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Current Question (ID: 19688)

Question:
$\text{A monoatomic gas performs a work of } \frac{Q}{4} \text{ where } Q \text{ is the heat}$ $\text{supplied to it. During this transformation, the molar heat capacity of}$ $\text{the gas will be: } (R \text{ is the gas constant.})$
Options:
  • 1. $R$
  • 2. $2R$
  • 3. $3R$
  • 4. $4R$
Solution:
$\text{Hint: } C = \frac{Q}{n \Delta T}$ $\text{Step: Apply the first law of thermodynamics.}$ $Q = W + \Delta U$ $Q = \frac{Q}{4} + \frac{3}{2} n R \Delta T$ $n \Delta T = \frac{Q}{2R} \quad \text{(i)}$ $\text{Step 2: Find the molar heat capacity of the gas.}$ $C = \frac{Q}{n \Delta T} \quad \text{(ii)}$ $\text{from equations (i) \& (ii), we get-}$ $C = \frac{Q}{\frac{Q}{2R}}$ $= 2R$

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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
}