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.
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A-136.5 °C
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B-136.5 K
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C-273 °C
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D0 K
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E-68.25 °C
Answer
Correct Answer: -136.5 °C
Explanation
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:
T1 = 273 K.V2/V1 = 1/2 ⇒ T2/T1 = 1/2 ⇒ T2 = 273 * 1/2 = 136.5 K.Convert to Celsius: t2(°C) = T2 − 273 = 136.5 − 273 = −136.5 °C.Hence, the final temperature is −136.5 °C.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