Question

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?

Answer 1

Answer:

ΔS = 7.618 J/K

Explanation:

The entropy change associated to an isothermal process is:

ΔS = $\frac{Q}{T}$

We can calculate Q (heat) how:

Q = nRTln($\frac{Vf}{Vi}$)

   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 = $\frac{nRTln(2.5)}{T}$

ΔS = (1)(8.314)ln(2.5)

ΔS = 7.618 J/K