Question
Ethanol has a heat of vaporization of 38.56 kJ>mol and a normal boiling point of 78.4 C. What is the vapor pressure of ethanol at 15 C?
Step 1
So we can write: \[P_1 = 760 \, \text{torr}\] \[T_1 = 78.4 \, ^\circ \text{C} = 78.4 + 273.15 = 351.55 \, \text{K}\] where \(P_1\) is the vapor pressure at temperature \(T_1\). Show more…
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Ethanol has a heat of vaporization of 38.56 $\mathrm{kJ} / \mathrm{mol}$ and a normal boiling point of $78.4^{\circ} \mathrm{C}$ . What is the vapor pressure of ethanol at $15^{\circ} \mathrm{C} ?$
The vapor pressure of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, at $50.0{ }^{\circ} \mathrm{C}$ is $233 \mathrm{~mm} \mathrm{Hg}$, and its normal boiling point at $1 \mathrm{~atm}$ is $78.3^{\circ} \mathrm{C}$. What is the $\Delta H_{\text {vap }}$ of ethanol?
The vapor pressure of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, at $50.0{ }^{\circ} \mathrm{C}$ is $233 \mathrm{~mm} \mathrm{Hg}$, and its normal boiling point at $1 \mathrm{~atm}$ is $78.3^{\circ} \mathrm{C} .$ Calculate the $\Delta H_{\mathrm{vap}}$ of ethanol.
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