Question

A balloon filled with an air sample occupies 3 L volume at 35°C. On lowering the temperature to T, the volume decreases to 2.5 L. The temperature T is? [Assume p-constant]

• A. 16°C
• B. -16°C
• C. 24°C
• D. -20°C

Hint:

Key Exam Tip:
When using gas laws like Charles's Law, always convert temperatures to the absolute Kelvin scale. Remember that K = °C + 273.15.

Answer 1

The Correct Option is B

According to Charles's Law, at constant pressure, the volume of a gas is directly proportional to its absolute temperature ($V/T$ = constant).
Initial conditions: $V_1 = 3$ L, $T_1 = 35\text{°C}$.
Final conditions: $V_2 = 2.5$ L, $T_2 = T$ (°C).
Convert temperatures to Kelvin:
$T_1 = 35\text{°C} + 273.15 = 308.15\text{ K}$.
Apply Charles's Law: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$.
$\frac{3 \text{ L}}{308.15 \text{ K}} = \frac{2.5 \text{ L}}{T_2}$
Solve for $T_2$ in Kelvin:
$T_2 = \frac{2.5 \text{ L} \times 308.15 \text{ K}}{3 \text{ L}} \approx 256.79 \text{ K}$.
Convert $T_2$ back to Celsius:
$T_2(\text{°C}) = T_2(\text{K}) - 273.15$
$T_2(\text{°C}) = 256.79 - 273.15 \approx -16.36\text{°C}$.
The closest option is -16°C.
Final Answer: \(\boxed{B}\)