together to get

    \[ \left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{RT}{V} - p \label{eq5}\]

    And since also becuase \(p = RT/V\), then Equation \ref{eq5} simplifies to

    \[ \left( \dfrac{\partial U}{\partial V} \right)_T = p -p = 0\]

    So while Joule’s observation was consistent with limiting ideal behavior, his result was really an artifact of his experimental uncertainty masking what actually happened.

    Appliation to a van der Waals Gas

    For a van der Waals gas,

    \[ p = \dfrac{RT}{V-b} - \dfrac{a}{V^2} \label{eqV1}\]


    \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{R}{V-b} \label{eqV2}\]


    \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T\dfrac{R}{V-b} - p \label{eqV3}\]

    Substitution of the expression for \(p\) (Equation \ref{eqV1}) into this Equation \ref{eqV3}

    \[ \left( \dfrac{\partial U}{\partial V} \right)_T = \dfrac{a}{V^2}\]

    In general, it can be shown that

    \[\left( \dfrac{\partial p}{\partial T} \right)_V = \dfrac{\alpha}{\kappa_T}\]

    And so the internal pressure can be expressed entirely in terms of measurable properties

    \[ \left( \dfrac{\partial U}{\partial V} \right)_T = T \dfrac{\alpha}{\kappa_T}-p\]

    and need not apply to only gases (real or ideal)!

    Contributors and Attributions

    • Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)