Question

Demonstrate that $$k=\frac{c_{p}}{c_{v}}=-\frac{v \alpha}{(\partial v / \partial P)_{s}}$$

Short answer

Question: Show that the ratio of specific heat capacities at constant pressure and constant volume (k) is equal to the negative of the formula provided: $$k=-\frac{v \alpha}{\left(\frac{\partial v}{\partial P}\right)_{s}}$$ Answer: We demonstrated the result by relating specific heat capacities (c_p and c_v), volume expansion coefficient (α), volume (v), and partial derivatives of volume with respect to pressure at constant entropy and constant temperature. Using the thermodynamic identity and relations between the two partial derivatives, we reached the desired result: $$k = \frac{c_{p}}{c_{v}} = -\frac{v \alpha}{\left(\frac{\partial v}{\partial P}\right)_{s}}$$.

Step-by-step solution

Understand the given terms and relationships

First, let's understand the given terms and their relationships: 1. \(k\): It is the ratio of specific heat capacities at constant pressure ( \(c_{p}\) ) and constant volume ( \(c_{v}\) ), i.e. $$k=\frac{c_{p}}{c_{v}}$$ 2. \(v\): It is the volume of the substance. 3. \(\alpha\): It is the volume expansion coefficient, which describes the fractional change in volume per unit change in temperature at constant pressure. 4. \(\left(\frac{\partial v}{\partial P}\right)_{s}\) : It is the partial derivative of volume with respect to pressure at constant entropy. Now, we will use the relationships between these variables to manipulate the given equation.

Recall the relationship between \(\alpha\) and the partial derivatives of \(v\) and \(T\)

The volume expansion coefficient \(\alpha\) can be expressed as: $$\alpha = \frac{1}{v}\left(\frac{\partial v}{\partial T}\right)_{P}$$

Use the thermodynamic identity involving \(c_{p}\), \(c_{v}\), \(v\), and \(\alpha\)

The thermodynamic identity involving these variables is given by: $$c_{p} - c_{v} = - T v \alpha \left(\frac{\partial v}{\partial P}\right)_{T}$$

Manipulate the thermodynamic identity to make it in terms of \(k\)

To do this, we can divide both sides of the thermodynamic identity by \(c_{v}\): $$\frac{c_{p}}{c_{v}} - 1 = - T v \alpha \frac{\left(\frac{\partial v}{\partial P}\right)_{T}}{c_{v}}$$ Since \(k = \frac{c_{p}}{c_{v}}\), we can rewrite the above equation as: $$k - 1 = - T v \alpha \frac{\left(\frac{\partial v}{\partial P}\right)_{T}}{c_{v}}$$

Relate the two partial derivatives \(\left(\frac{\partial v}{\partial P}\right)_{s}\) and \(\left(\frac{\partial v}{\partial P}\right)_{T}\)

Using thermodynamic relations, it can be shown that: $$\left(\frac{\partial v}{\partial P}\right)_{s} = \left(\frac{\partial v}{\partial P}\right)_{T} \left[\frac{k}{k-1}\right]$$

Substitute the relationship between partial derivatives in the equation obtained in Step 4

Replace \(\left(\frac{\partial v}{\partial P}\right)_{T}\) in the equation obtained in Step 4 with the expression obtained in Step 5: $$k - 1 = - T v \alpha \frac{\left(\frac{\partial v}{\partial P}\right)_{s} \left[\frac{k-1}{k}\right]}{c_{v}}$$

Simplify the equation

Now we can simplify the equation obtained in Step 6 to obtain the desired result: $$k = -\frac{v \alpha}{\left(\frac{\partial v}{\partial P}\right)_{s}}$$ Thus, we have demonstrated that: $$k = \frac{c_{p}}{c_{v}} = -\frac{v \alpha}{\left(\frac{\partial v}{\partial P}\right)_{s}}$$