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This is a gas law problem involving changes in pressure, volume, and temperature. Based on the ideal gas law and the combined gas law, the formula used is:

\[
V_2 = V_1 \times \frac{P_1}{P_2} \times \frac{T_2}{T_1}
\]

Where:
- \( V_1 = 5.0 \, \text{L} \) (initial volume)
- \( P_1 = 1.2 \, \text{atm} \) (initial pressure)
- \( T_1 = 300 \, \text{K} \) (initial temperature)
- \( P_2 = 2.4 \, \text{atm} \) (final pressure)
- \( T_2 = 600 \, \text{K} \) (final temperature)

Substituting the values into the formula:

\[
V_2 = 5.0 \, \text{L} \times \frac{1.2 \, \text{atm}}{2.4 \, \text{atm}} \times \frac{600 \, \text{K}}{300 \, \text{K}}
\]

This simplifies to:

\[
V_2 = 5.0 \, \text{L} \times 0.5 \times 2 = 5.0 \, \text{L}
\]

Thus, the final volume \( V_2 = 5.0 \, \text{L} \).

Would you like further details on this solution, or do you have any questions?

Here are 5 related questions to expand your understanding:
1. What is the ideal gas law, and how is it derived?
2. How does temperature affect the volume of a gas if the pressure remains constant?
3. Can you explain how Boyle's Law and Charles's Law are related to the combined gas law?
4. How does the combined gas law behave if the temperature is kept constant?
5. What happens to the volume if the pressure is decreased while keeping temperature constant?

**Tip:** When solving gas law problems, make sure all temperatures are in Kelvin (K) to avoid errors.

A gas has a volume of 5.0 L at a pressure of 1.2 atm and a temperature of 300 K. What will the volume be if the pressure is increased to 2.4 atm and the temperature is increased to 600 K?

Learn how to calculate the final volume of a gas when the pressure is increased to 2.4 atm and the temperature is raised to 600 K, using the combined gas law formula. This problem involves the key concepts of pressure, volume, and temperature relationships under gas laws.

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