JEE Main & Advanced Chemistry States of Matter Question Bank Kinetic molecular theory of gases and Molecular collisions

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    Ratio of \[{{C}_{p}}\] and \[{{C}_{v}}\] of a gas 'X' is 1.4. The number of atoms of the gas 'X' present in 11.2 litres of it at N.T.P. is [CBSE PMT 1989]

    A)                 \[6.02\times {{10}^{23}}\]           

    B)                 \[1.2\times {{10}^{24}}\]

    C)                 \[3.01\times {{10}^{23}}\]           

    D)                 \[2.01\times {{10}^{23}}\]

    Correct Answer: A

    Solution :

               Since \[\frac{{{C}_{P}}}{{{C}_{V}}}=1.4\], the gas should be diatomic.                    If volume is 11.2 lt then,  no. of moles = \[\frac{1}{2}\]                    \ no. of molecules = \[\frac{1}{2}\]´ Avagadro?s No.                    no. of atoms = 2 ´ no. of molecules                    2 ´ \[\frac{1}{2}\]´ Avagadro?s No.                                 \[=6.0223\times {{10}^{23}}\]

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