QJEEM21Che6ff: JEE Main 2021 Chemistry - Class 11 JEEM21Che6ff Solution: An empty LPG cylinder weighs 14.8 kg. When full, it weighs 29.0 …

Weight of empty LPG cylinder = 14.8 kg

Weight of full LPG cylinder = 29 kg

$\therefore$ Weight of gas = 29 $-$ 14.8 = 14.2 kg

If weight of full LPG cylinder = 23 kg

then weight of gas used = 29 $-$ 23 = 6 kg at ambient temperature.

From ideal gas equation, pV = nRT

or $pV = {{Weight\,of\,solute} \over {Molecular\,mass\,of\,solute}} \times RT$

or $pV = {W \over M} \times RT$

Applying ideal gas to LPG cylinder when gas is full,

$pV = nRT$

$3.47\,atm \times V = {{14.2kg} \over M} \times RT$ .... (i)

Applying ideal gas to LPG cylinder when gas is reduced to 23 kg at ambient temperture,

$pV = nRT$

$p \times V = {{8.2kg} \over M} \times RT$ .... (ii)

Divide Eq. (i) by (ii)

${{3.47 \times V} \over {p \times V}} = {{{{14.2kg\,RT} \over M}} \over {{{8.2kg\, \times RT} \over M}}}$

${{3.47} \over p} = {{14.2} \over {8.2}}$

$\Rightarrow p = {{3.47 \times 41} \over {71}} = 2.003$ atm

Hence, answer is 2.