question_answer

Two flasks A and B have equal volumes. A is maintained at \[400\text{ }K\] and B at \[800\text{ }K\]. While A contains \[{{H}_{2}}\] gas, B has an equal mass of \[C{{H}_{4}}\] gas. Assuming ideal behaviour for both the gases (Given: (i) \[2{{\lambda }_{A}}={{\lambda }_{B}}\] (\[{{\lambda }_{A}}\] and \[{{\lambda }_{B}}\]are mean free path of molecules in flask A and B respectively.) \[{{Z}_{11}}(A)\] and \[{{Z}_{11}}(B)\] are the total number of bimolecular collisions per unit volume per unit time in flask A and B respectively. Select the CORRECT statement

A) \[{{Z}_{11}}(A)=32\,\,{{Z}_{11}}(B)\]    

B) \[32\,{{Z}_{11}}(A)=\,\,{{Z}_{11}}(B)\]   

C) \[{{Z}_{11}}(A)=\,16\,{{Z}_{11}}(B)\]    

D) \[16\,{{Z}_{11}}(A)=\,{{Z}_{11}}(B)\]

Correct Answer: A

Solution :

[a] Since \[{{Z}_{1}}=\frac{{{\mu }_{av}}}{{{\lambda }_{A}}},\] \[{{Z}_{1A}}=\frac{{{\mu }_{av}}(A)}{{{\lambda }_{A}}}\] and \[{{Z}_{1B}}=\frac{{{\mu }_{av(A)}}}{{{\lambda }_{A}}}\] \[\frac{{{\mu }_{av}}(A)}{{{\mu }_{av}}(B)}\,\,\,\sqrt{\frac{{{T}_{A}}}{{{M}_{A}}}\times \frac{{{M}_{B}}}{{{T}_{B}}}}=\sqrt{\frac{400}{2}\times \frac{16}{800}}=2\] \[\frac{{{\lambda }_{A}}}{{{\lambda }_{B}}}=\frac{1}{2}\]  (Given) \[\frac{{{Z}_{1}}_{A}}{{{Z}_{1}}_{B}}=\left( \frac{{{\mu }_{av(A)}}}{{{\mu }_{av(B)}}} \right)\left( \frac{{{\lambda }_{B}}}{{{\lambda }_{A}}} \right)=2\times 2=4\] To calculate \[\left( N_{A}^{*}/N_{B}^{*} \right):\] Let \[{{n}_{A}}\] and \[{{n}_{B}}\] are the amount of \[{{H}_{2}}\] and \[C{{H}_{4}}\] \[{{n}_{A}}=\frac{m}{2},\]    \[{{n}_{B}}=\frac{m}{16},\]    \[\frac{{{n}_{A}}}{{{n}_{B}}}=8\] The number of molecules in flasks A and B are: \[N_{A}^{*}=\frac{m}{2}{{N}_{A}},\]   \[N_{B}^{*}=\frac{m}{16}{{N}_{A}}\] \[\frac{N_{A}^{*}}{N_{B}^{*}}=8\] Since \[{{Z}_{11}}=\frac{1}{2}N*{{Z}_{1}}\] \[\frac{{{Z}_{11(A)}}}{{{Z}_{11(B)}}}=\left( \frac{N_{A}^{*}{{Z}_{1A}}/2}{N_{B}^{*}{{Z}_{1B}}/2} \right)=\left( \frac{N_{A}^{*}}{N_{B}^{*}} \right)\left( \frac{{{Z}_{1A}}}{{{Z}_{1B}}} \right)\] \[=8\times 4=32\] \[{{Z}_{11(A)}}=32{{Z}_{11(B)}}\]