Determine the pressure in the 20 l vessel containing 0.05 kg of water in at 110 celsius
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Expert's answer
2017-01-05T08:31:14-0500
Water at 110 celsius is in a gaseous state (steam). According to the ideal gas law PV = nRT or PV = (m/M)RT. So, using SI units: P = mRT/MV = (0.05 × 8.31 × (110 + 273.15)) / ((18 × 0.001) × (20 × 0.001)) = 442218 Pa = 442.22 kPa
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