Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2006

  • question_answer
    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   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,                 \[\mu =\frac{pV}{RT}\] 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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