Question

Calculate the entropy change when Argon (C_v,m = 3/2 R) at 25ºC and 1 atmis heated to   89 ∘C    in a closed container of volume 500 cm^3 (constant volume)

Answer 1

Formula to calculate entropy change for 1 mole of gas n=1,

∆S = [Cv ln T2/T1 + R ln (V2/V1)]

At constant volume V2 = V1 V2 / V1 = 1 ln(V2/V1) = 0

Hence formula reduced to,

∆S = [Cv ln (T2/T1)]

Given data:

R = 8.314 J/K.mole

Cv = (3/2) R = 1.5 R = 1.5 x 8.314

T2 = 89 oC = 89 + 273 = 362 K

T1 = 25 oC = 25 + 273 = 298 K

Using these values in above formula we write,

∆S = 1.5 x 8.314 ln (362/298)

∆S = 2.43 JK^-1.

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