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Current Question (ID: 19858)

Question:
$N$ \text{ moles of non-linear polyatomic gas (degree of freedom } \delta) \text{ is mixed with } 2 \text{ moles of monoatomic gas. The resultant mixture has molar-specific heat equal to that of a diatomic gas, then the number of moles } (N) \text{ is:}$
Options:
  • 1. $4$
  • 2. $5$
  • 3. $6$
  • 4. $3$
Solution:
$\text{Hint: } C_{V_{\text{mix}}} = \frac{n_1 \times C_{V_1} + n_2 \times C_{V_2}}{n_1 + n_2}$ $\frac{n_1 \frac{1}{2} R + n_2 \frac{5}{2} R}{n_1 + n_2} = \frac{5}{2} R$ $\frac{2 \times \frac{3}{2} R + N \times \frac{6}{2} R}{N + 2} = \frac{5}{2} R$ $\frac{6 + 6N}{N + 2} = 5$ $6 + 6N = 5N + 10$ $N = 4$

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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
}