Question

Consider Maxwell's relation \( \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V \). The equation of state of a thermodynamic system is given as \( P = \frac{AT}{\sqrt{V}} + \frac{BT^3}{V} \), where A and B are constants of appropriate dimensions. Then \( \left(\frac{\partial C_V}{\partial V}\right)_T \) of the system varies with temperature as (\(C_V\) is the heat capacity at constant volume)

• A. \(T^2\)
• B. \(T\)
• C. \(T^{-1}\)
• D. \(T^3\)

Hint:

A useful identity to remember for problems like this is \( \left(\frac{\partial C_V}{\partial V}\right)_T = T \left(\frac{\partial^2 P}{\partial T^2}\right)_V \). Knowing this identity allows you to directly proceed to differentiating the equation of state, saving valuable time during an exam. This identity itself is derived using a Maxwell relation, as shown in the solution.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The problem asks for the temperature dependence of the variation of heat capacity \(C_V\) with volume at constant temperature, i.e., \( \left(\frac{\partial C_V}{\partial V}\right)_T \). We can find this relationship by starting with the definition of \(C_V\) in terms of entropy and using the given Maxwell relation and equation of state.
Step 2: Key Formula or Approach
1. Definition of heat capacity at constant volume: \( C_V = T \left(\frac{\partial S}{\partial T}\right)_V \).
2. The goal is to find \( \left(\frac{\partial C_V}{\partial V}\right)_T \). We start by differentiating the definition of \(C_V\) with respect to V at constant T.
3. We will use the fact that entropy \(S\) is a state function, which means its mixed second partial derivatives are equal: \( \frac{\partial^2 S}{\partial V \partial T} = \frac{\partial^2 S}{\partial T \partial V} \).
4. The given Maxwell relation is \( \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V \).
Step 3: Detailed Explanation
Let's find an expression for \( \left(\frac{\partial C_V}{\partial V}\right)_T \).
Start with the definition of \(C_V\): \[ C_V = T \left(\frac{\partial S}{\partial T}\right)_V \] Differentiate both sides with respect to \(V\) at constant \(T\): \[ \left(\frac{\partial C_V}{\partial V}\right)_T = \left[ \frac{\partial}{\partial V} \left( T \left(\frac{\partial S}{\partial T}\right)_V \right) \right]_T \] Since \(T\) is constant in this differentiation, we can move it and the \(\frac{\partial}{\partial V}\) operator around: \[ \left(\frac{\partial C_V}{\partial V}\right)_T = T \frac{\partial}{\partial V} \left(\frac{\partial S}{\partial T}\right)_V = T \frac{\partial^2 S}{\partial V \partial T} \] Since S is a state function, we can switch the order of differentiation: \[ \frac{\partial^2 S}{\partial V \partial T} = \frac{\partial^2 S}{\partial T \partial V} = \frac{\partial}{\partial T} \left( \left(\frac{\partial S}{\partial V}\right)_T \right)_V \] Substituting this back, we get: \[ \left(\frac{\partial C_V}{\partial V}\right)_T = T \left[ \frac{\partial}{\partial T} \left( \left(\frac{\partial S}{\partial V}\right)_T \right) \right]_V \] Now, use the given Maxwell's relation \( \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V \): \[ \left(\frac{\partial C_V}{\partial V}\right)_T = T \left[ \frac{\partial}{\partial T} \left( \left(\frac{\partial P}{\partial T}\right)_V \right) \right]_V = T \left(\frac{\partial^2 P}{\partial T^2}\right)_V \] This is a standard thermodynamic identity that connects the volume dependence of \(C_V\) to the equation of state.
Now, we use the given equation of state: \( P = \frac{AT}{\sqrt{V}} + \frac{BT^3}{V} \).
First, calculate the first partial derivative of \(P\) with respect to \(T\) at constant \(V\): \[ \left(\frac{\partial P}{\partial T}\right)_V = \frac{\partial}{\partial T} \left( \frac{AT}{\sqrt{V}} + \frac{BT^3}{V} \right) = \frac{A}{\sqrt{V}} + \frac{3BT^2}{V} \] Next, calculate the second partial derivative: \[ \left(\frac{\partial^2 P}{\partial T^2}\right)_V = \frac{\partial}{\partial T} \left( \frac{A}{\sqrt{V}} + \frac{3BT^2}{V} \right) = 0 + \frac{6BT}{V} = \frac{6BT}{V} \] Step 4: Final Answer
Substitute this result back into our derived expression for \( \left(\frac{\partial C_V}{\partial V}\right)_T \): \[ \left(\frac{\partial C_V}{\partial V}\right)_T = T \left(\frac{6BT}{V}\right) = \frac{6BT^2}{V} \] From this expression, we can see the temperature dependence: \[ \left(\frac{\partial C_V}{\partial V}\right)_T \propto T^2 \] This corresponds to option (A).