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Standard enthalpy of vapourisation $\Delta_{vap}H^\ominus$ for water at $100^\circ C$ is 40.66 kJ $mol^{-1}$. The internal energy of vapourisation of water at $100^\circ C$ (in kJ $mol^{-1}$) is
A
+40.66
B
+37.56
C
−43.76
D
+43.76
Detailed Solution
$H_2O(l) \rightarrow H_2O(g)$
$V_1$ of liquid water = 18 mL
$V_2$ of water vapour $= \frac{nRT}{P} = \frac{1\times0.0821\times373}{1} = 30.6233$ L
$\Delta V = 30623.3 - 18 = 30605.3$ mL = 30.605 L
$\Delta E = \Delta H - P\Delta V = 40.66\times10^3$ J $- 1\times30.605\times101.325$ J
$= 40660 - 3101 = 37559$ J ≈ 37.56 kJ $mol^{-1}$
(Equivalently, $\Delta U = \Delta H - \Delta n_gRT = 40.66 - 1\times8.314\times10^{-3}\times373 = 37.56$ kJ $mol^{-1}$.)
$V_1$ of liquid water = 18 mL
$V_2$ of water vapour $= \frac{nRT}{P} = \frac{1\times0.0821\times373}{1} = 30.6233$ L
$\Delta V = 30623.3 - 18 = 30605.3$ mL = 30.605 L
$\Delta E = \Delta H - P\Delta V = 40.66\times10^3$ J $- 1\times30.605\times101.325$ J
$= 40660 - 3101 = 37559$ J ≈ 37.56 kJ $mol^{-1}$
(Equivalently, $\Delta U = \Delta H - \Delta n_gRT = 40.66 - 1\times8.314\times10^{-3}\times373 = 37.56$ kJ $mol^{-1}$.)