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Current Question (ID: 11010)

Question:
$\text{A closed hollow insulated cylinder is filled with gas at } 0^{\circ}\text{C and also contains an insulated piston of negligible weight and negligible thickness at the middle point. The gas on one side of the piston is heated to } 100^{\circ}\text{C. If the piston moves } 5\text{ cm, the length of the hollow cylinder will be:}$
Options:
  • 1. $13.65 \text{ cm}$
  • 2. $27.3 \text{ cm}$
  • 3. $38.6 \text{ cm}$
  • 4. $64.6 \text{ cm}$ (Correct)
Solution:
$\text{Using Charle's law, we have } \frac{V}{T} = \text{constant}$ \n\n $\text{Let the original length of the cylinder be } l \text{ and the cross-sectional area be } A \text{.}$ \n\n $\text{The initial volume of the gas on each side is } V_1 = V_2 = A \cdot \frac{l}{2}.$ \n\n $\text{The initial temperatures are } T_1 = 0^{\circ}\text{C} = 273\text{ K} \text{ and } T_2 = 0^{\circ}\text{C} = 273\text{ K}.$ \n\n $\text{After heating one side, the new temperatures are } T_1' = 100^{\circ}\text{C} = 373\text{ K} \text{ and } T_2' = 0^{\circ}\text{C} = 273\text{ K}.$ \n\n $\text{As the piston moves } 5 \text{ cm, the new lengths of the two sides will be } (\frac{l}{2} + 5) \text{ and } (\frac{l}{2} - 5).$ \n\n $\text{The new volumes are } V_1' = A(\frac{l}{2} + 5) \text{ and } V_2' = A(\frac{l}{2} - 5).$ \n\n $\text{Applying Charle's law to the two sides of the piston (assuming pressure remains constant), we get:}$ \n\n $\frac{V_1'}{T_1'} = \frac{V_2'}{T_2'}$ \n\n $\frac{A(\frac{l}{2}+5)}{373} = \frac{A(\frac{l}{2}-5)}{273}$ \n\n $\frac{\frac{l}{2}+5}{373} = \frac{\frac{l}{2}-5}{273}$ \n\n $\text{Cross-multiplying:}$ \n\n $273(\frac{l}{2}+5) = 373(\frac{l}{2}-5)$ \n\n $136.5l + 1365 = 186.5l - 1865$ \n\n $186.5l - 136.5l = 1365 + 1865$ \n\n $50l = 3230$ \n\n $l = \frac{3230}{50} = 64.6 \text{ cm}.$ \n\n $\text{The length of the hollow cylinder is } 64.6 \text{ cm}.$ \n\n $\text{The correct option is 4.}$

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{
  "question": "The mass of carbon present in 0.5 mole of $\\mathrm{K}_4[\\mathrm{Fe(CN)}_6]$ is:",
  "options": [
    {
      "id": 1,
      "text": "1.8 g"
    },
    {
      "id": 2,
      "text": "18 g"
    },
    {
      "id": 3,
      "text": "3.6 g"
    },
    {
      "id": 4,
      "text": "36 g"
    }
  ],
  "solution": "\\begin{align}\n&\\text{Hint: Mole concept}\\\\\n&1 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\text{ moles of carbon atom}\\\\\n&0.5 \\text{ mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6] = 6 \\times 0.5 \\text{ mol} = 3 \\text{ mol}\\\\\n&1 \\text{ mol of carbon} = 12 \\text{ g}\\\\\n&3 \\text{ mol carbon} = 12 \\times 3 = 36 \\text{ g}\\\\\n&\\text{Hence, 36 g mass of carbon present in 0.5 mole of } \\mathrm{K}_4[\\mathrm{Fe(CN)}_6].\n\\end{align}",
  "correct_answer": 4
}