Gases possess characteristic critical temperature which depends upon the magnitude of intermolecualr forces between the gas particles, critical temperatures of ammonia and carbon dioxide are 405.5 K and 304.10 K respectively. Which of these gases will liquefy first when you start cooling from 500 K to their critical temperature ?

Options

A. `NH_(3)`
B. `CO_(2)`
C. both
D. none

Answer

To determine which gas will liquefy first when cooling from 500 K to their respective critical temperatures, we need to analyze the critical temperatures of both gases and their cooling process.

### Step-by-Step Solution:

1. **Identify the Critical Temperatures:**
- The critical temperature of ammonia (NH₃) is 405.5 K.
- The critical temperature of carbon dioxide (CO₂) is 304.10 K.

2. **Determine the Cooling Range:**
- We start cooling from 500 K.
- We need to find out how far we need to cool each gas to reach its critical temperature.

3. **Calculate the Temperature Difference for Each Gas:**
- For ammonia (NH₃):
\[ \text{Cooling required} = 500 \, \text{K} - 405.5 \, \text{K} = 94.5 \, \text{K} \]
- For carbon dioxide (CO₂):
\[ \text{Cooling required} = 500 \, \text{K} - 304.1 \, \text{K} = 195.9 \, \text{K} \]

4. **Compare the Cooling Requirements:**
- Ammonia requires a cooling of 94.5 K to reach its critical temperature.
- Carbon dioxide requires a cooling of 195.9 K to reach its critical temperature.

5. **Conclusion:**
- Since ammonia requires less cooling (94.5 K) compared to carbon dioxide (195.9 K), ammonia will liquefy first when cooling from 500 K.

### Final Answer:
Ammonia (NH₃) will liquefy first when cooling from 500 K to their critical temperatures.

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