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Current Question (ID: 21054)

Question:
$\text{A mixture contains one mole of a monoatomic gas and one mole of a diatomic gas.}$ $\text{What is the ratio of the heat capacities at constant volume } (C_V) \text{ to the heat capacities at constant pressure } (C_P) \text{ for the mixture } \left( \text{i.e. } \frac{C_V}{C_P} \right) ?$
Options:
  • 1. $\frac{2}{3}$
  • 2. $\frac{7}{5}$
  • 3. $\frac{5}{7}$
  • 4. $\frac{3}{5}$
Solution:
$\text{Hint: For monoatomic gases, the heat capacities are:}$ $C_V = \frac{3}{2}R; \quad C_P = \frac{5}{2}R.$ $C_v = \frac{1}{2} \left( \frac{3R}{2} \right) + \frac{1}{2} \left( \frac{5R}{2} \right)$ $= \frac{8R}{4} = 2R$ $C_p = \frac{1}{2} \left( \frac{5R}{2} \right) + \frac{1}{2} \left( \frac{7R}{2} \right)$ $= \frac{12R}{4} = 3R$ $\frac{C_v}{C_p} = \frac{2R}{3R}$ $= \frac{2}{3}$

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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
}