Question

In this lab, you will determine the vapour pressure (P) of water at various temperatures and use this data to find its enthalpy of vaporization (ΔvapH).

More generally, for an unknown liquid: if a plot of ln P vs. 1/T gives a slope of -2669 K, what is its ΔvapH in kJ/mol? Do not worry about how realistic the value is. Assume R = 8.314 J mol⁻¹ K⁻¹.

Answer 1

To calculate the enthalpy of vaporization (ΔvapH) for an unknown liquid with a slope of -2669 K on a ln P vs 1/T plot, use the Clausius-Clapeyron equation with the universal gas constant (R). This results in ΔvapH = -22.189786 kJ/mol.

Step-by-step explanation:

For an unknown liquid, if a plot of ln P versus 1/T gives a slope of -2669 K, we can use the Clausius-Clapeyron equation to find the enthalpy of vaporization (ΔvapH). The Clausius-Clapeyron equation relates the vapor pressure of a liquid with its temperature and can be expressed in a linear logarithmic form: ln P = -ΔvapH/R · (1/T) + C, where R is the universal gas constant and C is a constant specific to the substance.

From the given slope of -2669 K and knowing the value of R (8.314 J mol⁻¹ K⁻¹), the calculation of ΔvapH is straightforward. The slope of the ln P versus 1/T plot is equivalent to -ΔvapH/R. Thus, ΔvapH = slope · R.

By plugging in the values, we obtain the following: ΔvapH = -2669 K · 8.314 J mol⁻¹ K⁻¹ = -22189.786 J/mol, which can be converted to kilojoules per mole by dividing by 1000, giving ΔvapH = -22.189786 kJ/mol.