Use the Gibbs-Duhem equation to show that the partial molar volume (or any partial molar property) of a component \(\mathrm{B}\) can be obtained if the partial molar volume (or other property) of A is known for all compositions up to the one of interest. Do this by proving that $$ V_{\mathrm{B}}=V_{\mathrm{B}}^{*} \int_{V_{\mathrm{A}}}^{v_{\mathrm{A}}} \frac{x_{\mathrm{A}}}{1-x_{\mathrm{A}}} \mathrm{d} V_{\mathrm{A}} $$ Use the following data (which are for \(298 \mathrm{~K}\) ) to evaluate the integral graphically to find the partial molar volume of acetone at \(x=0.500\). $$ \begin{array}{llllllll} x\left(\mathrm{CHCl}_{3}\right) & 0 & 0.194 & 0.385 & 0.559 & 0.788 & 0.889 & 1.000 \\ V_{\mathrm{m}} /\left(\mathrm{cm}^{3} \mathrm{~mol}^{-1}\right) & 73.99 & 75.29 & 76.50 & 77.55 & 79.08 & 79.82 & 80.67 \end{array} $$

The Gibbs-Duhem equation is a fundamental relationship in thermodynamics that provides valuable insights into how the various components of a mixture contribute to its overall molar properties. In particular, this equation allows us to understand how the partial molar properties of each substance in a binary mixture are related. To put it simply, the Gibbs-Duhem equation states that the sum of the changes in the partial molar volumes multiplied by the respective mole numbers of all components must equal zero.

This condition arises from the fact that the total volume of a mixture is a constant at a given state, meaning as one substance's concentration increases, it must be at the expense of the other component's concentration. In equation form, it's expressed as:
\[ n_A d\bar{V}_A + n_B d\bar{V}_B = 0 \]
Let's apply this concept to solve the problem at hand. To find the partial molar volume of component B, we start by integrating the expression after isolating it for component B's volume. Through integration, we can determine \(V_{\text{B}}\) as long as we know \(V_{\text{A}}\) and the composition of each component at the starting and ending points.

A binary mixture consists of two different substances that, when combined, form a solution. The partial molar volume is a way to quantify how much volume one mole of a substance contributes to the overall volume of the mixture. It's important to note that the partial molar volume is not necessarily constant. It can vary with the composition of the mixture, as each component interacts with the others.

In the context of our exercise, we're interested in finding out how the partial molar volume of acetone changes in a mixture with chloroform, especially when they are present in equal proportions. To find this, we use the Gibbs-Duhem equation, which helps us calculate the partial molar volume of one component if we have the data for the other over a range of compositions. \(V_{\text{B}}\) is then dependent on the integral we derive from our equation, involving the composition of \(V_{\text{A}}\) primarily.

Molar quantities, like partial molar volume, are intrinsic properties that provide a deep understanding of the behavior of substances in mixtures. When dealing with solutions, it's often useful to speak in terms of 'per mole' quantities because they allow for standardizing measurements and make it easier to compare different substances or mixtures on a common scale.

In our exercise, the molar volume \(V_{\text{m}}\) given in the table is the volume occupied by one mole of the binary mixture at various compositions. Knowing how to deduce partial molar volumes from molar quantities gives us powerful insights into the properties of individual components and how they contribute to the total volume of the mixture. This knowledge is not just academic; it has real-world applications in fields like material science, chemistry, and chemical engineering, where understanding the specific roles of components in a mixture is crucial.

Graphical integration is a practical approach to evaluate integrals, especially when the functional relationship is not easily integrable algebraically, or when working with empirical data. In the case of our exercise, graphical integration helps convert the qualitative data from a graph into a quantitative estimate of the integral.

By plotting the molar volume against the composition, as provided in the exercise, we can visualize how the molar volume changes with varying compositions of chloroform and acetone. Drawing a tangent to this curve gives us the slope, which corresponds to the derivative needed in our integral. The area under this curve between two points gives us the change in partial molar volume.
Students can benefit from learning this technique, not only to solve the current problem but also to handle real-world situations where they may have to estimate values from graphical data. Making accurate estimates from graphs is a skill that becomes invaluable in various technical and scientific disciplines.