RAJASTHAN PMT Rajasthan - PMT Solved Paper-2009

  • question_answer
    42.          Two containers of equal volume contain the same gas at pressure\[{{P}_{1}}\]and\[{{P}_{2}}\]and absolute                 temperature\[{{T}_{1}}\]and\[{{T}_{2}}\]respectively. On joining the vessels the gas reaches a common pressure p and common temperature T. The      ratio p/T is equal to

    A)  \[\frac{{{P}_{1}}}{{{T}_{1}}}+\frac{{{P}_{2}}}{{{T}_{2}}}\]                             

    B)  \[\frac{{{p}_{1}}{{T}_{1}}+{{p}_{2}}{{T}_{2}}}{{{({{T}_{1}}+{{T}_{2}})}^{2}}}\]

    C)  \[\frac{{{p}_{1}}{{T}_{2}}+{{p}_{2}}{{T}_{1}}}{{{({{T}_{1}}+{{T}_{2}})}^{2}}}\]      

    D)         \[\frac{{{p}_{1}}}{2{{T}_{1}}}\,+\frac{{{p}_{2}}}{2{{T}_{2}}}\]

    Correct Answer: D

    Solution :

    From ideal gas equation,                 \[pV=\mu RT\] \[\therefore \]  \[\mu =\frac{pV}{RT}\] \[\therefore \]Number of moles of gas in first container,                 \[{{\mu }_{1}}=\frac{{{p}_{1}}V}{R{{T}_{1}}}\] Number of moles of gas in second container,                 \[{{\mu }_{2}}=\frac{{{p}_{2}}V}{R{{T}_{2}}}\] Number of moles in containers when joined with each other,                 \[\mu =\frac{{{p}_{2}}V}{R{{T}_{2}}}\] But,        \[\mu ={{\mu }_{1}}+{{\mu }_{2}}\]                 \[\frac{p(2V)}{RT}=\frac{{{p}_{1}}V}{R{{T}_{1}}}+\frac{{{p}_{2}}V}{R{{T}_{2}}}\]                 \[\frac{2p}{T}=\frac{{{p}_{1}}}{{{T}_{1}}}+\frac{{{p}_{2}}}{{{T}_{2}}}\]                 \[\frac{P}{T}=\frac{{{p}_{1}}}{2{{T}_{1}}}+\frac{{{p}_{2}}}{2{{T}_{2}}}\]  


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