Page 43 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch03 Elementary Physical Chemistry

28 Elementary Physical Chemistry

3.4.2. Entropy change in (Ideal) Gas Expansion

i) At constant T, ∆U = 0. Hence,

q = −w PV = − (−P ext dV )= PdV (3.22)

= nRT dV/V = nRT ln V f /V i (3.23)

∆S = nR ln V f /V i (3.24)

ii) For variable T, ∆S = dq/T = CdT/T and if C is independent of T ,
we have at constant pressure,

∆S = C P dT/T = C p ln T f /T i (3.25)

and at constant volume

∆S = C V dT/T = C V ln T f /T i (3.26)
Example 3.2.

a) One mole of an ideal gas expands at T = 298 K from 24.79 L to 49.58 L
in a reversible process. Calculate the change in entropy.

Solution
∆S = R ln V f /V i =8.3145 J K −1 mol −1 × ln 2

=5.76J K −1 mol −1
b) One mole of an ideal gas expands at T = 298 K into a vacuum from an
initial volume of 24.79 L to a final volume of 49.58 L. Calculate ∆S.

Solution
Here, w =0(since P ext =0) and so q = 0. One cannot use dS =dq/T
because q is not q rev . However, we know that the initial and final volumes
and temperatures are the same as in Part (a). The entropy, being a state
function depends only on the initial and final states, which are known from
Part (b). Thus the value of ∆S is the same as in Part (a).