_ vapH^ ⊖ for water is +40.79 kJ mol^-1 at 1 bar and 100^ C. Change in internal energy for this vapourisation under same condition is kJmol^-1. (Integer answer) (Given R=8.3 JK^-1 mol^-1 )

Question

Accepted Answer

Correct answer: 38. (38) H_2O (l ) ⇌H_2O (g ) H_ vap^0=40.79 kJ/ mole H_ vap^0= U_ vap^0+ n_g RT40.79= U_ vap^0+ 1 8.3 373.151000 U_ vap^0=40.79-3.0971 =37.6929 U_ vap^0 ≃38

$Δ_{vap} H^{⊖} for water is + 40.79 kJ mol^{- 1} at 1 bar and 100^{\circ} C . Change in internal energy for this vapourisation under same condition is kJmol^{- 1} . (Integer answer) (Given R = 8.3 JK^{- 1} mol^{- 1})$

Solution

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