1.

A cylinder of gas supplied by Bharat Petroleum is assumed to contain 14 kg of butane. If a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder lasts (Delta_(C ) H^(@) of C_(4) H_(10) = - 2658 kJ mol^(-1))

Answer»

15 DAYS
20 days
50 days
32 days

Solution :Calorific value of butan = `(DeltaH_(c))/("MOL. wt.")=(2658)/(58)=45.8 KJ//gm`
Cylinder consist 14 Kg of butane means 14000 gm of butane
`because` 1 gm gives 45.8 kJ
`therefore` 14000 gm gives`14000xx45.8 = 641200 kJ`
So gas full fill the requirement for `(641200)/(20,000)=32.06` days.


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