Question

An LPG (Liquefied Petroleum Gas) cylinder weighs 15.0 kg when empty. When full, it weighs 30.0 kg and shows a pressure of 3.0 atm. In the course of usage at $27^\circ$C, the mass of the full cylinder is reduced to 24.2 kg. The volume of the used gas in the normal usage condition (1 atm and $27^\circ$C) is (assume LPG to be normal butane and it behaves ideally):

• A. $24.6 \text{ m}^3$
• B. $246 \text{ m}^3$
• C. $0.246 \text{ m}^3$
• D. $2.46 \text{ m}^3$

Hint:

For ideal gas problems: \begin{itemize} \item First convert mass $\rightarrow$ moles \item Use $PV = nRT$ directly at required conditions \item Same temperature simplifies calculations \end{itemize}

Answer 1

The Correct Option is D

Concept: The problem involves: \begin{itemize} \item Mass difference to find gas used \item Conversion of mass to moles using molar mass \item Ideal gas equation: \[ PV = nRT \] \item Volume comparison at different pressures but same temperature \end{itemize} Step 1: {\color{red}Find mass of LPG initially and after usage.} Empty cylinder mass = 15 kg Full cylinder mass = 30 kg So, initial LPG mass: \[ 30 - 15 = 15 \text{ kg} \] After usage, cylinder mass = 24.2 kg Remaining LPG: \[ 24.2 - 15 = 9.2 \text{ kg} \] Gas used: \[ 15 - 9.2 = 5.8 \text{ kg} \] Step 2: {\color{red}Convert mass of used gas into moles.} LPG is assumed to be butane ($C_4H_{10}$). Molar mass: \[ 4(12) + 10(1) = 58 \text{ g/mol} \] Convert 5.8 kg to grams: \[ 5.8 \text{ kg} = 5800 \text{ g} \] Number of moles: \[ n = \frac{5800}{58} = 100 \text{ mol} \] Step 3: {\color{red}Use ideal gas equation at normal usage conditions.} Given: \[ P = 1 \text{ atm}, \quad T = 27^\circ C = 300 \text{ K} \] \[ R = 0.0821 \text{ L·atm/mol·K} \] Using: \[ V = \frac{nRT}{P} \] \[ V = \frac{100 \times 0.0821 \times 300}{1} \] \[ V = 2463 \text{ L} \] Step 4: {\color{red}Convert volume to cubic meters.} \[ 1000 \text{ L} = 1 \text{ m}^3 \] \[ V = \frac{2463}{1000} = 2.463 \text{ m}^3 \] \[ V \approx 2.46 \text{ m}^3 \]