A sample of gas occupies 100 ml at 27oC and 740 mm pressure. When its volume is changed to 80 ml at — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

# Solution Since pressure is constant (740 mm), this is an **isobaric process**. Use **Charles's Law**: $$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$ **Step 1:** Convert initial temperature to Kelvin $$T_1 = 27°C + 273 = 300 \text{ K}$$ **Step 2:** Identify given values - $V_1 = 100$ ml - $V_2 = 80$ ml - $T_1 = 300$ K - $T_2 = ?$ **Step 3:** Substitute into Charles's Law $$\frac{100}{300} = \frac{80}{T_2}$$ **Step 4:** Solve for $T_2$ $$T_2 = \frac{80 \times 300}{100} = \frac{24000}{100} = 240 \text{ K}$$ **Step 5:** Convert back to Celsius (if needed) $$T_2 = 240 - 273 = -33°C$$ **Answer:** The temperature decreases to **240 K** (or **-33°C**) because volume decreased while pressure remained constant, requiring proportionally lower temperature.