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Current Question (ID: 11144)

Question:

$\text{The root mean square velocity of the molecules of a gas is 300 m/s. What will be the root mean square speed of the molecules if the atomic weight is doubled and the absolute temperature is halved?}$

Options:

1. $300 \text{ m/s}$
2. $150 \text{ m/s}$ (Correct)
3. $600 \text{ m/s}$
4. $75 \text{ m/s}$

Solution:

$\text{Hint: } v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ $\text{Step: Find the root mean square speed of the molecules.}$ $\text{Use the formula for RMS speed.}$ $v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$ $v_{\text{rms}} \propto \sqrt{\frac{T}{M}}$ $\frac{v_2}{v_1} = \sqrt{\frac{T_2}{T_1} \times \frac{M_1}{M_2}} = \frac{1}{2}$ $\Rightarrow v_2 = \frac{300}{2} = 150 \text{ m/s}$ $\text{Hence, option (2) is the correct answer.}$

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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
}