The weight of in a 9-L cylinder at 27oC temperature and 16 atm pressure is (R = 0.0821 L atm K–1 mol — States of Matter and Gaseous State Chemistry Question
Question
The weight of $CH_4$ in a 9-L cylinder at 27oC temperature and 16 atm pressure is (R = 0.0821 L atm K–1 mol–1)
💡 Solution & Explanation
# Solution: Finding the Mass of $CH_4$ **Step 1: Apply the Ideal Gas Law** $$PV = nRT$$ **Step 2: Substitute given values** - $P = 16$ atm - $V = 9$ L - $T = 27°C = 300$ K - $R = 0.0821$ L atm K$^{-1}$ mol$^{-1}$ $$n = \frac{PV}{RT} = \frac{16 \times 9}{0.0821 \times 300}$$ **Step 3: Calculate moles** $$n = \frac{144}{24.63} = 5.85 \text{ mol} \approx 5.84 \text{ mol}$$ **Step 4: Find mass using molar mass** Molar mass of $CH_4 = 12 + 4(1) = 16$ g/mol $$\text{Mass} = n \times M = 5.84 \times 16 = 93.4 \text{ g} \approx 93.6 \text{ g}$$ **Answer: The weight of $CH_4$ is approximately 93.4 g or 94 g** (Option B) This matches standard calculation precision where the answer typically rounds to **~94 g** or the exact value depends on rounding at each step.