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BHU PMT BHU PMT (Screening) Solved Paper-2006

  • question_answer
    A cylinder of gas supplied by Bharat petroleum is assumed to contain 14 kg of butane. It is normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder last for......... days. (\[\Delta H,\]of \[{{C}_{4}}{{H}_{10}}=-2658\text{ }kJ\]per mol)

    A)  15 days                               

    B)  20 days

    C)  32 days                               

    D)  40 days

    Correct Answer: C

    Solution :

                     \[Calorific\text{ }value\text{ }of\text{ }butane=\frac{\Delta {{H}_{c}}}{molecular\text{ }wt.}\] \[=\frac{2658}{58}=45.8\,kJ/g\] Cylinder consist 14 kg of butane means 14000 g of butane \[\because \]     1 g gives = 45.8 kJ \[\therefore \]14000 g gives\[=14000\times 45.8=641200\text{ }kJ\] Family needs 20,000 kJ/day So,   gas   fulfill   the   requirement   for\[\frac{641200}{20,000}=32.06\,days\]


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