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NEET CHEMISTRYMedium

2 mole of an ideal gas at 27∘C27^\circ\text{C} temp. is expanded reversibly from 2 lit.2\text{ lit.} to 20 lit.20\text{ lit.} Find entropy change (R=2 cal/mol KR = 2\text{ cal/mol K}):

A

92.1

B

0

C

4

D

9.2

Step-by-Step Solution

  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 (ΔS\Delta S) for an ideal gas undergoing isothermal expansion is given by: ΔS=nRln⁡(V2V1)\Delta S = nR \ln\left(\frac{V_2}{V_1}\right)
  3. Substitute Values:
  • Number of moles (nn) = 2 mol2\text{ mol}
  • Gas Constant (RR) = 2 cal/mol K2\text{ cal/mol K}
  • Initial Volume (V1V_1) = 2 lit2\text{ lit}
  • Final Volume (V2V_2) = 20 lit20\text{ lit} ΔS=2×2×ln⁡(202)\Delta S = 2 \times 2 \times \ln\left(\frac{20}{2}\right) ΔS=4×ln⁡(10)\Delta S = 4 \times \ln(10)
  1. Calculate: Convert natural log (ln⁡\ln) to base-10 log (log⁡\log): ln⁡(10)=2.303log⁡10(10)=2.303\ln(10) = 2.303 \log_{10}(10) = 2.303. ΔS=4×2.303=9.212 cal/K\Delta S = 4 \times 2.303 = 9.212\text{ cal/K} Rounding to one decimal place gives 9.29.2.
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