?The mass density of water vapour at 327.6 atm and 776.4 K is \(133.2 \mathrm{~kg}
Chapter 1, Problem P1C.7(choose chapter or problem)
The mass density of water vapour at 327.6 atm and 776.4 K is \(133.2 \mathrm{~kg} \mathrm{m}^{-3}\). Given that for water \(a=5.464 \mathrm{dm}^{6} \text { atm } \mathrm{mol}^{-2}\), \(b=0.03049 \mathrm{dm}^{3} \mathrm{mol}^{-1}\), and \(M=18.02 \mathrm{gmol}^{-1}\), calculate (a) the molar volume. Then calculate the compression factor (b) from the data, and (c) from the virial expansion of the van der Waals equation.
Text Transcription:
133.2 kg m^−3
a = 5.464 dm^6 atm mol^−2
b = 0.03049 dm^3 mol^−1
M = 18.02 g mol^−1,
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