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

Concept: Free expansion describes gas expanding into an absolute vacuum environment. Because there is no opposing external pressure (\(P_{\text{ext}} = 0\)), the mechanical work performed by the system is zero: \[ W = -P_{\text{ext}}\Delta V = 0 \] For an ideal gas, internal energy (\(U\)) depends purely on temperature. Given that the process is explicitly isothermal (\(T = 300\text{ K} = \text{constant}\)), the change in internal energy must be zero: \[ \Delta U = nC_v\Delta T = 0 \] Applying the First Law of Thermodynamics (\(\Delta U = Q + W\)), we find that the heat exchanged with the surroundings is also zero: \[ 0 = Q + 0 \quad \Rightarrow \quad Q = 0 \]

Step 1: Calculating \(\Delta S_{\text{surroundings}}\)
Entropy changes in the surroundings are governed exclusively by the actual heat transferred across the boundary in a reversible framework, given by: \[ \Delta S_{\text{surroundings}} = \frac{Q_{\text{surr}}}{T} = \frac{-Q_{\text{actual}}}{T} \] Since the heat exchanged during a free expansion process is exactly \(Q_{\text{actual}} = 0\), no thermal energy enters or leaves the surroundings: \[ \Delta S_{\text{surroundings}} = \frac{0}{300} = 0 \]

Step 2: Calculating \(\Delta S_{\text{system}}\)
Entropy is a fundamental state function. Even though the process occurs irreversibly, we can calculate the entropy change of the system by integrating along an equivalent reversible isothermal pathway connecting the exact same initial and final states: \[ \Delta S_{\text{system}} = nR\ln\left(\frac{V_2}{V_1}\right) + nC_v\ln\left(\frac{T_2}{T_1}\right) \] Since temperature is invariant (\(T_1 = T_2 = 300\text{ K}\)), the temperature component drops to zero: \[ \Delta S_{\text{system}} = nR\ln\left(\frac{V_2}{V_1}\right) \] Converting the natural logarithm into a base-10 logarithm (\(\ln x \approx 2.303 \log_{10} x\)): \[ \Delta S_{\text{system}} = 2.303 \cdot n \cdot R \cdot \log_{10}\left(\frac{V_2}{V_1}\right) \]

Step 3: Substituting the numerical values into the system equation
We are provided with:

• Number of moles, \(n = 2\)

• Initial volume, \(V_1 = 10\text{ L}\)

• Final volume, \(V_2 = 100\text{ L}\)
Plugging these parameters into our formulated expression gives: \[ \Delta S_{\text{system}} = 2.303 \times 2 \times R \times \log_{10}\left(\frac{100}{10}\right) \] \[ \Delta S_{\text{system}} = 4.606 \times R \times \log_{10}(10) \] Knowing that \(\log_{10}(10) = 1\): \[ \Delta S_{\text{system}} = 4.606 \cdot R \cdot 1 = 4.606\text{ R} \] Thus, we determine that \(\Delta S_{\text{system}} = 4.606\text{ R}\) and \(\Delta S_{\text{surroundings}} = 0\).