Question

A balloon filled with an air sample occupies \( 3 \, \text{L} \) volume at \( 35^\circ \text{C} \). On lowering the temperature to \( T \), the volume decreases to \( 2.5 \, \text{L} \). The temperature \( T \) is: [Assume \( P \)-constant]

• A. \( 16^\circ \text{C} \)
• B. \( -16^\circ \text{C} \)
• C. \( 24^\circ \text{C} \)
• D. \( -20^\circ \text{C} \)

Hint:

Charles's law relates the volume and temperature of a gas when pressure is constant. Remember to always convert temperatures to Kelvin before applying the formula.

Answer 1

The Correct Option is B

Applying Charles's law: \[\frac{V_1}{T_1} = \frac{V_2}{T_2}.\] Given \( T_1 = 35^\circ \text{C} = 308 \, \text{K} \), and \( V_1 = 3 \), \( V_2 = 2.5 \). Substitute these values: \[\frac{3}{308} = \frac{2.5}{T_2}.\] Solving for \( T_2 \): \[\ T_2 = \frac{2.5 \cdot 308}{3} = 256.6 \, \text{K}.\] Convert \( T_2 \) to Celsius: \[\ T_2 = 256.6 - 273 = -16.4^\circ \text{C} \approx -16^\circ \text{C}.\] Final Answer: \[\ \boxed{-16^\circ \text{C}}\ \]