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Current Question (ID: 11145)

Question:

$\text{The molecules of a given mass of gas have rms velocity of 200 ms}^{-1} \text{ at 27°C and 1.0 × 10}^5 \text{ Nm}^{-2} \text{ pressure. When the temperature and pressure of the gas are increased to, respectively, 127°C and 0.05 × 10}^5 \text{ Nm}^{-2}\text{, rms velocity of its molecules in ms}^{-1} \text{ will become:}$

Options:

1. $\frac{400}{\sqrt{3}}$ (Correct)
2. $\frac{100\sqrt{2}}{3}$
3. $\frac{100}{3}$
4. $\frac{100}{\sqrt{2}}$

Solution:

$\text{It is given that,}$ $v_{\text{rms}} = 200 \text{ ms}^{-1}, T_1 = 300\text{K}, P_1 = 10^5 \text{ N/m}^2$ $T_2 = 400\text{K}, P_2 = 0.05 \times 10^5 \text{ N/m}^2$ $v_{\text{rms}} = \sqrt{\frac{3RT}{m}}$ $\therefore v_{\text{rms}} \propto \sqrt{T}$ $\text{For two different cases} \Rightarrow$ $\frac{v_{\text{rms}_1}}{v_{\text{rms}_2}} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{400}}$ $v_{\text{rms}_2} = \frac{400}{\sqrt{3}} \text{ m/s}$

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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
}