A gas initially at 0 °C is cooled at constant pressure until its volume becomes one-half of the original. What is the final temperature of the gas (in °C)? Assume ideal-gas behavior.

Introduction / Context:
Ideal-gas relations connect temperature and volume at constant pressure (Charles’s law). This problem checks your ability to convert Celsius to Kelvin, apply proportionality of V to T, and convert back to Celsius correctly—a common exam trap lies in mishandling the absolute temperature scale.

Given Data / Assumptions:

• Initial temperature T1 = 0 °C = 273 K.
• Pressure constant; ideal-gas behavior assumed.
• Final volume V2 = V1/2.

Concept / Approach:
At constant pressure for an ideal gas: V ∝ T (in Kelvin). Therefore, V2/V1 = T2/T1. Halving volume halves absolute temperature. Always work in Kelvin for gas-law calculations, then convert to °C if needed.

Step-by-Step Solution:

Verification / Alternative check:
If a further halving to V/4 occurred, T would be 68.25 K (−204.75 °C), showing the linear proportionality and reinforcing correct unit handling.

Why Other Options Are Wrong:

• −136.5 K: Kelvin values cannot be negative; also the number is actually in °C.
• −273 °C / 0 K: correspond to zero absolute temperature, which would imply zero volume—too large a cooling step.
• −68.25 °C: would correspond to 204.75 K, not 136.5 K.

Common Pitfalls:
Using Celsius directly in Charles’s law or forgetting to convert back properly from Kelvin, leading to sign or magnitude errors.

Final Answer:
−136.5 °C