Question

A cylinder of gas supplied by Bharat Petroleum is assumed to contain 14 kg of butane. It a normal family requires 20,000 kJ of energy per day for cooking, butane gas in the cylinder last for … Days
`(DeltaH_(c)" of "C_(4)H_(10)=-2658 "KJ per mole")`
A. 15 days
B. 20 days
C. 50 days
D. 32 days

Answer 1

Correct Answer - D
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.