Question

During an isothermal expansion of an ideal gas, how does the entropy of the system change?

• A. The entropy decreases.
• B. The entropy remains constant.
• C. The entropy increases.
• D. The entropy initially increases then decreases.

Hint:

Entropy Change (Ideal Gas). \(\Delta S = nC_v \ln(T_f/T_i) + nR \ln(V_f/V_i)\). For isothermal (\(T_f=T_i\)) expansion (\(V_f>V_i\)), \(\Delta S\) is positive (entropy increases).

Answer 1

The Correct Option is C

For an ideal gas undergoing a reversible isothermal process, the change in entropy (\(\Delta S\)) is given by: $$ \Delta S = \frac{Q_{rev}}{T} $$ where \(Q_{rev}\) is the heat transferred reversibly and \(T\) is the constant absolute temperature.
From the first law, for an isothermal process (\(\Delta U = 0\)), \(Q_{rev} = W_{rev}\).
During expansion, the gas does work on the surroundings (\(W_{rev}>0\)).
To maintain constant temperature, the gas must absorb an equivalent amount of heat from the surroundings (\(Q_{rev}>0\)).
Since \(Q_{rev}\) is positive and \(T\) is positive, the entropy change \(\Delta S = Q_{rev}/T\) must be positive.
Alternatively, for an ideal gas, \(\Delta S = nC_v \ln(T_f/T_i) + nR \ln(V_f/V_i)\).
For an isothermal process, \(T_f=T_i\), so \(\ln(T_f/T_i)=0\).
For expansion, \(V_f>V_i\), so \(\ln(V_f/V_i)>0\).
Thus, \(\Delta S = nR \ln(V_f/V_i)>0\).
The entropy of the system increases during isothermal expansion.