If water vapour is assumed to be an ideal gas and the molar enthalpy change for vaporization of one mole of water at 1 bar and 100° C is 41 kJ / mol. The internal energy change when 1 mol of water is vaporised

Question

If water vapour is assumed to be an ideal gas and the molar enthalpy change for vaporization of one mole of water at 1 bar and 100° C is 41 kJ / mol. The internal energy change when 1 mol of water is vaporised (A) 37.904 kJ / mol (B) 35 kJ / mol (C) 40.5 kJ / mol (D) 39 kJ / mol

Tardigrade Admin · Accepted Answer

H 2 O (l) arrow H 2 O (g) ; Δ H=41 kJ / mol Δ H=Δ E+Δ ng R T ; Δ E =Δ H-Δ ng R T Δ ng=1-0=1, R=8.314, T=373 K Δ E=41-(1 × 8.314 × 373 × 10-3) =41-3.1=37.9 kJ / mol

Q. If water vapour is assumed to be an ideal gas and the molar enthalpy change for vaporization of one mole of water at $1$ bar and $10 0^{\circ} C$ is $41 k J / m o l$ . The internal energy change when $1$ mol of water is vaporised

$37.904 k J / m o l$

$35 k J / m o l$

$40.5 k J / m o l$

$39 k J / m o l$

$H_{2} O_{(l)} \to H_{2} O_{(g)}; Δ H = 41 k J / m o l$

$Δ H = Δ E + Δ n_{g} RT; Δ E$

$= Δ H - Δ n_{g} RT$

$Δ n_{g} = 1 - 0 = 1, R = 8.314, T = 373 K$

$Δ E = 41 - (1 \times 8.314 \times 373 \times 1 0^{- 3})$

$= 41 - 3.1 = 37.9 k J / m o l$

Q. If water vapour is assumed to be an ideal gas and the molar enthalpy change for vaporization of one mole of water at 1 bar and 100∘C is 41kJ/mol. The internal energy change when 1 mol of water is vaporised

37.904kJ/mol

35kJ/mol

40.5kJ/mol

39kJ/mol