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Sagar kumar biswal Asked a Question
December 8, 2020 9:07 ampts 30 pts
Consider free expansion of one mole of an ideal gas in an adiabatic container from volume V1 to V2. The entropy change of the gas, calculated by considering a reversible process between the original state (V, T) to the final stale (V2, T) where Tis the temperature of the system, is denoted by AS1. The corresponding change in the entropy of the surrounding is AS2. Which of the following combinations is correct? (A) AS= RIn (V/V2), AS2= - Rn (V1/V2) - Rln (V,/V2), AS>= Rln (V,/V2) (B) AS = (C) AS = Rln (V2/V), AS2 = 0 (D) AS = -RIn (V2/V), AS2-
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  • Lekhika thankyou
    see attachment
    • cropped6296870660283817715.jpg
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    Sagar kumar biswal
    but mam in a reversible process the total entropy change of the universe should be zero. So I think option B is correct. Although there is no heat exchange between system and surro...
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