A) \[4{{\lambda }_{A}}={{\lambda }_{B}}\] (Where \[{{\lambda }_{A}}\] and \[{{\lambda }_{B}}\] are mean free Path of molecules. The collision diameter of \[{{O}_{2}}\] may be assumed to be twice as that of \[{{H}_{2}}\]and \[{{p}_{A}}=8{{p}_{B}}\])
B) \[2{{\eta }_{A}}^{2}={{\eta }_{B}}^{2}\](Where \[{{\eta }_{A}}\] and \[{{\eta }_{B}}\] are the viscosity of gases)
C) \[2{{(KE\,mo{{l}^{-1}})}_{A}}={{(KE\,mo{{l}^{-1}})}_{B}}\] (Where \[{{(KE\,mo{{l}^{-1}})}_{A}}\] and \[{{(KE\,mo{{l}^{-1}})}_{B}}\] are the kinetic energies of .4 and B respectively.
D) \[{{Z}_{A}}=2{{Z}_{B}}\] (Where \[{{Z}_{A}}\] and \[{{Z}_{B}}\] are the compressibility factor of gas A and B respectively)
Correct Answer: D
Solution :
[d] Statement (1) is CORRECT Since, \[{{\lambda }_{A}}=\frac{k{{T}_{A}}}{\sqrt{2}\pi \sigma _{A}^{2}{{p}_{A}}}\]and \[{{\lambda }_{B}}=\frac{k{{T}_{B}}}{\sqrt{2}\,\pi \sigma _{B}^{2}{{p}_{B}}}\] \[\frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\left( \frac{{{T}_{A}}}{{{T}_{B}}} \right)\left( \frac{\sigma _{B}^{2}}{\sigma _{A}^{2}} \right)\left( \frac{{{p}_{B}}}{{{p}_{A}}} \right)\] \[=\left( \frac{200}{400} \right){{\left( \frac{2}{1} \right)}^{2}}\left( \frac{1}{8} \right)=\frac{1}{4}\] \[4\,{{\lambda }_{A}}={{\lambda }_{B}}\] Statement (2) is CORRECT Since \[{{\eta }_{A}}=\frac{{{({{m}_{A}}k{{T}_{A}})}^{\frac{1}{2}}}}{{{\pi }^{\frac{3}{2}}}\sigma _{A}^{2}}\] and \[{{\eta }_{B}}=\frac{{{({{m}_{B}}k{{T}_{B}})}^{\frac{1}{2}}}}{{{\pi }^{\frac{3}{2}}}\sigma _{BA}^{2}}\] \[\frac{{{\eta }_{A}}}{{{\eta }_{B}}}={{\left( \frac{{{m}_{A}}}{{{m}_{B}}} \right)}^{\frac{1}{2}}}{{\left( \frac{{{T}_{A}}}{{{T}_{B}}} \right)}^{\frac{1}{2}}}{{\left( \frac{{{\sigma }_{B}}}{{{\sigma }_{A}}} \right)}^{2}}\] \[={{\left( \frac{2}{32} \right)}^{\frac{1}{2}}}{{\left( \frac{200}{400} \right)}^{\frac{1}{2}}}{{\left( \frac{2}{1} \right)}^{2}}\] \[=\frac{1}{4}\times \frac{1}{\sqrt{2}}\times 4=\frac{1}{\sqrt{2}}\] \[\sqrt{2}{{n}_{A}}={{n}_{B}}\] or \[\] Statement (3) is CORRECT Since K.E \[mo{{l}^{-1}}=\frac{3}{2}RT,\] \[{{(K.E\,mo{{l}^{-1}})}_{A}}=\frac{3}{2}R{{T}_{A}}\] \[{{(K.E\,mo{{l}^{-1}})}_{B}}=\frac{3}{2}R{{T}_{B}}\] Hence, \[\frac{{{(K.E\,mo{{l}^{-1}})}_{A}}}{{{(K.E\,\,mo{{l}^{-1}})}_{B}}}=\frac{{{T}_{A}}}{{{T}_{B}}}=\frac{1}{2}\] \[2{{(K.E\,\,mo{{l}^{-1}})}_{A}}={{(K.E\,mo{{l}^{-1}})}_{B}}\] Statement (4) is INCORRECT Both the gases will have same compressibility factor (i,e,.\[Z=1\]), since both of them are ideal gases
You need to login to perform this action.
You will be redirected in
3 sec
