(0)
TU
48. Two moles of an ideal gas undergo reversible
isothermal expansion from 2,303 L to 23.03 L at
27°C. The change in entropy is
(R = 8.314 J mol-1K-1)
(1) 38.3 J K-1
2.303
(2) 16.6 JK-1
28.03
4374.6 JK-
1 12
(4) 4.6 J K-1-
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answer : option (1) 38.3 J/K
in isothermal process, change in internal energy , ∆U = 0
so from 1st law of thermodynamics,
Q = ∆U + W ⇒Q = 0 + W
⇒Q = W
as W = -P_(ex)dV = nRT ∫dV/V
= nRT ln[V1/V2]
so, Q = W = nRTln[V1/V2]
entropy , S = Q/T = nRln[V1/V2]
here, n = 2 mol , R = 8.314 J/mol.K, T = 27° C = 300K , V1 = 2.303l and V2 = 23.03l
now, S = 2 × 8.314 ln[23.03/2.303]
= ln(10)
= 16.628 ln(10)
= 16.628 × 2.303
= 38.294 ≈ 38.3 J/K
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