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BHU PMT BHU PMT (Mains) Solved Paper-2011

  • question_answer
    Heat absorbed by a mass of 1 g of helium, when its temperature rises from\[11{}^\circ C\]to \[131{}^\circ C\]at constant volume, is H. Heat absorbed by 7 g of nitrogen, when its temperature rises from\[11{}^\circ C\]to\[71{}^\circ C\]at constant volume, is\[{{H}_{N}}\]. The ratio of H to\[{{H}_{N}}\] is

    A) \[\frac{3}{2}\]                                   

    B)  \[\frac{4}{3}\]

    C)  \[\frac{6}{5}\]  

    D)  \[\frac{8}{7}\]  

    Correct Answer: C

    Solution :

                     Heat absorbed at constant volume \[H=n{{C}_{V}}\Delta T\] where,\[n=\]number of moles\[=\frac{m}{M}\] Heat absorbed by helium \[H=\frac{1}{4}\times \frac{3}{2}R(131-11)\]                 \[=\frac{1}{4}\times \frac{3}{2}R\times 120\]                 \[=45R\] [For the (monoatomic gas),\[{{C}_{V}}=\frac{3}{2}R\]] Heat absorbed by nitrogen                 \[{{H}_{N}}=\frac{7}{28}\times \frac{5}{2}R(71-11)\] \[=\frac{1}{4}\times \frac{5}{2}R\times 60\]                 \[=\frac{75}{2}R\] [For\[{{H}_{2}}\](diatomic gas)\[{{C}_{V}}=\frac{5}{2}R\]] \[\frac{H}{{{H}_{N}}}=\frac{45R}{(75/2)R}=\frac{6}{5}\]


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