When the pressure increased by 2 times and the temperature increased by 300 K
| April 19, 2021 | education
When the pressure increased by 2 times and the temperature increased by 300 K, the gas density decreased by 2 times. Determine the initial ideal gas temperature.
Data: P2 (final pressure) = 2P1 (initial pressure); ΔT (temperature increase) = 300 K; ρ2 (final density) = 0.5ρ1 (initial density).
1) We transform the equation of state of an ideal gas: P * V = m / M * R * T; M / R = m * T / (P * V) = ρ * T / P.
2) Determine the initial temperature: M / R = ρ1 * T1 / P1 = ρ2 * T2 / P2.
ρ1 * T1 / P1 = 0.5 ρ1 * (T1 + 300) / 2P1.
T1 = (T1 + 300) / 4.
4T1 = T1 + 300.
3T1 = 300 and T1 = 300/3 = 100 K.
Answer: D. The ideal gas had a temperature of 100 K.
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