Question

Change of entropy of a perfect gas for isochoric process is

• A. \( C_v \log_e \frac{V_2}{V_1} \)
• B. \( C_v \log_e \frac{P_2}{P_1} \)
• C. \( C_p \log_e \frac{V_2}{V_1} \)
• D. \( (C_p - C_v) \log_e \left(\frac{V_2}{V_1}\right) \)

Hint:

You can derive the entropy change formulas for all basic processes (isothermal, isobaric, isochoric) from the general expression \( \Delta S = C_V \ln(T_2/T_1) + R \ln(V_2/V_1) \) for 1 mole of an ideal gas.
{Isochoric (V=const):} The second term is zero, \( \Delta S = C_V \ln(T_2/T_1) = C_V \ln(P_2/P_1) \).
{Isobaric (P=const):} \( \Delta S = C_P \ln(T_2/T_1) = C_P \ln(V_2/V_1) \).
{Isothermal (T=const):} The first term is zero, \( \Delta S = R \ln(V_2/V_1) \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Entropy is a measure of the disorder or randomness of a system. The change in entropy (\(\Delta S\)) for a reversible process is defined as \( dS = dQ_{rev}/T \). We need to find the expression for the total entropy change for an isochoric (constant volume) process involving a perfect gas.
Step 2: Key Formula or Approach:
From the first law of thermodynamics, the heat added to a system is \( dQ = dU + dW \).
For a perfect gas, the change in internal energy is \( dU = nC_V dT \) and the work done is \( dW = P dV \).
Therefore, for a reversible process:
\[ TdS = nC_V dT + P dV \] Dividing by T, we get the general expression for entropy change:
\[ dS = nC_V \frac{dT}{T} + \frac{P}{T} dV \] We will apply the condition for an isochoric process to this equation.
Step 3: Detailed Explanation:
An isochoric process is one that occurs at constant volume.
This means \( V_2 = V_1 \), and the change in volume \( dV = 0 \).
Substituting \( dV = 0 \) into the general entropy equation:
\[ dS = nC_V \frac{dT}{T} + 0 \] To find the total change in entropy from state 1 to state 2, we integrate this expression:
\[ \Delta S = \int_{T_1}^{T_2} nC_V \frac{dT}{T} = nC_V \ln\left(\frac{T_2}{T_1}\right) \] For a perfect gas at constant volume, the pressure and temperature are related by Gay-Lussac's Law:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \implies \frac{T_2}{T_1} = \frac{P_2}{P_1} \] Substituting this into our entropy equation (assuming 1 mole, or that \(C_V\) represents the total heat capacity):
\[ \Delta S = C_V \ln\left(\frac{P_2}{P_1}\right) \] Step 4: Final Answer:
The change of entropy for a perfect gas during an isochoric process is \( C_V \log_e \frac{P_2}{P_1} \).