The van der Waals parameters for gases W, X, Y and Z are
`{:("Gas",a("atm"L^(2)mol^(-2)),b(L"mol"^(-1))),(W," "4.0," "0.027),(X," "8.0," "0.030),(Y," "6.0," "0.032),(Z," "12.0," "0.027):}`
Which one of these gases has the highest critical temperature?

Options

A. W
B. X
C. Y
D. Z

Answer

To determine which gas has the highest critical temperature, we can use the formula for critical temperature (\(T_c\)) given by:

\[ T_c = \frac{8a}{27Rb} \]

Where:
- \(a\) is the van der Waals parameter related to the attractive forces between molecules.
- \(b\) is the van der Waals parameter related to the volume occupied by the gas molecules.
- \(R\) is the universal gas constant.

Since \(R\) is a constant, we can simplify our comparison by focusing on the ratio \(\frac{a}{b}\). The gas with the highest \(\frac{a}{b}\) ratio will have the highest critical temperature.

### Step-by-step Solution:

1. **Calculate \(\frac{a}{b}\) for each gas:**
- For gas W:
\[ a = 4.0, \quad b = 0.027 \]
\[ \frac{a}{b} = \frac{4.0}{0.027} \approx 148.14 \]

- For gas X:
\[ a = 8.0, \quad b = 0.030 \]
\[ \frac{a}{b} = \frac{8.0}{0.030} \approx 266.67 \]

- For gas Y:
\[ a = 6.0, \quad b = 0.032 \]
\[ \frac{a}{b} = \frac{6.0}{0.032} \approx 187.50 \]

- For gas Z:
\[ a = 12.0, \quad b = 0.027 \]
\[ \frac{a}{b} = \frac{12.0}{0.027} \approx 444.44 \]

2. **Compare the \(\frac{a}{b}\) values:**
- Gas W: \(\frac{a}{b} \approx 148.14\)
- Gas X: \(\frac{a}{b} \approx 266.67\)
- Gas Y: \(\frac{a}{b} \approx 187.50\)
- Gas Z: \(\frac{a}{b} \approx 444.44\)

3. **Identify the highest value:**
- The highest \(\frac{a}{b}\) value is for gas Z, which is approximately 444.44.

4. **Conclusion:**
- Therefore, gas Z has the highest critical temperature.

### Final Answer:
Gas Z has the highest critical temperature.