A cylinder contains 64 g of an ideal gas () at 27°C and 3 atm. In transportation, the cylinder fell — States of Matter and Gaseous State Chemistry Question
Question
A cylinder contains 64 g of an ideal gas ($M = 64$) at 27°C and 3 atm. In transportation, the cylinder fell down and a dent was created, i.e., the effective volume of the cylinder decreases. But the valve attached to the cylinder cannot keep the pressure greater than 3 atm, so 8 g of gas leaked out. The volume of the cylinder before and after the dent was $a$ l and $b$ l, respectively. If the valve were pretty strong, the pressure of gas in the cylinder after dent were $cd/7$ atm. Temperature remained constant during this process. The value of $abcd$ is
💡 Solution & Explanation
Using given $R = 0.08\text{ L-atm/K-mol}$. Initial moles $n_1 = 64/64 = 1\text{ mol}$. Initial volume $a = \frac{nRT}{P} = \frac{1 \times 0.08 \times 300}{3} = 8.0\text{ L}$, so $a=8$. After the dent and leakage, 8 g leaked, leaving $64-8=56\text{ g}$ ($0.875\text{ mol}$). The pressure settled at the valve's max of $3\text{ atm}$. Final volume $b = \frac{0.875 \times 0.08 \times 300}{3} = 7.0\text{ L}$, so $b=7$. If no gas leaked, 1 mole would be squeezed into 7 L. Pressure would be $P' = \frac{1 \times 0.08 \times 300}{7} = \frac{24}{7}\text{ atm}$. Thus $cd = 24$. The concatenated digits $abcd$ are 8724. Therefore, correct answer is 8724.