2 mole of an ideal gas at $27^\circ ext{C}$ temp. is expanded reversibly from $2 ext{ lit.}$ to $20 ext{ li

Question

2 mole of an ideal gas at $27^\circ	ext{C}$ temp. is expanded reversibly from $2	ext{ lit.}$ to $20	ext{ lit.}$ Find entropy change ($R = 2	ext{ cal/mol K}$):

Accepted Answer

1. **Identify the Process:** The expansion is reversible. Given a single temperature and volume change, we calculate the entropy change for the system using the isothermal expansion formula.
2. **Formula:** The change in entropy ($\Delta S$) for an ideal gas undergoing isothermal expansion is given by:
   $$ \Delta S = nR \ln\left(\frac{V_2}{V_1}ight) $$
3. **Substitute Values:**
   - Number of moles ($n$) = $2	ext{ mol}$
   - Gas Constant ($R$) = $2	ext{ cal/mol K}$
   - Initial Volume ($V_1$) = $2	ext{ lit}$
   - Final Volume ($V_2$) = $20	ext{ lit}$
   $$ \Delta S = 2 	imes 2 	imes \ln\left(\frac{20}{2}ight) $$
   $$ \Delta S = 4 	imes \ln(10) $$
4. **Calculate:**
   Convert natural log ($\ln$) to base-10 log ($\log$): $\ln(10) = 2.303 \log_{10}(10) = 2.303$.
   $$ \Delta S = 4 	imes 2.303 = 9.212	ext{ cal/K} $$
   Rounding to one decimal place gives $9.2$.

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