The volumes of two systems are: V1 and V2.
From the given data:
V1 + V2 = V = 0.025 m³
Both systems are in a closed container in adiabatic condition, we have:
T1 = 175 K
T2 = 400 K
PV = nRT
nRT1/P1 + nRT2/P2 = nRT/P
2R×175/P + 1.5R×400/P = 3.5RT/P
2 × 175 + 1.5 × 400 = 3.5T
350 + 600 = 3.5T
950 = 3.5T
T = 950/3.5 = 271.42 K
Hence the final temeperature is 271.42 K.
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