When a student plotted ln [vapor pressure of a gas] vs. inverse Kelvin temperature, she obtained a straight line with a slope equal to -8.000 x 10^3 K.

According to the Clausius Clapeyron equation, Hvap is ?
kJ/mol.

To determine the value of Hvap (the enthalpy of vaporization) in kJ/mol using the given information, we need to understand the Clausius-Clapeyron equation.

The Clausius-Clapeyron equation relates the natural logarithm of the vapor pressure of a substance (ln P) to its enthalpy of vaporization (Hvap), the molar gas constant (R), and the temperature (T) in Kelvin:

ln P = -Hvap / (R * T) + C

In this equation, ln P is the logarithm of the vapor pressure of the gas, Hvap is the enthalpy of vaporization, R is the gas constant (8.314 J/(mol*K) or 0.008314 kJ/(mol*K)), T is the temperature in Kelvin, and C is a constant.

We are given that when the student plotted ln (vapor pressure) vs. 1/T, the resulting line had a slope of -8.000 x 10^3 K. From this slope, we can determine the value of -Hvap / (R).

Slope = -Hvap / (R * T)

Rearranging the equation, we can isolate Hvap:

-Hvap = slope * (R * T)

Substituting the values, we get:

-Hvap = -8.000 x 10^3 K * (0.008314 kJ/(mol*K) * T)

The negative sign means that Hvap will be positive. Now, we have the equation for Hvap:

Hvap = 8.000 x 10^3 K * (0.008314 kJ/(mol*K) * T)

To find Hvap in kJ/mol, we need to multiply the temperature (T) by the constant 0.008314 kJ/(mol*K) and then multiply the product by 8.000 x 10^3.

Remember to convert the temperature to Kelvin if it is given in Celsius or Fahrenheit.

Plug in the given temperature value in Kelvin, and calculate Hvap using the above equation.