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Current Question (ID: 11136)

Question:

$\text{If at a pressure of } 10^{6} \ \text{dyne/cm}^{2}\text{, one gram of nitrogen occupies } 2 \times 10^{4} \ \text{c.c. volume, then the average energy of a nitrogen molecule in erg is:}$

Options:

1. $\text{14} \times 10^{-13}$ (Correct)
2. $\text{10} \times 10^{-12}$
3. $10^{6}$
4. $\text{2} \times 10^{6}$

Solution:

$\text{Hint: Use the relation, } \text{v}_{\text{rms}} = \sqrt{\frac{3\text{P}}{\rho}} \\ \\ \text{Step 1: Find the RMS speed.} \\ \text{Here, } \\ \text{P} = 10^{6} \ \text{dyne/cm}^{2} \\ \text{Now as, } \\ \text{m} = \frac{\text{M}}{\text{N}_{\text{A}}} = \frac{28}{6 \times 10^{23}} \text{ g} \\ \text{And, } \\ \rho = \frac{\text{Mass}}{\text{Volume}} = \frac{1}{2 \times 10^{4}} \ \text{g/cm}^{3} \\ \text{Where, } \\ \text{v}_{\text{rms}} = \sqrt{\frac{3\text{P}}{\rho}} = \sqrt{\frac{3 \times 10^{6}}{\frac{1}{2 \times 10^{4}}}} = \sqrt{3 \times 10^{6} \times 2 \times 10^{4}} = \sqrt{6 \times 10^{10}} = 10^{5}\sqrt{6} \\ \text{Step 2: Find the average energy using, } \\ \text{K} = \frac{1}{2}\text{m}v_{\text{rms}}^{2} \\ \text{So,} \\ \text{K} = \frac{1}{2} \times \frac{28}{6 \times 10^{23}} \times \left(10^{5}\sqrt{6}\right)^{2} \\ = \frac{1}{2} \times \frac{28}{6 \times 10^{23}} \times 6 \times 10^{10} \\ = \frac{28}{2 \times 10^{13}} = \frac{14}{10^{13}} = 14 \times 10^{-13} \text{erg.} \\ \text{Hence, the average energy of a nitrogen molecule is } 14 \times 10^{-13} \text{ erg. Therefore, option 1 is correct.}$

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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
}