A) 44 g
B) 33 g
C) 22 g
D) 5g
Correct Answer: C
Solution :
For \[\mu \] moles of a real gas. Van der Waals equation is given \[p+\frac{{{\mu }^{2}}a}{{{V}^{2}}}(V-\mu b)+\mu RT\] \[\Rightarrow \] \[p=\frac{\mu RT}{V-\mu b}-\frac{{{\mu }^{2}}a}{{{V}^{2}}}\] On comparing this equation with given equation. \[\mu =\frac{1}{2}\] But \[\mu =\frac{m}{M}\] Mass of gas, \[m=\mu M=\frac{1}{2}\times 44=22\,g\]You need to login to perform this action.
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