At what temperature was the gas if the volume tripled under isobaric heating at 250 K?
| July 14, 2021 | education
Initial data: isobaric process (P = const); the gas was heated at 250 K (T2 = 250 + T1); the gas volume increased 3 times (V2 = 3V1).
For isobaric heating, the following equality will be true: V1 / T1 = V2 / T2, whence T1 = V1 * T2 / V2 = V1 * (250 + T1) / 3V1 = (250 + T1) / 3.
3T1 = 250 + T1.
3T1 – T1 = 250.
2T1 = 250.
T1 = 250/2 = 125 K.
Answer: Before isobaric heating, the gas was at a temperature of 125 K.
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