Standard enthalpy of vapourisation Δ text vap H Θ for water at 100° C is 40.66 kJ mol -1. The internal energy of vaporisation of water at 100° C (in kJmol -1) is: (Assume water vapour to behave like an ideal gas)

Question

Standard enthalpy of vapourisation Δ text vap H Θ for water at 100° C is 40.66 kJ mol -1. The internal energy of vaporisation of water at 100° C (in kJmol -1) is: (Assume water vapour to behave like an ideal gas) (A) +43.76 (B) +40.66 (C) +37.56 (D) -43.76

Tardigrade Admin · Accepted Answer

H 2 O (1) arrow H 2 O ( g ) Δ H =Δ E +Δ nRT 40.66=Δ E +1 × (8.314/1000) × 373 Δ E =37.5 kJ

Q. Standard enthalpy of vapourisation $Δ_{vap} H^{Θ}$ for water at $10 0^{\circ} C$ is $40.66 k J m o l^{- 1}$ . The internal energy of vaporisation of water at $10 0^{\circ} C$ (in $k J m o l^{- 1}$ ) is :
(Assume water vapour to behave like an ideal gas)

+43.76

+40.66

+37.56

-43.76

$H_{2} O_{(1)} \to H_{2} O_{(g)}$

$Δ H = Δ E + Δ n RT$

$40.66 = Δ E + 1 \times \frac{8.314}{1000} \times 373$

$Δ E = 37.5 k J$

Q. Standard enthalpy of vapourisation Δvap ​HΘ for water at 100∘C is 40.66kJmol−1. The internal energy of vaporisation of water at 100∘C (in kJmol−1) is : (Assume water vapour to behave like an ideal gas)