NEET Physics Isothermal & Adiabatic Processes MCQ Question
For an isothermal expansion of 1 mole of an ideal gas from volume V_i to volume V_f at a temperature of 300 K, which of the following correctly represents the work done by the gas?
w = -nRT ln(V_f/V_i)
w = nRT ln(V_f/V_i)
w = -2.303 nRT log(V_f/V_i)
w = 2.303 nRT log(V_f/V_i)
Correct Answer
Detailed Explanation
For isothermal reversible expansion, work done w = -2.303 nRT log(V_f/V_i) as derived in NCERT equation 5.5.
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