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For water at $25^{o} C, d p_{s} / d T_{s} = 0.189 k P a / K$ ( $p_{s}$ is the saturation pressure in $k P a$ an $T_{s}$ is the saturation temperature in $K$ ) and the specific volume of dry saturated vapour is $43.38 m^{3} / k g$ . Assume that the specific volume of liquid is negligible in comparison with that of vapour. Using the Clausius-Clapeyron equation, an estimate of the enthalpy of evaporation of water at $25^{o} C$ (in $k J / k g$ ) is

2443.2478

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Solution

The correct option is A 2443.2478
$At 25^{o} C, \frac{d p_{s}}{d T_{s}} = 0.189 k P a / K$

Specific volume of dry saturated vapour,

$v_{s} = 43.38 m^{3} / k g$

From Clausius-Clapeyron equation, neglecting specific volume of liquid

$(v_{2} >> v_{f})$

$\frac{d p_{s}}{d T_{s}} = \frac{h_{f g}}{T (v_{δ} - v_{δ})} = \frac{h_{f g}}{T V_{g}}$

$\frac{d p_{s}}{d T_{s}} = \frac{h_{f g}}{T_{s} \times v_{s}}$

$h_{g f} = Enthalpy of vapourisation$

$0.189 = \frac{h_{f g}}{(25 + 273) \times 43.38}$

$∴, h_{f g} = 2443.2478 k J / k g$

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