Question

For a sample of perfect gas when its pressure is changed isothermally from $p_i$ to $p_f$, the entropy change is given by

• A. $\Delta S = nR \, ln \bigg( \frac{P_f}{p_i}\bigg)$
• B. $\Delta S = nR \, ln \bigg( \frac{p_i}{p_f}\bigg)$
• C. $\Delta S = nRT\, ln \bigg( \frac{p_f}{p_i}\bigg)$
• D. $\Delta S = RT \, ln \bigg( \frac{p_i}{p_f}\bigg)$

Answer 1

The Correct Option is B

 To determine the entropy change \((\Delta S)\) of a perfect gas undergoing an isothermal process where the pressure changes from \(p_i\) to \(p_f\), let's explore the basic principles involved.

For an isothermal process, the process occurs at a constant temperature. The entropy change for an ideal gas under an isothermal condition is given as:

• \(\Delta S = nR \, \ln \bigg( \frac{V_f}{V_i}\bigg)\)

Where:

• \(n\) is the number of moles,
• \(R\) is the universal gas constant,
• \(V_i and\)

Using the ideal gas law, \(PV = nRT\), we can substitute \(V\) in terms of \(P\):

• \(V_i = \frac{nRT}{p_i}\) and \(V_f = \frac{nRT}{p_f}\)

Substituting back into the entropy change equation, we get:

• \(\Delta S = nR \, \ln \bigg( \frac{\frac{nRT}{p_f}}{\frac{nRT}{p_i}} \bigg)\)

This simplifies to:

• \(\Delta S = nR \, \ln \bigg( \frac{p_i}{p_f} \bigg)\)

Hence, the correct answer is:

\(\Delta S = nR \, \ln \bigg( \frac{p_i}{p_f} \bigg)\)

This option represents the correct formula for the change in entropy for an isothermal process in an ideal gas when pressure changes from \(p_i\) to \(p_f\).