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Current Question (ID: 19836)

Question:
$\text{For a given at 1 atm pressure, the rms speed of the molecules is } 200 \text{ m/s at } 127^\circ\text{C.}$ $\text{At 2 atm pressure and at } 227^\circ\text{C, the rms speed of the molecules will be:}$
Options:
  • 1. $100 \text{ m/s}$
  • 2. $80\sqrt{5} \text{ m/s}$
  • 3. $100\sqrt{5} \text{ m/s}$
  • 4. $80 \text{ m/s}$
Solution:
$\text{Hint: The RMS speed depends on the temperature but not on the pressure.}$ $\text{Step 1: Find the relation between the RMS speed and the temperature.}$ $\text{The RMS velocity of a gas molecule is given by:}$ $v_{\text{rms}} = \sqrt{\frac{3RT}{m}}$ $\Rightarrow v_{\text{rms}} \propto \sqrt{T}$ $\text{Step 2: Find the final RMS speed of the molecules.}$ $\text{For two different cases:}$ $\Rightarrow \frac{v_{\text{rms1}}}{v_{\text{rms2}}} = \sqrt{\frac{T_1}{T_2}}$ $\frac{200}{v_{\text{rms2}}} = \sqrt{\frac{400}{500}} = \sqrt{\frac{4}{5}}$ $\Rightarrow v_{\text{rms2}} = \frac{\sqrt{5}}{2} \times 200$ $v_{\text{rms2}} = 100\sqrt{5} \text{ ms}^{-1}$ $\text{Hence, option (3) is the correct answer.}$

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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
}