Question

Physical Chemistry

118.0 g acetylene (ethyne) exists as a liquid T=-84C which is the liquids boiling point. The liquid boils. The resulting gas (ideal) is heated to 10C and is isothermally compressed from 200.0 L to 125.0Lwgst is the total deltaS for this process

Heat of vaporization =16.7 kJ/mole

Answer 1

Given:-

weight of ethyne (C₂H₂)= 118.0 g

molar mass of ethyne (C₂H₂) (m) = 26 g / mol

boiling point of ethyne(C₂H₂) = - 84⁰C = - 84 + 273 = 189 K

initial volume of ethyne(C₂H₂) (V₁) = 200.0 L

final volume of ethyne(C₂H₂) (V₂) = 125.0 L

heat of vaporization of ethyne(C₂H₂) (H_v) = 16.7 KJ /mol = 16700 J /mol

gas constant (R) =8.314 JK^-1mol^-1

As we know that

no. of moles of  of ethyne(C₂H₂) (n) = weight of ethyne (C₂H₂) / molar mass of ethyne (C₂H₂) (m)

no. of moles of  of ethyne(C₂H₂) (n) = 118.0 / 26.0 g / mol

no. of moles of  of ethyne(C₂H₂) (n) = 4.54 mol

Also we know that

1 mole of  ethyne(C₂H₂) has heat of vaporization =  16700 J /mol

4.54 mole of  ethyne(C₂H₂) has heat of vaporization = 4 16700 J /mol

4.54 mole of  ethyne(C₂H₂) has heat of vaporization = 66800 J /mol

therefore

entropy of vaporization (S_v) = heat of vaporization of 4.54 mole of  ethyne(C₂H₂)(H_v) / boiling point of ethyne(C₂H₂) (T)

entropy of vaporization (S_v) = 66800 J mol^-1 / 189 K

entropy of vaporization (S_v) = 353.44 J mol^-1 K^-1

We also know that entropy change at isothermal process

entropy change (S) = 2.303nRlog(V₂ /V₁)

entropy change (S) = 2.303 4.54 mol 8.314 JK^-1mol^-1log(125.0 L / 200.0 L)

entropy change (S) = 86.928 JK^-1 log(0.625)

entropy change (S) = 86.928 JK^-1 - 0.20412

entropy change (S) = - 17.744 JK^-1

therefore

total entropy change  (S_total) = entropy of vaporization (S_v) + entropy change (S)

total entropy change  (S_total) = 353.44 J mol^-1 K^-1 +  ( - 17.744 JK^-1 )

total entropy change  (S_total) = 353.44 J mol^-1 K^-1 - 17.744 JK^-1

total entropy change  (S_total) = 335.696 J mol^-1 K^-1 (i.e the answer)