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 daysB. 20 daysC. 50 daysD. 32 days

A cylinder of gas supplied by Bharat Petroleum is assumed to contain 1

1.	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 daysB. 20 daysC. 50 daysD. 32 days
Answer» 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.

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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 daysB. 20 daysC. 50 daysD. 32 days

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