States of Matter (Gaseous and Liquid States)

Weight of empty LPG cylinder = 14.8 kg Weight of full LPG cylinder = 29 kg ∴ 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 equaiton, pV = nRT or $p V = \frac{\text{Weight of solute}}{\text{Molecular mass of solute}} \times R T$ or$p V = \frac{W}{M} \times R T$ Applying ideal gas to LPG cylinder when gas is full, $p V = n R T$ $3.47 \text{atm} \times V = \frac{14.2 \text{kg}}{M} \times RT$ ...(i) Applying ideal gas to LPG cylinder when gas is reduced to 23 kg at ambient temperature, $p V = n R T$ $p \times V = \frac{8.2 \text{kg}}{M} \times RT$ ...(ii) Divide Eq. (i) by (ii) $\frac{3.47 \times V}{p \times V} = \frac{\frac{14.2 \text{kg} RT}{M}}{\frac{8.2 \text{kg} \times RT}{M}}$ $\frac{3.47}{p} = \frac{14.2}{8.2}$ ⇒ $p = \frac{3.47 \times 41}{71} = 2.003 \text{atm}$ Hence, answer is 2.