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5. Thermodynamics and thermochemistry — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry5. Thermodynamics and thermochemistry1 Mark
MCQ
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 $\Delta H$ for combustion of butane is $ -2658\, kJ$)....$days$
  • A
    $20$
  • B
    $25$
  • ✓
    $26 $
  • D
    $24 $
✓

Answer

Correct option: C.
$26 $
c
Molar mass of butane $=58 \,g / mol$

$58 g$ of butane gives $-2658 \,kJ$ of heat energy.

$\therefore 11.2 \,kg$ of butane will give heat energy $=\frac{2658}{58} \times 11200=513.27 \times 10^3 \,\,kJ$

Daily energy required for cooking $=20000 \,kJ =2 \times 10^4\, kJ$ per day

$\therefore$ Number of days cylinder will last $=\frac{513.27 \times 10^3}{2 \times 10^4} \approx 26\, days$

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