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What is the entropy change associated with the expansion of one mole of an ideal gas from an initial volume of v to a final volume of v of 2.50v at constant temperature?

Respuesta :

Answer:

ΔS = 7.618 J/K

Explanation:

The entropy change associated to an isothermal process is:

ΔS = [tex]\frac{Q}{T}[/tex]

We can calculate Q (heat) how:

Q = nRTln([tex]\frac{Vf}{Vi}[/tex])

   Where Vf = 2.5v

               Vi = v

               R = 8.314J/mol.k  (Constant of ideal gases)

               n: Number of moles = 1 mol

Then

Q = nRTln(2.5)

ΔS = [tex]\frac{nRTln(2.5)}{T}[/tex]

ΔS = (1)(8.314)ln(2.5)

ΔS = 7.618 J/K

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