A sample of gas has a volume of 0.2 lit measured at 1 atm pressure and 0oC. At the same pressure, bu — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

# Solution **Given Data:** - Initial volume: $V_1 = 0.2$ L - Initial temperature: $T_1 = 0°C = 273$ K - Initial pressure: $P_1 = 1$ atm - Final pressure: $P_2 = 1$ atm (same) - Final temperature: $T_2 = 273$ K **Key Observation:** The initial temperature is $0°C = 273$ K, which is **identical** to the final temperature of 273 K. **Applying Gay-Lussac's/Combined Gas Law:** Since pressure is constant (isobaric process): $$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$ $$\frac{0.2}{273} = \frac{V_2}{273}$$ $$V_2 = 0.2 \text{ L}$$ **Conclusion:** Because the temperature remains unchanged at 273 K and pressure is constant, the volume **remains 0.2 L**. No change occurs in the volume. *Option B is correct because it states the volume remains **0.2 L** (unchanged).*