A) \[Z=1-\frac{Pb}{RT}\]
B) \[Z=1+\frac{Pb}{RT}\]
C) \[Z=1+\frac{RT}{Pb}\]
D) \[Z=1-\frac{a}{VRT}\]
Correct Answer: D
Solution :
[d] \[\left( P+\frac{{{n}^{2}}a}{{{V}^{2}}} \right)(V-nb)=nRT\] For 1 mole, \[\left( P+\frac{a}{{{V}^{2}}} \right)(V-b)=RT\] \[PV=RT+Pb-\frac{a}{V}+\frac{ab}{{{V}^{2}}}\] At low pressure, terms \[pb\] and \[\frac{ab}{{{V}^{2}}}\]will be negligible as compared to RT. So,\[PV=RT-\frac{a}{V},Z=1-\frac{a}{VRT}\]
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