At a given temperature T, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. The

Q: Why did we divide the entire equation by $RT$?

We divided by $RT$ to forcefully create the term $\frac{pV}{RT}$. This specific ratio is the definition of the compressibility factor, $Z$. By doing this, we transformed an equation about $p$, $V$, and $T$ into an equation about $Z$ and $p$.

Q: What does the constant $b$ physically mean in the real world?

In the ideal gas law, we pretend gas molecules are infinitely small points. In reality, they are tiny spheres that take up space. The constant $b$ represents this "excluded volume"—the actual physical space the molecules occupy, which cannot be compressed.

Q: Why does the graph start at $Z=1$ on the y-axis?

The y-intercept of our derived equation $Z = 1 + \left(\frac{b}{RT}\right)p$ is 1. Physically, this means that as the pressure $p$ approaches zero, the gas behaves perfectly ideally, and its compressibility factor $Z$ becomes exactly 1.

Q: Why does Xenon have the largest value of $b$?

As you move down the noble gas group in the periodic table (Neon $\rightarrow$ Argon $\rightarrow$ Krypton $\rightarrow$ Xenon), the atoms gain more electron shells. This makes the Xenon atom physically the largest, meaning it has the greatest excluded volume, and thus the largest $b$.

Q: What happened to the attractive forces (the constant $a$)?

The equation provided in the question, $p = \frac{RT}{V-b}$, intentionally omits the $\frac{a}{V^2}$ term. This implies that at this specific high temperature $T$, the kinetic energy of the molecules is so high that intermolecular attractive forces are negligible, and only their physical size ($b$) causes deviation from ideal behavior.

Question

At a given temperature $T$, gases Ne, Ar, Xe and Kr are found to deviate from ideal gas behaviour. Their equation of state is given as,
$p = \frac{RT}{V - b}$ at $T$.
Here, $b$ is the van der Waals' constant. Which gas will exhibit steepest increase in the plot of $Z$ (compression factor) vs $p$?

Options:

(A) Xe

(B) Ar

(C) Kr

(D) Ne

Sigma Percentile · Accepted Answer

Imagine you are trying to squeeze a balloon. At first, it is easy, but as it gets smaller, the gas molecules inside start fighting back. They take up physical space!

This is the essence of the van der Waals constant $b$. It represents the actual volume occupied by the gas molecules themselves.

In this problem, we are exploring how this physical size affects the compressibility of different noble gases.

### The Modified Equation of State

We are given a modified version of the ideal gas law. The equation is:

$$p = \frac{RT}{V-b}$$

Notice that the volume term is $V-b$ instead of just $V$. This subtraction accounts for the fact that the molecules are not point masses; they have a finite size.

Our goal is to understand the behavior of the **compressibility factor**, $Z$. By definition, $Z$ is the ratio of the actual molar volume of a gas to the molar volume of an ideal gas at the same temperature and pressure. Mathematically, it is defined as:

$$Z = \frac{pV}{RT}$$

To see how $Z$ behaves, we need to manipulate our given equation of state to isolate this exact expression.

### Unveiling the Compressibility Factor

Let us start by cross-multiplying the denominator to get rid of the fraction.

$$p(V-b) = RT$$

Expanding the bracket gives us a clearer view of the terms.

$$pV - pb = RT$$

Now, we want to create the term $\frac{pV}{RT}$. To do this, we divide the entire equation by $RT$.

$$\frac{pV}{RT} - \frac{pb}{RT} = 1$$

The first term is exactly our definition of $Z$! Let us substitute $Z$ into the equation and move the other term to the right side.

$$Z = 1 + \left(\frac{b}{RT}ight)p$$

This is a beautiful result. It is a linear equation, perfectly matching the form of a straight line, $y = mx + c$.

### The Geometry of the Gas

If we plot $Z$ on the y-axis and $p$ on the x-axis, the y-intercept is $1$. This makes perfect physical sense: at zero pressure, all real gases behave ideally.

The slope of this line is given by the term $\frac{b}{RT}$. Since the temperature $T$ is constant, the steepness of the line depends entirely on the constant $b$.

**The greater the value of $b$, the steeper the slope of the $Z$ versus $p$ graph.**

Now, we must connect the math back to the physical reality of the noble gases: Neon, Argon, Krypton, and Xenon.

The constant $b$ is directly proportional to the physical size of the atoms. As we move down the noble gas group in the periodic table, the atoms gain more electron shells.

Therefore, the atomic size increases in the order: $	ext{Ne} < 	ext{Ar} < 	ext{Kr} < 	ext{Xe}$.

Because Xenon is the largest atom, it has the largest excluded volume, and thus the largest value of $b$.

Consequently, Xenon will have the maximum slope. **Xenon (Xe) will exhibit the steepest increase in the plot of $Z$ versus $p$.**

Select Answer:

Visualized Solution

$p = \frac{R T}{V - b}$

$Definition of Z$

$Rearranging the Equation$

$Creating the Z Term$

$Equation of the Line$

$Analyzing the Slope$

$Atomic Size and b$

$Conclusion$

The Sigma Insight: Gaseous State

The Modified Equation of State

Unveiling the Compressibility Factor

The Geometry of the Gas

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If $Z$ is a compressibility factor, van der Waals' equation at low pressure can be written as

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p=V−bRT​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Definition of Z

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging the Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Creating the Z Term

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of the Line

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Slope

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Atomic Size and b

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion