Question

One mole of an ideal monatomic gas at absolute temperature \( T \) undergoes free expansion to double its original volume, so that the entropy change is \( \Delta S_1 \). An identical amount of the same gas at absolute temperature \( 2T \) undergoes isothermal expansion to double its original volume, so that the entropy change is \( \Delta S_2 \). The value of \( \frac{\Delta S_1}{\Delta S_2} \) (in integer) is:

Hint:

For entropy changes during volume expansion, remember that for an ideal gas undergoing free expansion, the entropy change is calculated using \( nR \ln \left( \frac{V_f}{V_i} \right) \), and for isothermal expansion, it follows the same formula.

Answer 1

To solve this, we need to calculate the entropy changes for both the free expansion and the isothermal expansion. 
Step 1: Entropy Change for Free Expansion (\( \Delta S_1 \)) 
For free expansion, the process is irreversible, and since the gas expands into a vacuum, there is no heat exchange with the surroundings. Therefore, the entropy change \( \Delta S_1 \) for the gas during free expansion is given by: \[ \Delta S_1 = nR \ln \left( \frac{V_f}{V_i} \right), \] where \( V_f \) and \( V_i \) are the final and initial volumes of the gas, respectively, and \( n \) is the number of moles. Since the volume doubles, \( \frac{V_f}{V_i} = 2 \), and we get: \[ \Delta S_1 = nR \ln(2). \] Step 2: Entropy Change for Isothermal Expansion (\( \Delta S_2 \)) 
For the isothermal expansion, the process is reversible, and the gas is at a constant temperature. The entropy change \( \Delta S_2 \) for the gas during an isothermal expansion is given by: \[ \Delta S_2 = nR \ln \left( \frac{V_f}{V_i} \right), \] where \( V_f \) and \( V_i \) are the final and initial volumes of the gas, respectively. Since the volume also doubles in this case, \( \frac{V_f}{V_i} = 2 \), and we get: \[ \Delta S_2 = nR \ln(2). \] Step 3: Comparing the Two Entropy Changes. 
We are asked to find the ratio \( \frac{\Delta S_1}{\Delta S_2} \). Since both entropy changes are equal: \[ \frac{\Delta S_1}{\Delta S_2} = \frac{nR \ln(2)}{nR \ln(2)} = 1. \] Thus, the value of \( \frac{\Delta S_1}{\Delta S_2} \) is \( 1 \).