A) \[{{N}_{A}}=8{{N}_{B}}\] (where \[{{N}_{A}}\]and \[{{N}_{B}}\] are the number of molecules in flask A and B respectively).
B) \[{{p}_{A}}=4{{p}_{B}}\] (where \[{{p}_{A}}\] and \[{{p}_{B}}\] are the pressure in flask A and B respectively)
C) \[{{({{\mu }_{av}})}_{A}}=4{{({{\mu }_{av}})}_{B}}\] and \[{{({{\mu }_{av}})}_{A}}\] are \[{{({{\mu }_{av}})}_{B}}\] average speed of molecules in flask A and B respectively)
D) \[{{X}_{A}}=16\,{{X}_{B}}\] (where \[{{X}_{A}}\] and \[{{X}_{B}}\] are the number of collisions per unit area per unit time in flask A and B respectively).
Correct Answer: C
Solution :
[c] Statement (1) is CORRECT Let \[{{n}_{A}}\] and \[{{n}_{B}}\] are the amount of \[{{H}_{2}}\] and \[C{{H}_{4}}\] respectively. \[{{n}_{A}}=\frac{m}{2},\,\,\,\,\,\,{{n}_{B}}=\frac{m}{16}\] Where m is the mass of the gas in flask A and B (which are equal). \[\frac{{{n}_{A}}}{{{n}_{B}}}=8\] The number of molecules in flasks A and B are: \[{{N}_{A}}=\frac{m}{2}N,\] \[{{N}_{B}}=\frac{m}{16}N,\] (Where N is Avogadro number) \[\frac{{{N}_{A}}}{{{N}_{B}}}\] \[{{N}_{A}}=8{{N}_{B}}\] Statement (2) is CORRECT Since, \[{{p}_{A}}={{n}_{A}}R{{T}_{A}}\] and \[{{p}_{B}}={{n}_{B}}R{{T}_{B}}\] \[\frac{{{p}_{A}}}{{{p}_{B}}}=\frac{{{n}_{A}}{{T}_{A}}}{{{n}_{B}}{{T}_{B}}}=8\times \frac{300}{600}=4\] \[{{p}_{A}}=4{{p}_{B}}\] Statement (3) is INCORRECT Since, \[{{({{\mu }_{av}})}_{A}}=\sqrt{\frac{8R{{T}_{A}}}{\pi {{M}_{A}}}}\] \[{{({{\mu }_{av}})}_{B}}=\sqrt{\frac{8R{{T}_{B}}}{\pi {{M}_{B}}}}\] \[\frac{{{({{\mu }_{av}})}_{A}}}{{{({{\mu }_{av}})}_{B}}}=\sqrt{\frac{{{T}_{A}}{{M}_{B}}}{{{T}_{B}}{{M}_{A}}}}=\sqrt{\frac{300K(0.016\,kg\,mo{{l}^{-1}})}{600K\,(0.002\,kg\,mo{{l}^{-1}})}}\] \[=\sqrt{\frac{300\times 16}{600\times 2}}=2\] \[{{({{\mu }_{av}})}_{A}}=2{{({{\mu }_{av}})}_{B}}\] Statement (4) is CORRECT Number of collisions per unit area per unit time, \[X=\left( \frac{1}{4} \right)N\,\,{{\mu }_{av}}\] \[{{X}_{A}}=\left( \frac{1}{4} \right){{N}_{A}}\,{{({{\mu }_{av}})}_{A}}\] \[{{X}_{B}}=\left( \frac{1}{4} \right){{N}_{B}}\,{{({{\mu }_{av}})}_{B}}\] Hence, \[\frac{{{X}_{A}}}{{{X}_{B}}}=\frac{{{N}_{A}}{{({{\mu }_{av}})}_{A}}}{{{N}_{B}}{{({{\mu }_{av}})}_{B}}}=8\times 2=16\] \[{{X}_{A}}=16\times {{X}_{B}},\]
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