9.0 , g of H_2O is vaporized at 100 , ^circ C | vedclass

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(D) The change in entropy for a phase transition is given by the formula: $\Delta S = \frac{\Delta H_{vap}}{T_b}$ Given: Mass of $H_2O = 9.0 \, g$ Lat

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$\frac{1}{2} 	imes \frac{18x}{373}$

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(D) The change in entropy for a phase transition is given by the formula: $\Delta S = \frac{\Delta H_{vap}}{T_b}$ Given: Mass of $H_2O = 9.0 \, g$ Latent heat of vaporization $= x \, J / g$ Total enthalpy of vaporization $\Delta H_{vap} = 9.0 \, g 	imes x \, J / g = 9x \, J$ Boiling point $T_b = 100 \, ^\circ C = 100 + 273 = 373 \, K$ Substituting these values into the formula: $\Delta S = \frac{9x}{373} \, J / K$ Since $\frac{9}{373} = \frac{1}{2} 	imes \frac{18}{373}$,we can write: $\Delta S = \frac{1}{2} 	imes \frac{18x}{373} \, J / K$ Thus,the correct option is $D$.

$9.0 \, g$ of $H_2O$ is vaporized at $100 \, ^\circ C$ and $1 \, atm$ pressure. If the latent heat of vaporization of water is $x \, J / g$,then $\Delta S$ is given by :-

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