Question

the diagram below shows the ln(P) vs 1/T for four substances . The Clausius Claperyon equation describes the relation . According to the graph , at high tempreature , which substance has the greatest vapour pressure?

Answer 1

D

Answer 2

The relationship between vapour pressure (P) and temperature (T) is described by the Clausius-Clapeyron equation:

$\ln(P) = -\frac{\Delta H_{vap}}{R} \frac{1}{T} + C$

where $\Delta H_{vap}$ is the enthalpy of vaporization, R is the ideal gas constant, and C is a constant related to the entropy of vaporization.

The given graph plots $\ln(P)$ on the y-axis and $1/T$ on the x-axis. This is a linear relationship of the form $y = mx + c$, where $y = \ln(P)$, $x = 1/T$, the slope $m = -\frac{\Delta H_{vap}}{R}$, and the y-intercept $c = C$.

The question asks which substance has the greatest vapour pressure at high temperature.

High temperature corresponds to a small value of $1/T$. On the graph, this corresponds to the left side of the x-axis.

To find the substance with the greatest vapour pressure at high temperature, we need to look at the left side of the graph and identify which substance has the highest value of $\ln(P)$. Since $\ln(P)$ is plotted on the y-axis, the substance with the highest line on the left side of the graph has the greatest $\ln(P)$, and thus the greatest vapour pressure (P).

Looking at the graph, on the left side (representing high temperature, small $1/T$), the lines for the four substances A, B, C, and D are arranged from bottom to top as A, B, C, and D. Substance D is the topmost line among the four.

This means that for any given high temperature (small $1/T$ value) within the range shown on the graph, the value of $\ln(P)$ for Substance D is greater than the values for substances C, B, and A.

$\ln(P)_D > \ln(P)_C > \ln(P)_B > \ln(P)_A$ at high temperature.

Since the natural logarithm function is monotonically increasing, a higher value of $\ln(P)$ corresponds to a higher value of P. Therefore, the vapour pressure of Substance D is the greatest at high temperature.

The slope of the line is $m = -\frac{\Delta H_{vap}}{R}$. The slopes are negative. The magnitude of the slope is $|\frac{\Delta H_{vap}}{R}| = \frac{\Delta H_{vap}}{R}$.

From the graph, the slopes are ordered as follows (from steepest negative to least steep negative): A, B, C, D.

Magnitude of slopes: $|m_A| > |m_B| > |m_C| > |m_D|$.

Since $m = -\frac{\Delta H_{vap}}{R}$ and R is positive, $\Delta H_{vap} = -mR$.

So, $\Delta H_{vap, A} > \Delta H_{vap, B} > \Delta H_{vap, C} > \Delta H_{vap, D}$.

Substance D has the smallest enthalpy of vaporization. Substances with smaller enthalpies of vaporization have weaker intermolecular forces and tend to have higher vapour pressures at a given temperature, which is consistent with our finding from the graph.

The y-intercept is $C$. The y-intercept represents $\ln(P)$ as $1/T \to 0$ ($T \to \infty$). From the graph, the y-intercepts are ordered as $C_D > C_C > C_B > C_A$.

At high temperature (small $1/T$), the term $-\frac{\Delta H_{vap}}{R} \frac{1}{T}$ becomes smaller in magnitude. The value of $\ln(P)$ approaches the y-intercept C. Since $C_D$ is the largest, $\ln(P)_D$ is the largest at very high temperatures.

Based on the graph, Substance D has the highest $\ln(P)$ value in the high-temperature region (left side of the graph). Therefore, Substance D has the greatest vapour pressure at high temperature.