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Chemistry · JEE | Thermodynamics (Chemical Thermodynamics)

हिन्दी

For an isothermal reversible expansion of \( n \) moles of an ideal gas from initial volume \( V_i \) to final volume \( V_f \), the work done (\( w_{rev} \)) is given by:

  1. A
    \( w_{rev} = -2.303 nRT \log (\frac{V_f}{V_i}) \)
  2. B
    \( w_{rev} = -p_{ex}(V_f - V_i) \)
  3. C
    \( w_{rev} = nRT \ln(V_i V_f) \)
  4. D
    \( w_{rev} = -2.303 RT \log (\frac{V_i}{n} ) \)
View Answer & Solution

Correct answer · Option A

\( w_{rev} = -2.303 nRT \log (\frac{V_f}{V_i}) \)

Explanation

Integrating \( -p dV \) using \( p = \frac{nRT}{V} \) yields \( w_{rev} = -2.303 nRT \log (\frac{V_f}{V_i}) \).
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