Question

One of the Maxwell equations is expressed as \(\left( \dfrac{\partial s}{\partial v} \right)_{T = \left( \dfrac{\partial p}{\partial T} \right)_{v}\), where \(s\) is the entropy per unit mass, \(v\) is the mass specific volume, \(p\) is the pressure, and \(T\) is the temperature. In this expression, \(s\) is a continuous function of \(T\) and \(v\). Let \(c_v\) be the specific heat capacity at constant volume for a gas. Then, \(\left( \dfrac{\partial c_v}{\partial v} \right)_{T}\) can be written as}

• A. ( \dfrac{p}{T} \left( \dfrac{\partial^2 p}{\partial T^2} \right)_{v} \)
• B. ( \dfrac{v}{T} \left( \dfrac{\partial^2 p}{\partial v^2} \right)_{T} \)
• C. ( T \left( \dfrac{\partial^2 p}{\partial T^2} \right)_{v} \)
• D. ( \dfrac{1}{T} \left( \dfrac{\partial^2 p}{\partial v^2} \right)_{T} \)

Hint:

Maxwell relations help convert entropy derivatives into pressure–temperature derivatives, making thermodynamic identities easier to evaluate.

Answer 1

The Correct Option is C

Start with the definition of specific heat at constant volume: \[ c_v = T \left( \frac{\partial s}{\partial T} \right)_{v} \] Differentiate with respect to \(v\) at constant \(T\): \[ \left( \frac{\partial c_v}{\partial v} \right)_{T} = T \left( \frac{\partial}{\partial v} \frac{\partial s}{\partial T} \right) \] Since the order of partial derivatives can be interchanged: \[ \left( \frac{\partial c_v}{\partial v} \right)_{T} = T \left( \frac{\partial}{\partial T} \frac{\partial s}{\partial v} \right) \] Using the Maxwell relation: \[ \left( \frac{\partial s}{\partial v} \right)_{T} = \left( \frac{\partial p}{\partial T} \right)_{v} \] Thus: \[ \left( \frac{\partial c_v}{\partial v} \right)_{T} = T \left( \frac{\partial}{\partial T} \left( \frac{\partial p}{\partial T} \right)_{v} \right) \] \[ = T \left( \frac{\partial^2 p}{\partial T^2} \right)_{v} \] Hence the correct expression is: \[ \boxed{T \left( \dfrac{\partial^2 p}{\partial T^2} \right)_{v}} \]