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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 25

  • question_answer
    \[28\text{ }g\]of \[{{N}_{2}}\] gas is contained in a flask at a pressure of \[10\text{ }atm\]and at a temperature of\[57{}^\circ \]. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature reduced to \[27{}^\circ C\]. The quantity of \[{{N}_{2}}\]gas that is leaked out is

    A) \[11/20\text{ }g\]          

    B)       \[20/11\text{ }g\]           

    C)   \[5/63\text{ }g\]              

    D)        \[63/5\text{ }g\]

    Correct Answer: D

    Solution :

    [d] \[PV=\frac{m}{M}RT\] \[P\propto m.T\] \[\frac{10}{5}=\frac{28}{m}\times \frac{330}{300}\] \[\Rightarrow \,\,\,\,m=\frac{28\times 330\times 5}{10\times 300}=\frac{231}{15}gm=\frac{77}{5}gm\] Gas leaked \[=28-\frac{77}{5}=\frac{63}{5}gm\]


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