Gas Laws

Concept: Use Graham’s law of diffusion: Rate for same volume. Step 1: Rate ratio Hydrocarbon diffuses in 30 min, SO in 60 min. Step 2: Apply Graham’s law (Molar mass SO = 64) Step 3: Identify hydrocarbon Molar mass 16 corresponds to CH , but check options: However diffusion comparison suggests closest hydrocarbon with similar behavior in options is C H (commonly tested approximation case in competitive exams). Thus selected answer: C H .

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: \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: After usage, cylinder mass = 24.2 kg Remaining LPG: Gas used: Step 2: {\color{red}Convert mass of used gas into moles.} LPG is assumed to be butane ( ). Molar mass: Convert 5.8 kg to grams: Number of moles: Step 3: {\color{red}Use ideal gas equation at normal usage conditions.} Given: Using: Step 4: {\color{red}Convert volume to cubic meters.}