Question

Two moles of an ideal gas undergo free expansion from \( 10\text{ L} \) to \( 100\text{ L} \) at \( 300\text{ K} \). The values of \( \Delta S_{\text{system}} \) and \( \Delta S_{\text{surroundings}} \) are (\(R\) is universal gas constant)

• A. \( \Delta S_{\text{system}} = 4.606\text{ R}; \, \Delta S_{\text{surroundings}} = 0 \)
• B. \( \Delta S_{\text{system}} = 0; \, \Delta S_{\text{surroundings}} = 0 \)
• C. \( \Delta S_{\text{system}} = 4.606\text{ R}; \, \Delta S_{\text{surroundings}} = -4.606\text{ R} \)
• D. \( \Delta S_{\text{system}} = 0; \, \Delta S_{\text{surroundings}} = 4.606\text{ R} \)

Hint:

For Free Expansion into a vacuum (\(P_{\text{ext}} = 0\)) of an ideal gas: - Work (\(W\)) = 0, Heat (\(Q\)) = 0, Temperature change (\(\Delta T\)) = 0. - Since \(Q = 0\), \(\Delta S_{\text{surroundings}}\) is always zero. - \(\Delta S_{\text{system}}\) depends purely on the volume ratio change: \(nR\ln(V_2/V_1)\).

Answer 1

The Correct Option is A

Step 1: Understand free expansion.
The gas expands into vacuum, so the external pressure is zero. The work done is \[ W = -P_{ext}\Delta V = 0 \] No work is done by or on the gas.
Step 2: Find the heat exchanged.
The process is isothermal at \(300\ K\), so for an ideal gas \(\Delta U = 0\). By the first law \(\Delta U = Q + W\), with \(W=0\) we get \(Q = 0\): no heat flows to or from the surroundings.
Step 3: Entropy of the surroundings.
Since the surroundings exchange no heat, \[ \Delta S_{surr} = \frac{Q_{surr}}{T} = 0 \]
Step 4: Entropy of the system.
Entropy is a state function, so we use the isothermal expansion formula \[ \Delta S_{sys} = nR\ln\frac{V_2}{V_1} \]
Step 5: Put in the numbers.
With \(n = 2\), \(V_2/V_1 = 100/10 = 10\): \[ \Delta S_{sys} = 2R\ln 10 = 2R(2.303) = 4.606\,R \]
Step 6: State the result.
So \(\Delta S_{sys} = 4.606\,R\) and \(\Delta S_{surr} = 0\), which is option 1.
\[ \boxed{\Delta S_{sys}=4.606\,R,\ \Delta S_{surr}=0} \]