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: Concept Definition:
Entropy quantifies system disorder. For a reversible process, the entropy change is \( dS = dQ_{rev}/T \). We aim to determine the total entropy change for a perfect gas undergoing an isochoric (constant volume) process.
Step 2: Fundamental Equations:
The first law of thermodynamics states \( dQ = dU + dW \). For a perfect gas, \( dU = nC_V dT \) and \( dW = P dV \). Thus, for a reversible process:
\[ TdS = nC_V dT + P dV \]Dividing by T yields the general entropy change expression:
\[ dS = nC_V \frac{dT}{T} + \frac{P}{T} dV \]We will apply the isochoric condition to this equation.
Step 3: Derivation:
An isochoric process maintains constant volume (\( dV = 0 \)).
Substituting \( dV = 0 \) into the general entropy equation:
\[ dS = nC_V \frac{dT}{T} + 0 \]Integrating from state 1 to state 2 gives the total entropy change:
\[ \Delta S = \int_{T_1}^{T_2} nC_V \frac{dT}{T} = nC_V \ln\left(\frac{T_2}{T_1}\right) \]According to Gay-Lussac's Law for a perfect gas at constant volume, \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \), which implies \( \frac{T_2}{T_1} = \frac{P_2}{P_1} \).
Substituting this into the entropy equation (assuming 1 mole, or \(C_V\) as total heat capacity):
\[ \Delta S = C_V \ln\left(\frac{P_2}{P_1}\right) \]Step 4: Concluding Result:
The entropy change for a perfect gas in an isochoric process is \( C_V \log_e \frac{P_2}{P_1} \).