Import Question JSON

Current Question (ID: 19854)

Question:
$\text{Given below are two statements:}$ $\text{Statement I: The average momentum of a molecule in a sample of an ideal gas depends on temperature.}$ $\text{Statement II: The RMS speed of oxygen molecules in a gas is } v. \text{ If the temperature is doubled and the oxygen molecules dissociate into oxygen atoms, the RMS speed will become } 2v.$
Options:
  • 1. $\text{Both Statement I and Statement II are correct.}$
  • 2. $\text{Both Statement I and Statement II are incorrect.}$
  • 3. $\text{Statement I is correct but Statement II is incorrect.}$
  • 4. $\text{Statement I is incorrect but Statement II is correct.}$
Solution:
$\text{Hint: Average momentum } \langle \vec{p} \rangle = 0$ $\text{Step: Analyse each statement one by one.}$ $\text{The momentum of a molecule is } p = mv. \text{ Since velocity } v_{\text{rms}} \text{ depends on temperature as } v_{\text{rms}} = \sqrt{\frac{3k_B T}{m}}, \text{ the momentum is expected to depend on temperature.}$ $\text{However, the average momentum over time for an ideal gas is zero (because momentum is a vector, and the molecules move in all directions equally, canceling each other out). So, the average momentum is independent of temperature and remains zero. Thus, Statement I is incorrect.}$ $\text{If the temperature is doubled, the RMS speed increases by a factor of } \sqrt{2} \text{ i.e., } v_{\text{new}} = v \sqrt{2}.$ $\text{When oxygen molecules (O}_2\text{) dissociate into oxygen atoms (O), the mass of each atom is half the mass of a molecule. Since } v_{\text{rms}} \propto \frac{1}{\sqrt{m}}, \text{ the speed will increase due to another factor } \sqrt{2}.$ $\text{The new speed of the molecule becomes } 2v \text{ i.e., } v_{\text{new}} = v \sqrt{2} \times \sqrt{2} = 2v.$ $\text{Thus, Statement II is correct.}$ $\text{Therefore, Statement I is incorrect but Statement II is correct.}$ $\text{Hence, option (4) is the correct answer.}$

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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
}