Answer
Clausius-Clapeyron equation:
$ln\frac{P_{1}}{P_{2}} = \frac{\Delta H_{vap}}{R}(\frac{1}{T_{2}} - \frac{1}{T_{1}})$
From this equation, we can see that $T_{1}, T_{2}$ and $R$ are constant. So $\Delta H_{vap}$ will be directly proportional to $\frac{P_{1}}{P_{2}}$. The vapor pressure($P_{1}$) of liquid X is lower than vapor pressure($P_{1}$) of liquid Y at $20^{\circ}C$, but the vapor pressure ($P_{2}$) of liquid X is lower than the vapor pressure ($P_{2}$) of liquid Y at $60^{\circ}C$. The value of $\frac{P_{1}}{P_{2}}$ for liquid X will be lower than that of liquid Y. Therefore, molar heat of vaporization of X will be lower than molar heat of vaporization of Y.
$\Delta H_{vap}$ of X < $\Delta H_{vap}$ of Y
Work Step by Step
Clausius-Clapeyron equation:
$ln\frac{P_{1}}{P_{2}} = \frac{\Delta H_{vap}}{R}(\frac{1}{T_{2}} - \frac{1}{T_{1}})$
From this equation, we can see that $T_{1}, T_{2}$ and $R$ are constant. So $\Delta H_{vap}$ will be directly proportional to $\frac{P_{1}}{P_{2}}$. The vapor pressure($P_{1}$) of liquid X is lower than vapor pressure($P_{1}$) of liquid Y at $20^{\circ}C$, but the vapor pressure ($P_{2}$) of liquid X is lower than the vapor pressure ($P_{2}$) of liquid Y at $60^{\circ}C$. The value of $\frac{P_{1}}{P_{2}}$ for liquid X will be lower than that of liquid Y. Therefore, molar heat of vaporization of X will be lower than molar heat of vaporization of Y.
$\Delta H_{vap}$ of X < $\Delta H_{vap}$ of Y
