Question

The change in entropy involved in heating a gram-atomic weight of silver at constant volume from 0°C to 30°C is ? [Given: C v = 5.85 cal deg -1 mol -1 ]

Answer 1

We have to find change in entropy involved in hearing a gram - atomic weight of silver at constant volume from 0°C to 30°C.

To find : one gram atomic weight of silver means 1 mole of silver atom.

Entropy change , ∫dS = ∫dQ/T

we know, dQ = nCpdT

so, ∆S = ∫nCpdT/T = nCp ∫dT/T

= nCp ln[T₂/T₁]

here n = 1 , Cp = Cv + R = 5.85 + 2 = 7.85 cal/°C/mol , T₂ = 30° = 303K and T₁ = 0°C = 273K

now ∆S = 1 × 7.85 × ln[303/273]

= 7.85 × 0.10426

≈ 0.821 cal/°C

Therefore change in entropy would be 0.821 Cal/°C