Question

An ideal gas in equilibrium at temperature T expands isothermally to twice its initial volume. If ΔS, ΔU and ΔF denote the changes in its entropy, internal energy and Helmholtz free energy respectively, then

• A. ΔS < 0, ΔU > 0, ΔF < 0
• B. ΔS > 0, ΔU = 0, ΔF < 0
• C. ΔS < 0, ΔU = 0, ΔF > 0
• D. ΔS > 0, ΔU > 0, ΔF = 0

Answer 1

The Correct Option is B

To solve this problem related to the thermodynamics of an ideal gas, let's consider the concepts involved:

Isothermal Process: An isothermal process is one in which the temperature \(T\) of the system remains constant. For an ideal gas undergoing an isothermal process, the internal energy change \(\Delta U\) is zero because internal energy of an ideal gas depends only on temperature, which is unchanged. Hence, \(\Delta U = 0\).

Entropy Change (\(\Delta S\)): The entropy change for an ideal gas undergoing an isothermal expansion from volume \(V_i\) to volume \(V_f = 2V_i\) is given by:

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

Since \(V_f = 2V_i\), this becomes \(\Delta S = nR \ln 2\). Since \( \ln 2 > 0 \), it follows that \(\Delta S > 0\).

Helmholtz Free Energy Change (\(\Delta F\)): The Helmholtz free energy change is defined as:

\(\Delta F = \Delta U - T\Delta S\)

With \(\Delta U = 0\), the equation simplifies to \(\Delta F = -T\Delta S\). Since \(\Delta S > 0\), it follows that \(\Delta F < 0\).

Therefore, the correct statements are: \(\Delta S > 0\), \(\Delta U = 0\), and \(\Delta F < 0\).

Thus, the correct option is:

\(\Delta S > 0\), \(\Delta U = 0\), \(\Delta F < 0\)