हिंदी

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 last for ___ Days. -

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प्रश्न

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 last for ______ Days. (Δ Hc of C4 H10 = - 2658 JK per mole)

विकल्प

  • 20.69

  • 10.05

  • 25.06

  • 32.06

MCQ
रिक्त स्थान भरें

उत्तर

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 last for 32.06 Days.

Explanation:

Calorific value of butane `= (Delta "H"_"c")/("mol." " wt.")`

`= 2658/58`

= 45.8 kJ/g

Cylinder consist 14 kg of butane means 14000 g of butane

∴ 1 g gives = 45.8 kJ/g

∴ 14000 g gives = 14000 × 45.8 = 641200 kJ

Family need 20,000 kJ/day

So gas full fill the requirement for `641200/20000` = 32.06 days

shaalaa.com
Enthalpy Change, ∆_rH of a Reaction - Reaction Enthalpy - Thermochemical Equations
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