Animated Solution for Chemistry - Solutions: For the solution of the gases $w,x,y$ and $z$ in water at $298\text{ K}$, the Henry's law constants ($K_H$) are $0.5,2,35$ and $40\text{ K bar}$, respectively. The correct plot for the given data is

Henry's Law

Let's embark on a journey to understand how gases dissolve in liquids. The fundamental rule governing this phenomenon is Henry's Law. It elegantly states that at a constant temperature, the partial pressure of a gas ($p_{\text{gas}}$) above a liquid is directly proportional to its mole fraction (${\chi }_{\text{gas}}$) within the solution. Mathematically, this is expressed as:

Here, $K_H$ is the Henry's Law constant, a unique fingerprint for every gas-solvent pair at a specific temperature.

Mole Fraction of Water

The question presents us with a slight twist. Instead of plotting the partial pressure against the mole fraction of the gas, we need to plot it against the mole fraction of the solvent, which is water (${\chi }_{{\text{H}}_2\text{O}}$).

Since we are dealing with a binary mixture containing only the dissolved gas and water, the sum of their mole fractions must perfectly equal one:

By rearranging this simple relation, we can express the mole fraction of the gas entirely in terms of the mole fraction of water:

Now, let's substitute this new expression back into our original Henry's Law equation. This will give us the exact mathematical function we need to graph:

Expanding the bracket reveals a beautiful linear equation:

This equation is in the classic straight-line format, $y=mx+c$. Here, our y-axis represents the partial pressure ($p_{\text{gas}}$), and our x-axis represents the mole fraction of water (${\chi }_{{\text{H}}_2\text{O}}$).

Notice two critical features of this equation: 1. The Slope ($m$): It is $-K_H$. Because $K_H$ is always positive, the slope is inherently negative. This means as the mole fraction of water increases, the partial pressure of the gas must decrease. 2. The y-intercept ($c$): It is exactly $+K_H$. This is the point where the line crosses the y-axis (when ${\chi }_{{\text{H}}_2\text{O}}=0$).

Let's find the x-intercept. Where do these lines hit the x-axis? We set $p_{\text{gas}}=0$:

This is a profound realization! All the lines, regardless of their individual $K_H$ values, must converge and intersect at exactly the same point on the x-axis: $(1,0)$.

Finally, let's look at the y-intercepts. The problem provides the $K_H$ values for four gases: $w=0.5$, $x=2$, $y=35$, and $z=40$. Since the y-intercept is simply $K_H$, the intercepts must follow the order:

Gas $z$ will have the highest intercept and the steepest negative slope, while gas $w$ will have the lowest intercept and the gentlest slope. Comparing these derived properties with the given options, only the first graph perfectly captures all these mathematical truths.