Q. A cylinder of gas is assumed to contain 11.2 kg of butane $(C_{4}H_{10})$. If a normal family needs 20000 kJ of energy per day. The cylinder will last (Given that $\mathrm{\Delta}H$ for combustion of butane is $- 2658$ kJ)
A) 20 days
B) 25 days
C) 26 days
D) 24 days
Explanation
(c)
Cylinder contains 11.2 kg or 193.10 mole butane. $(\because$ molecular mass of butane =58)
$\because$ Energy released by 1 mole of butane $= - 2658$
$\therefore\$Energy released by 193.10 mole of butane
$= - 2658 \times 193.10$
$= 5.13 \times 10^{5}$ kJ
$\therefore\frac{5.13 \times 10^{5}}{20000} = 25.66\ or\ 26\ days$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.