Consider the van der Waals' constants, a and b, for the following gases.

Gas: Ar, Ne, Kr, Xe
a / (a

Q: Why do we ignore the $10^{-2}$ in the values of 'b'?

Since we are only comparing the relative values of $T_c$ across different gases, any constant multiplier present in all the values of 'b' will cancel out in the comparison. It saves time to just use the base numbers!

Q: Does a larger molecule always have a higher critical temperature?

Not necessarily! While larger molecules often have stronger London dispersion forces (higher 'a'), they also have a larger volume (higher 'b'). The critical temperature depends on the *ratio* of $a$ to $b$, so it's a balance between stickiness and size.

Q: What exactly happens above the critical temperature?

Above $T_c$, the kinetic energy of the gas molecules is so high that it completely overpowers the intermolecular attractive forces. No matter how much pressure you apply, the molecules won't condense into a liquid. It becomes a supercritical fluid!

Q: Why is the formula for $T_c$ specifically $\frac{8a}{27Rb}$?

This formula is derived from the van der Waals equation of state by finding the point of inflection on the critical isotherm (where the first and second derivatives of pressure with respect to volume are zero).

Q: Why is Krypton's 'a' value so high compared to Argon and Neon?

Krypton is a larger atom with more electrons. This makes its electron cloud more polarizable, leading to stronger temporary dipoles (London dispersion forces), which is reflected in a significantly higher 'a' value.

Question

Consider the van der Waals' constants, $a$ and $b$, for the following gases.

\begin{array}{|c|c|c|c|c|}
\hline
	extbf{Gas} & 	extbf{Ar} & 	extbf{Ne} & 	extbf{Kr} & 	extbf{Xe} \
\hline
\mathbf{a / (atm\ dm^6\ mol^{-2})} & 1.3 & 0.2 & 5.1 & 4.1 \
\hline
\mathbf{b / (10^{-2}\ dm^3\ mol^{-1})} & 3.2 & 1.7 & 1.0 & 5.0 \
\hline
\end{array}

Which gas is expected to have the highest critical temperature ?

Options:

(A) Kr

(B) Xe

(C) Ar

(D) Ne

Sigma Percentile · Accepted Answer

### The Essence of Critical Temperature

Imagine you are trying to force a wild, energetic gas to calm down and become a liquid. You push and push, increasing the pressure, hoping the molecules will finally stick together. But there is a catch! If the gas is too hot, its molecules possess so much kinetic energy that they simply refuse to be bound together, no matter how much pressure you apply.

This magical temperature threshold is known as the **Critical Temperature ($T_c$)**. Above this temperature, the gas cannot be liquefied. It becomes a supercritical fluid, a state of matter that is neither truly gas nor liquid. To understand which gas has the highest critical temperature, we need to look at the forces acting between its molecules.

### Decoding the van der Waals Constants

Real gases don't behave perfectly. They have intermolecular forces and they take up physical space. This is where the van der Waals constants, **$a$** and **$b$**, come into play.

The constant **$a$** represents the "stickiness" of the molecules. It measures the magnitude of the intermolecular attractive forces. A higher $a$ means the molecules pull on each other more strongly, making it easier for the gas to condense into a liquid.

The constant **$b$** represents the "chunkiness" or the effective volume of the gas molecules. A higher $b$ means the molecules are larger and take up more space, which slightly hinders them from getting close enough to feel the full effect of the attractive forces.

### The Master Equation

The critical temperature is a beautiful mathematical balance between this stickiness and chunkiness. It is given by the formula:

$$T_c = \frac{8a}{27Rb}$$

Here, $R$ is the universal gas constant. Notice something interesting? The numbers $8$, $27$, and $R$ are all constants. This means that when we are comparing different gases, we don't need to calculate the exact temperature. We can simply look at the proportionality:

$$T_c \propto \frac{a}{b}$$

The gas with the highest ratio of $a$ to $b$ will inevitably have the highest critical temperature!

### The Grand Comparison

Let's put our gases to the test by calculating the $\frac{a}{b}$ ratio for each one. Note that the $10^{-2}$ multiplier in the $b$ values is common to all gases, so we can safely ignore it for the sake of comparison.

**1. Argon (Ar):**
$$\frac{a}{b} = \frac{1.3}{3.2} \approx 0.41$$

**2. Neon (Ne):**
$$\frac{a}{b} = \frac{0.2}{1.7} \approx 0.12$$

**3. Krypton (Kr):**
$$\frac{a}{b} = \frac{5.1}{1.0} = 5.10$$

**4. Xenon (Xe):**
$$\frac{a}{b} = \frac{4.1}{5.0} = 0.82$$

### The Final Verdict

Look at those numbers! Krypton completely dominates the competition with a ratio of $5.10$. Even though Xenon is a larger atom, the specific values provided in this dataset show that Krypton has a massive intermolecular attraction ($a$) relative to its excluded volume ($b$).

Because Krypton has the highest $\frac{a}{b}$ ratio, it requires the most thermal energy to overcome its intermolecular forces. Therefore, **Krypton (Kr)** is expected to have the highest critical temperature.

Select Answer:

Visualized Solution

Formula for $T_{c}$

$T_{c} \propto \frac{a}{b}$

Ratio for $Ar$

Ratio for $Ne$

Ratio for $Kr$

Ratio for $Xe$

Maximum $T_{c}$

The Sigma Insight: Gaseous State

The Essence of Critical Temperature

Decoding the van der Waals Constants

The Master Equation

The Grand Comparison

The Final Verdict

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Select Answer:

Visualized Solution

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

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

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

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

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