\[\begin{matrix} Gas & Ar & Ne & Kr & Xe \\ a/(atm\,d{{m}^{6}}\,mo{{l}^{-2}}) & 1.3 & 0.2 & 5.1 & 4.1 \\ b/({{10}^{-2}}\,d{{m}^{3}}\,mo{{l}^{-1}}) & 3.2 & 1.7 & 1.0 & 5.0 \\ \end{matrix}\] |
A) Kr
B) Ne
C) Ar
D) Xe
Correct Answer: A
Solution :
\[{{T}_{c}}=\frac{8a}{27Rb}\] Greater value of\[\frac{a}{b}\Rightarrow \]higher is \['{{T}_{c}}'\] Gas a \[{}^{a}/{}_{b}\] Ar \[\frac{1.3}{3.2}=0.406\] Ne \[\frac{0.2}{1.7}=0.118\] Kr \[\frac{5.1}{1}=5.1\] Xe \[\frac{4.1}{5}=0.82\] \[\therefore \]\[{{T}_{C}}\] has order : \[Kr>Xe>Ar>Ne\] \[\therefore \] Ans. is [a]You need to login to perform this action.
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