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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 19

  • question_answer
    A \[2000\text{ }cc\]sample of \[{{O}_{2}}\] is confined to a cylinder. Initially, pressure of the gas is \[{{10}^{5}}\text{ }N/{{m}^{2}}\]and its temperature\[27{}^\circ C\]. The gas is subjected to a cyclic process. In the first step, its pressure is made twice at constant volume. In the second step, it expands to its initial pressure adiabatically and in the final step, the gas undergoes isobaric compression and finally attains its initial volume. Consider that the gas is ideal. Temperature of the gas at the ideal of the first step and volume of the gas at the end of the second step will be -

    A) \[177{}^\circ C,123cc\]     

    B)    \[327{}^\circ C,3272cc\]

    C) \[127{}^\circ C,1278cc\]    

    D)  \[148{}^\circ C,723cc\]

    Correct Answer: B

    Solution :

    \[P\propto T\] or \[\frac{{{P}_{2}}}{{{P}_{1}}}=\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\frac{2P}{P}=\frac{{{T}_{2}}}{300K}\] \[{{T}_{2}}=600K\] \[\gamma =7/5,\] \[{{P}_{1}}=2P,\] \[{{P}_{2}}=P,\] \[P{{V}^{\gamma }}=\] constant \[{{P}_{1}}{{V}_{1}}^{\gamma }={{P}_{2}}{{V}_{2}}^{\gamma }\]   or   \[2P{{[2000]}^{\gamma }}=P{{V}_{2}}^{\gamma }\] \[{{2}^{\gamma }}\times 2000=\frac{1}{2}\] \[{{V}_{2}}={{(2)}^{7/5}}\times 2000=3272cc\]


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