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Current Question (ID: 8171)

Question:

An open flask containing air is heated from 300 K to 500 K. The percentage of air that will escape into the atmosphere if pressure is kept constant is-

Options:

1. 80
2. 40
3. 60
4. 20

Solution:

**Step 1:** Understand the conditions for an open flask. - Since the flask is open, the pressure (\( P \)) remains constant, and the volume (\( V \)) adjusts to maintain equilibrium with the atmosphere. - The relationship between moles of gas and temperature is given by: \[ \frac{1}{n_1 T_1} = \frac{1}{n_2 T_2} \] where \( n_1 \) and \( n_2 \) are the initial and final moles of gas, and \( T_1 \) and \( T_2 \) are the initial and final temperatures, respectively. **Step 2:** Apply the given values. - Initial temperature (\( T_1 \)) = 300 K - Final temperature (\( T_2 \)) = 500 K - Let the initial moles of air be \( n \). Substitute into the equation: \[ \frac{1}{n \times 300} = \frac{1}{n_2 \times 500} \] \[ n_2 = \frac{300}{500} n = \frac{3}{5} n \] **Step 3:** Calculate the moles of air that escape. \[ \text{Escaped moles} = n - n_2 = n - \frac{3}{5}n = \frac{2}{5}n \] **Step 4:** Calculate the percentage of air that escapes. \[ \text{Percentage escaped} = \left( \frac{\frac{2}{5}n}{n} \right) \times 100 = 40\% \] Hence, **40%** of the air will escape into the atmosphere.

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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
}