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Current Question (ID: 19665)

Question:
$\text{An ideal gas goes through a reversible cycle } a \rightarrow b \rightarrow c \rightarrow d \text{ has the } V-T \text{ diagram as shown below.}$ $\text{Process } d \rightarrow a \text{ and } b \rightarrow c \text{ are adiabatic.}$
Options:
  • 1. $\text{Option 1: Diagram with } P \text{ and } V \text{ axes, cycle } d \rightarrow a \rightarrow b \rightarrow c$
  • 2. $\text{Option 2: Diagram with } P \text{ and } V \text{ axes, cycle } a \rightarrow b \rightarrow c \rightarrow d$
  • 3. $\text{Option 3: Diagram with } P \text{ and } V \text{ axes, cycle } a \rightarrow d \rightarrow c \rightarrow b$
  • 4. $\text{Option 4: Diagram with } P \text{ and } V \text{ axes, cycle } d \rightarrow c \rightarrow b \rightarrow a$
Solution:
$\text{Recall the thermodynamic processes.}$

Import JSON File

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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
}