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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2010

  • question_answer
        Oxygen gas is contained in a cylinder of volume \[1\times {{10}^{-3}}{{m}^{3}}\]at a temperature of 320 K and a pressure of\[3.53\times {{10}^{5}}N/{{m}^{2}}\]. After some of the oxygen is used at constant temperature, the pressure falls to\[2.7\times {{10}^{5}}N/{{m}^{2}}\]. The mass of oxygen used is

    A)  0.01 kg                                

    B)  0.001 kg

    C)  0.002 kg                              

    D)  0.003 kg

    Correct Answer: B

    Solution :

                    Gas equation is, \[pV=nRT\] or            \[pV=\frac{m}{M}RT\] \[3.53\times {{10}^{5}}\times 1\times {{10}^{-3}}\]                 \[=\frac{m}{32\times {{10}^{-3}}}\times R\times 320\]           ...(i) On solving, \[m=\frac{3.53}{R\times {{10}^{2}}}kg\] Let\[x\]kg mass of oxygen is used at constant temperature, so                 \[pV=\frac{m-x}{M}RT\] \[2.7\times {{10}^{5}}\times 1\times {{10}^{-3}}=\frac{m-x}{32\times {{10}^{-3}}}R\times 320\]...(ii)                 \[m-x=\frac{2.70}{R\times {{10}^{2}}}kg\] From Eqs. (i) and (ii)                 \[\frac{m-x}{m}=\frac{270}{353}\] \[\Rightarrow \]               \[x=\frac{83}{353}m\]                 \[=\frac{83}{353}.\frac{3.53}{R\times {{10}^{2}}}\]                 \[=\frac{83}{353}\times \frac{3.53}{8.3\times {{10}^{2}}}\] \[=0.001\text{ }kg\]


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