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Aakshi Gautam posted an Question
December 18, 2019 • 04:41 am 1 points
  • IIT JAM
  • Chemistry (CY)

I this due to expansion volume is changed but we applied here internal energy formula which is for constant volume but here is volume is change due to expansio

Th Va}i are either T and V or T and P (i) When T & V are Two Variables according to first law of thermodynamics q= dE + W for compression (VJ) dE - W for expansion (Vt) If the work involved is due to expansion (Vt) of the aac volume dV again a Pressure P -W = P.dV = dE + P.dV So The increase in entropy of the gas for an infinitesimallv small dS= T ds dE +P.dv So T Substituting the value of dE = C.dT and PRT V we have dS-,dT I RT dS- TV T dV V dS-C R Making the assumption that Cv is independent of temperature get AS-C, In + Rin V for n moles of the ideal gases, the above equation may be AS nc, In 2 + nRIn2 T V. It is evident that the entropy change for the change of state c initial and final volumes as well as on the initial and final temperature Contact Us: Webeite www.eduncle.com Email support@eduncle.com

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    Eduncle Best Answer

    Dear student,

    Greetings!!

    According to the first law of thermodynamic :

    q = dE + W (for expansion V ­increase)

    W = -pdV ( for compression V  decrease)

    If V is constant then,

    dV = 0

    So q = dE (W = 0)

    So the formula given in this derivation is correct.

    Thank you for asking your query.

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