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NEET AIPMT SOLVED PAPER MAINS 2010

  • question_answer
    The pressure exerted by 6.0 g of methane gas in a \[0.03\,\,{{m}^{3}}\] vessel at \[129{}^\circ \text{ }C\] is (Atomic masses: \[C=12.01,H=1.01\,\]and\[R=8.314J{{K}^{-1}}\,mo{{l}^{-1}}\]

    A)  215216 Pa          

    B)  13409 Pa            

    C)  41648 Pa            

    D)  31684 Pa

    Correct Answer: C

    Solution :

    Given, volume, \[V=0.03\,\,{{m}^{3}}\] temperature, T = 129 + 273 = 402 K mass of methane, w = 6.0 g mol. mass of methane, \[M=12.01+4\times 1.01=16.05\] From, ideal gas equation,                 \[pV=nRT\] \[p=\frac{6}{16.05}\times \frac{8.314\times 402}{0.03}=41648Pa\]


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