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Current Question (ID: 8141)

Question:

$\text{A sealed tube which can withstand a pressure of } 3 \text{ atmosphere is filled with air at } 27^{\circ}\text{C and } 760\text{ mm pressure. The tube will burst at -}$

Options:

1. $900^{\circ}\text{C}$
2. $627^{\circ}\text{C}$
3. $726^{\circ}\text{C}$
4. $1173^{\circ}\text{C}$

Solution:

$\text{Hint: Follow Gay Lussac's Law} \text{Step 1:} \text{Gay Lussac's law states that at constant volume, pressure of a fixed amount of a gas varies directly with the temperature.} $\frac{\text{P}_1}{\text{T}_1} = \frac{\text{P}_2}{\text{T}_2}$ \text{The given values are as follows:} \text{P}_1 = 760\text{ mm} = 1\text{ atm} \text{T}_1 = 300\text{ K} \text{P}_2 = 3\text{ atm} \text{Step 2:} \text{The tube will burst when the final pressure exceeds } 3\text{ atm. at constant volume. Calculate the final temperature as follows:} $\frac{\text{P}_1}{\text{T}_1} = \frac{\text{P}_2}{\text{T}_2}\text{ i.e.} \frac{1}{300}\text{ K} = \frac{3}{\text{T}_2} \text{T}_2 = 900\text{ K} = 627^{\circ}\text{C}$

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Upload a JSON file containing LaTeX/MathJax formatted question, options, and solution.

Expected JSON Format:

{
  "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
}