Import Question JSON

Current Question (ID: 19843)

Question:

$\text{Match the } \frac{C_P}{C_V} \text{ ratio for ideal gases with different type of molecules:}$ $\begin{array}{|c|c|} \hline \text{Column I} & \text{Column II} \\ \hline \text{(A) Monatomic} & \text{(I) } \frac{7}{5} \\ \text{(B) Diatomic rigid molecules} & \text{(II) } \frac{9}{7} \\ \text{(C) Diatomic non-rigid molecules} & \text{(III) } \frac{4}{3} \\ \text{(D) Triatomic rigid molecules} & \text{(IV) } \frac{5}{3} \\ \hline \end{array}$

Options:

1. $(\text{A}-(\text{III}), \text{B}-(\text{IV}), \text{C}-(\text{II}), \text{D}-(\text{I}))$
2. $(\text{A}-(\text{I}), \text{B}-(\text{III}), \text{C}-(\text{I}), \text{D}-(\text{IV}))$
3. $(\text{A}-(\text{IV}), \text{B}-(\text{II}), \text{C}-(\text{I}), \text{D}-(\text{III}))$
4. $(\text{A}-(\text{IV}), \text{B}-(\text{I}), \text{C}-(\text{II}), \text{D}-(\text{III}))$

Solution:

$\text{Hint: } \gamma = \frac{C_P}{C_V} = 1 + \frac{2}{f}$ $\text{(A) Monatomic } f = 3, \gamma = 1 + \frac{2}{3} = \frac{5}{3}$ $\text{(B) Diatomic rigid molecules, } f = 5, \gamma = 1 + \frac{2}{5} = \frac{7}{5}$ $\text{(C) Diatomic non-rigid molecules } f = 7, \gamma = 1 + \frac{2}{7} = \frac{9}{7}$ $\text{(D) Triatomic rigid molecules } f = 6, \gamma = 1 + \frac{2}{6} = \frac{4}{3}$

Import JSON File

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
}