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

ΔS = 7.618 J/K

Step-by-step explanation:

The entropy change associated to an isothermal process is:

ΔS =
`(Q)/(T)`

We can calculate Q (heat) how:

Q = nRTln(
`(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 =
`(nRTln(2.5))/(T)`

ΔS = (1)(8.314)ln(2.5)

ΔS = 7.618 J/K