The entropy of vaporization of a liquid is 58 JK-1 mol-1. If 100 g of its vapour condenses at its boiling point of 123°C, the value of entropy change for the process is (Molar mass of the liquid = 58 g mol-1)

Question

The entropy of vaporization of a liquid is 58 JK-1 mol-1. If 100 g of its vapour condenses at its boiling point of 123°C, the value of entropy change for the process is (Molar mass of the liquid = 58 g mol-1) (A)  - 100 JK-1 (B)  100 JK-1 (C)  - 123 JK-1 (D)  123 JK-1

Tardigrade Admin · Accepted Answer

Given entropy of vaporisation =58 J K -1 mol -1 i.e., entropy change when 1 mole of the liquid vapourises =58 JK -1 ∴ Entropy change when 1 mole of the liguid condenes =-58 JK -1 (∵ Condensation and vaporisation are opposite process). Moles of the liquid used in the reaction =(100/58)=1.72 mol ∴ Entropy change when 1.72 mole (or 100 g ) of the liquid condenses =(-58 × 1.72/1) =-100 JK -1

Q. The entropy of vaporization of a liquid is $58 J K^{- 1} m o l^{- 1}$ . If $100 g$ of its vapour condenses at its boiling point of $12 3^{\circ} C$ , the value of entropy change for the process is
(Molar mass of the liquid = $58 g m o l^{- 1})$

$- 100 J K^{- 1}$

$100 J K^{- 1}$

$- 123 J K^{- 1}$

$123 J K^{- 1}$

$1230 J K^{- 1}$

Q. The entropy of vaporization of a liquid is 58JK−1mol−1. If 100g of its vapour condenses at its boiling point of 123∘C, the value of entropy change for the process is (Molar mass of the liquid = 58gmol−1)

−100JK−1

100JK−1

−123JK−1

123JK−1