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Current Question (ID: 11153)

Question:

$\text{The root mean square speed of the molecules of a diatomic gas is } v. \text{ When the temperature is doubled, the molecules dissociate into two atoms. The new root mean square speed of the atom is:}$

Options:

1. $\sqrt{2}v$
2. $v$
3. $2v$ (Correct)
4. $4v$

Solution:

$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}. \text{ According to problem T will become 2T and M will become M/2 so the value of } v_{\text{rms}} \text{ will increase by } \sqrt{4} = 2 \text{ times i.e. new root mean square velocity will be 2v.}$ $\text{Note: This is video solution for a similar question}$

Import JSON File

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
}