The gas in the cylinder was heated from 260 K to 310 K. At the same time, its pressure increased by 400 kPa.
April 6, 2021 | education
| The gas in the cylinder was heated from 260 K to 310 K. At the same time, its pressure increased by 400 kPa. What was the initial gas pressure?
T 1 = 260 K.
T2 = 310 K.
ΔP = 400 kPa = 400 * 10 ^ 3 Pa.
P1 -?
The gas in the cylinder is heated at a constant volume, that is, isochoric. The isochoric process is described by Charles’s law: P1 / T1 = P2 / T2.
We write down the change in pressure by the formula: ΔP = P2 – P1.
P2 = ΔP + P1.
P1 / T1 = (ΔP + P1) / T2.
T1 * ΔP + T1 * P1 = P1 * T2.
T1 * ΔP = P1 * T2 – T1 * P1.
P1 = T1 * ΔP / (T2 – T1).
P1 = 260 K * 400 * 10 ^ 3 Pa / (310 K – 260 K) = 2080 * 10 ^ 3 Pa.
Answer: the pressure in the cylinder was P1 = 2080 * 10 ^ 3 Pa.
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