the vapor pressure of substance x is measured at several temperatures:\nuse this information to calculate…

Explanation:

Step1: Convert temperatures to Kelvin

The formula to convert Celsius to Kelvin is (T(K)=T(^{\circ}C)+273.15). For (T_1 = 10.^{\circ}C), (T_1=10 + 273.15=283.15\space K) For (T_2 = 23.^{\circ}C), (T_2=23+ 273.15 = 296.15\space K) For (T_3 = 36.^{\circ}C), (T_3=36+ 273.15=309.15\space K)

Step2: Use the Clausius - Clapeyron equation

The Clausius - Clapeyron equation is (\ln\left(\frac{P_2}{P_1}\right)=\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)), where (R = 8.314\space J/(mol\cdot K))

Let's take the first two data points: (P_1 = 0.171\space atm), (T_1 = 283.15\space K), (P_2=0.251\space atm), (T_2 = 296.15\space K)

(\ln\left(\frac{0.251}{0.171}\right)=\frac{\Delta H_{vap}}{8.314}\left(\frac{1}{283.15}-\frac{1}{296.15}\right))

First, calculate (\ln\left(\frac{0.251}{0.171}\right)\approx\ln(1.4678)\approx0.384)

Then calculate (\frac{1}{283.15}-\frac{1}{296.15}=\frac{296.15 - 283.15}{283.15\times296.15}=\frac{13}{283.15\times296.15}\approx1.57\times10^{-4})

So, (0.384=\frac{\Delta H_{vap}}{8.314}\times1.57\times 10^{-4})

(\Delta H_{vap}=\frac{0.384\times8.314}{1.57\times 10^{-4}})

(\Delta H_{vap}\approx2.0\times10^{4}\space J/mol)

Let's check with another pair (say (P_1 = 0.251\space atm), (T_1 = 296.15\space K), (P_2 = 0.357\space atm), (T_2=309.15\space K))

(\ln\left(\frac{0.357}{0.251}\right)=\frac{\Delta H_{vap}}{8.314}\left(\frac{1}{296.15}-\frac{1}{309.15}\right))

(\ln\left(\frac{0.357}{0.251}\right)\approx\ln(1.4223)\approx0.352)

(\frac{1}{296.15}-\frac{1}{309.15}=\frac{309.15 - 296.15}{296.15\times309.15}=\frac{13}{296.15\times309.15}\approx1.41\times10^{-4})

(0.352=\frac{\Delta H_{vap}}{8.314}\times1.41\times 10^{-4})

(\Delta H_{vap}=\frac{0.352\times8.314}{1.41\times 10^{-4}}\approx2.1\times 10^{4}\space J/mol)

Taking an average (since using two - point method with different pairs gives similar results)

Answer:

(2.1\times 10^{4}\space J/mol)