This is a simple update to the conventional instruction. This is motivated and reinforced to students in a series of simple plots. First, in figs. 2 and 3 we consider plots of ∆H vap /R and ∆Z vap versus T, and a plot of ∆H vap /R versus ∆Z vap . The plots span the temperature range from the triple point to the critical point. As expected, ∆H vap /R is positive, decreasing to a value of 0 at the critical point. Likewise, ∆Z vap ≈ 1 at low temperatures, decreasing to a value of 0 at the critical point. We therefore see that it is not reasonable to assume ∆Z vap = 1 over the entire temperature range, nor is it reasonable to assume ∆H vap is constant over the entire temperature range. However, the functional dependence of ∆H vap /R and ∆Z vap versus T are similar. Furthermore, the plot of ∆H vap /R versus ∆Z vap appears to be linear, except near the triple point. Near the triple point, the rate of decrease of ∆H vap is greater than ∆Z vap . Next, in figs. 4 and 5 we plot ∆H vap / (R∆Z vap ) versus T 1/T from the triple point to the critical point. We compare the case where ∆Z vap = 1, corresponding to the case of ∆H vap / (R∆Z vap ) = ∆H vap /R, to the case where we account for the temperature dependence of ∆Z vap . We find that the range of values for ∆H vap / (R∆Z vap ) is significantly smaller than for ∆H vap /R. While both ∆H vap and ∆Z vap decrease with increasing temperature, the effect in ∆H vap / (R∆Z vap ) appears to cancel. We see that it is more reasonable to assume ∆H vap / (R∆Z vap ) is constant as compared to assuming ∆H vap is constant. This is further emphasized by the Watson plot in figs. 6 and 7. In a Watson plot, we find that ln ∆H vap versus ln (1 -T r ) yields a straight line [15-17]. In fig. we consider the case of ln ∆H vap / (R∆Z vap ) and ln ∆H vap /R (where ∆Z vap = 1) versus ln (1 -T r ). While ln ∆H vap /R versus ln (1 -T r ) yields a straight line as expected, we find that ln ∆H vap / (R∆Z vap ) versus ln (1 -T r ) appears nearly constant. We do observe curvature near the triple point, but in general ln ∆H vap / (R∆Z vap ) appears constant over a wide range of temperatures. As our final example, we return to our starting point of the Clapeyron equation, eq. ( ), then separate and numerically integrate the resulting expression. Integrating from a reference temperature (T 0 ) and corresponding reference pressure (P sat 0 ):