Question

Which of following is correct form of first TdS equation ?

• A. \( C_v dT + T \left(\frac{\partial V}{\partial T}\right)_V dV \)
• B. \( C_v dT + T \left(\frac{\partial P}{\partial T}\right)_V dV \)
• C. \( C_v dT - T \left(\frac{\partial P}{\partial T}\right)_V dV \)
• D. \( C_p dT + T \left(\frac{\partial V}{\partial T}\right)_P dV \)

Hint:

There are two main TdS equations. Memorizing them is useful:
{1st TdS Equation (S in terms of T, V):} \( TdS = C_V dT + T \left(\frac{\partial P}{\partial T}\right)_V dV \)
{2nd TdS Equation (S in terms of T, P):} \( TdS = C_P dT - T \left(\frac{\partial V}{\partial T}\right)_P dP \)
Remembering that the first involves \(C_V\) and \(dV\), while the second involves \(C_P\) and \(dP\), helps to distinguish them.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The TdS equations relate the change in entropy (S) to other state variables like temperature (T), volume (V), and pressure (P). They are derived from the first law of thermodynamics combined with Maxwell's relations. The "first" TdS equation expresses dS in terms of dT and dV.
Step 2: Key Formula or Approach:
We start by considering entropy S as a function of temperature T and volume V, i.e., \( S(T, V) \). The total differential is:
\[ dS = \left(\frac{\partial S}{\partial T}\right)_V dT + \left(\frac{\partial S}{\partial V}\right)_T dV \] We need to evaluate the two partial derivatives.
Step 3: Detailed Explanation:
1. Evaluate \( (\partial S / \partial T)_V \):
From the definition of heat capacity at constant volume, \( C_V = T \left(\frac{\partial S}{\partial T}\right)_V \).
Rearranging this gives: \( \left(\frac{\partial S}{\partial T}\right)_V = \frac{C_V}{T} \).
2. Evaluate \( (\partial S / \partial V)_T \):
To evaluate this term, we use a Maxwell relation. The Maxwell relation derived from the Helmholtz free energy (\(F = U - TS\)) is:
\[ \left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V \] 3. Substitute back into the dS equation:
Substitute the expressions for the partial derivatives back into the total differential for dS:
\[ dS = \left(\frac{C_V}{T}\right) dT + \left(\frac{\partial P}{\partial T}\right)_V dV \] 4. Obtain the TdS form:
Multiply the entire equation by T to get the first TdS equation:
\[ TdS = C_V dT + T \left(\frac{\partial P}{\partial T}\right)_V dV \] Step 4: Final Answer:
The correct form of the first TdS equation is \( TdS = C_V dT + T \left(\frac{\partial P}{\partial T}\right)_V dV \). This matches the expression \( C_v dT + T \left(\frac{\partial P}{\partial T}\right)_V dV \).